Lessons 91–100,Investigation 10 Math...Investigation 10 Block: 90-Minute Class Day 1 Day 2 Day 3...
Transcript of Lessons 91–100,Investigation 10 Math...Investigation 10 Block: 90-Minute Class Day 1 Day 2 Day 3...
602A Saxon Algebra 1
S E C T I O N O V E R V I E W
10
Lesson Planner
Resources and Planner CD
for lesson planning support
Pacing Guide
45-Minute Class
Day 1 Day 2 Day 3 Day 4 Day 5 Day 6
Lesson 91 Lesson 92 Lesson 93 Lesson 94 Lesson 95 Cumulative Test 18
Day 7 Day 8 Day 9 Day 10 Day 11 Day 12
Lesson 96 Lab 9Lesson 97 Lesson 98 Lesson 99 Lesson 100 Cumulative Test 19
Day 13
Investigation 10
Block: 90-Minute Class
Day 1 Day 2 Day 3 Day 4 Day 5 Day 6
Lesson 91Lesson 92
Lesson 93Lesson 94
Lesson 95Cumulative Test 18
Lab 9Lesson 96Lesson 97
Lesson 98Lesson 99
Lesson 100Cumulative Test 19
Day 7
Investigation 10Lesson 101
Lesson New Concepts
91 Solving Absolute-Value Inequalities
92 Simplifying Complex Fractions
93 Dividing Polynomials
94 Solving Multi-Step Absolute-Value Equations
95 Combining Rational Expressions with Unlike Denominators
Cumulative Test 18, Performance Task 18
96 Graphing Quadratic Functions
LAB 9 Graphing Calculator Lab: Graphing Linear Inequalities
97 Graphing Linear Inequalities
98 Solving Quadratic Equations by Factoring
99 Solving Rational Equations
100 Solving Quadratic Equations by Graphing
Cumulative Test 19, Performance Task 19
INV 10 Investigation: Transforming Quadratic Equations
Resources for Teaching
• Student Edition• Teacher’s Edition• Student Edition eBook• Teacher’s Edition eBook • Resources and Planner CD • Solutions Manual• Instructional Masters• Technology Lab Masters• Warm Up and Teaching Transparencies• Instructional Presentations CD• Online activities, tools and homework help www.SaxonMathResources.com
Resources for Practice and Assessment
• Student Edition Practice Workbook• Course Assessments• Standardized Test Practice• College Entrance Exam Practice• Test and Practice Generator CD using
ExamView™
Resources for Differentiated Instruction
• Reteaching Masters• Challenge and Enrichment Masters• Prerequisite Skills Intervention• Adaptations for Saxon Algebra 1• Multilingual Glossary• English Learners Handbook• TI Resources
* For suggestions on how to implement Saxon Math in a block schedule, see the Pacing section at the beginning of the Teacher’s Edition.
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Section Overview 10 602B
SE
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0Le ssons 91–10 0, I nve st igat ion 10
Diff erentiated Instruction
Below Level Advanced Learners
Warm Up ............................. SE pp. 602, 609, 616, 624, 631, 638, 647, 655, 662, 669
Skills Bank ........................... SE pp. 846–883Reteaching Masters.............. Lessons 91–100,
Investigation 10Warm Up Transparencies .... Lessons 91–100Prerequisite Skills ................. Skills 61, 81 Intervention
Challenge ............................ TE pp. 608, 614, 622, 630, 637, 644, 654, 660, 668, 674
Extend the Example ............. TE pp. 605, 611, 620, 627, 629, 634, 641, 650, 657, 660, 665, 672
Extend the Exploration ........ TE pp. 648Extend the Problem ............. TE pp. 608, 613, 614, 622, 636, 642, 652,
653, 654, 666, 667, 668, 674, 675, 676Challenge and Enrichment ... Challenge: 91–100; Enrichment: 93 Masters
English Learners Special Needs
EL Tips ............................... TE pp. 603, 610, 617, 627, 636, 643, 648, 656, 664, 670, 677
Multilingual Glossary .......... Booklet and Online English Learners Handbook
Inclusion Tips ...................... TE pp. 604, 611, 619, 625, 632, 641, 649, 657, 663, 672
Adaptations for Saxon .......... Lessons 91–100, Cumulative Tests 18, 19 Algebra 1
For All Learners
Exploration .......................... SE pp. 648Caution ................................ SE pp. 603, 618, 627, 633,
649, 657, 664, 671Hints .................................... SE pp. 611, 617, 620, 624, 633,
639, 647, 648, 656, 663, 670, 672, 673
Alternate Method ................ TE pp. 605, 640Online Tools
Error Alert ........................... TE pp. 603, 606, 607, 610, 612, 615, 617, 620, 622, 626, 627, 630, 632, 634, 637, 641, 642, 648, 651, 652, 654, 658, 659, 660, 665, 666, 667, 670, 673, 674, 675, 676, 677
SE = Student Edition; TE = Teacher’s Edition
Math Vocabulary
Lesson New Vocabulary Maintained EL Tip in TE
91absolute-value inequality absolute value
absolute-value expressionabsolute
92complex fraction greatest common factor reciprocal
93coeffi cient rational expression
mono-, bi-, poly-
94 absolute value ring
95least common denominatorleast common multiple
cushion
96
standard form of a quadratic equation axis of symmetryvertexy-interceptzero of a function
diver
97
solution of a linear inequality coordinate plane linear equationordered pair
shaded
98 root of an equation quadratic function root
99
rational equationextraneous solution
cross productsleast common denominatorrational expression
extraneous
100
function notationparabola quadratic equation
related
INV 10
parent functionquadratic functiontransformation
parent function
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S E C T I O N O V E R V I E W 1 0
602C Saxon Algebra 1
Math Highlights
Enduring Understandings – The “Big Picture”
After completing Section 10, students will understand:
• How to solve absolute-value equations.
• How to simplify complex fractions and solve rational equations.
• How to divide polynomials.
• How to graph quadratic functions and solve quadratic equations by graphing or factoring.
Essential Questions
• How do solutions of absolute-value inequalities relate to compound inequalities?
• How can the methods of simplifying fractions be applied to simplifying complex fractions?
• How can long division be applied to dividing algebraic expressions?
• How can the methods of adding and subtracting fractions be applied to simplifying complex fractions?
• What are elements of a parabola and how can any quadratic function be graphed using these elements?
• What is the difference between a quadratic function and a quadratic equation?
• How can the solutions of a quadratic equation be found by graphing the related function?
Math Content Strands Math Processes
Polynomials• Lesson 93 Dividing Polynomials
Rational Expressions and Functions• Lesson 92 Simplifying Complex Fractions• Lesson 95 Combining Rational Expressions with Unlike
Denominators• Lesson 99 Solving Rational Equations
Systems of Equations and Inequalities• Lab 9 Graphing Calculator: Graphing Linear
Inequalities• Lesson 97 Graphing Linear Inequalities
Quadratic Equations
• Lesson 96 Graphing Quadratic Functions• Lesson 98 Solving Quadratic Equations by Factoring• Lesson 100 Solving Quadratic Equations by Graphing• Investigation 10 Transforming Quadratic Functions
Absolute-Value Equations and Inequalities• Lesson 91 Solving Absolute-Value Inequalities• Lesson 94 Solving Multi-Step Absolute-Value Equations
Connections in Practice Problems Lessons
Data Analysis 100Geometry 91, 92, 93, 94, 95, 96, 97, 98, 99, 100Measurement 91, 95, 96, 98, 99Probability 92, 93, 97, 99
Reasoning and Communication Lessons
• Analyze 91, 93, 94, 95, 96, 98, 99, Inv. 10• Error analysis 91, 92, 93, 94, 95, 96, 97, 98, 99, 100• Estimate 91• Formulate 94, 99• Generalize 91, 95, 96, 97, 98, 99, 100, Inv. 10• Justify 91, 92, 93, 94, 95, 96, 98, 99• Math Reasoning 91, 92, 94, 95, 96, 98, 99, 100,
Inv. 10• Multiple choice 91, 92, 93, 94, 95, 96, 97, 98, 99, 100• Multi-step 91, 92, 93, 94, 95, 96, 97, 98, 99, 100• Predict Inv. 10• Verify 94, 96, 97, 99, 100• Write 91, 92, 93, 94, 95, 96, 97, 98, 99, 100
Graphing Calculator 93, 96, 97, 98, 100, Inv. 10
Connections
In Examples: Archery, Baseball, Carnival, Gardening, Length of a Garden, Painting, Physics, Polling, Speed Walking, Traveling
In Practice problems: Ages, Archeology, Architecture, Art, Band, Banking, Baseball, Basketball, Bike riding, Canoeing, Carpeting, Census, Commuting, Cost, Diving, Exercise, Flooring, Football, Geography, Hardiness zones, Hiking, Horseback riding, House painting, Housekeeping, Industry, Investments, Lawn care, Olympic swimming pool, Pendulums, Physics, Population, Profit, Rate, Refurbishing, Running, Shopping, Skating, Soccer, Space, Swimming, Track, Travel, United States flag, Vertical motion
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Section Overview 10 602D
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0Le ssons 91–10 0, I nve st igat ion 10
Content Trace
LessonWarm Up:
Prerequisite Skills
New Concepts Where PracticedWhere
Assessed
Looking
Forward
91 Lessons 5, 74 Solving Absolute-Value Inequalities
Lessons 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 104, 106, 107
Cumulative Tests 19, 20, 21, 22, 23
Lessons 94, 101, 107
92 Lessons 11, 57, 72, 83
Simplifying Complex Fractions Lessons 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 105, 107, 108
Cumulative Tests 19, 20, 21, 22, 23
Lessons 95, 99
93 Lessons 38, 43, 53, 75, 83
Dividing Polynomials Lessons 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 106, 107, 108, 109
Cumulative Tests 19, 20, 21, 22, 23
Lessons 95, 99
94 Lessons 5, 26 Solving Multi-Step Absolute–Value Equations
Lessons 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 107, 108, 109, 110
Cumulative Tests 19, 20, 21, 22, 23
Lessons 101, 107
95 Lessons 2, 57, 72, 75
Combining Rational Expressions with Unlike Denominators
Lessons 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 108, 110, 111
Cumulative Tests 19, 20, 21, 22, 23
Lesson 99
96 Lessons 9, 89 Graphing Quadratic Functions Lessons 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 109, 110, 111, 112
Cumulative Test 20, 21
Lessons 98, 100, 102, 110
97 Lessons 49, 50 Graphing Linear Inequalities Lessons 98, 99, 100, 101, 102, 103, 105, 106, 107, 110, 112, 113, 115
Cumulative Test 20, 21, 22
Lesson 109
98 Lessons 72, 75, 83, 89
Solving Quadratic Equations by Factoring
Lessons 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 111, 113, 114
Cumulative Test 20, 21, 22
Lessons 100, 102, 104, 110, 113
99 Lessons 39, 57 Solving Rational Equations Lessons 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 112, 114, 115
Cumulative Test 20, 21, 22, 23
Lessons in other Saxon High School Math programs
100 Lessons 9, 84 Solving Quadratic Equations by Graphing
Lessons 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 113, 115, 116
Cumulative Test 20, 21
Lessons 102, 104, 110, 113
INV 10 N/A Transforming Quadratic Functions
Lessons 101, 103, 106, 112, 113, 114, 116, 119
Cumulative Test 22
Lessons 107, 114, 119
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S E C T I O N O V E R V I E W 1 0
602E Saxon Algebra 1
Ongoing Assessment
Type Feature Intervention *
BEFORE instruction Assess Prior Knowledge
• Diagnostic Test • Prerequisite Skills Intervention
BEFORE the lesson Formative • Warm Up • Skills Bank• Reteaching Masters
DURING the lesson Formative • Lesson Practice• Math Conversations with the Practice
problems
• Additional Examples in TE• Test and Practice Generator (for additional
practice sheets)
AFTER the lesson Formative • Check for Understanding (closure) • Scaffolding Questions in TE
AFTER 5 lessons Summative After Lesson 95• Cumulative Test 18 • Performance Task 18After Lesson 100• Cumulative Test 19• Performance Task 19
• Reteaching Masters• Test and Practice Generator (for additional
tests and practice)
AFTER 20 lessons Summative • Benchmark Tests • Reteaching Masters• Test and Practice Generator (for additional
tests and practice)
* for students not showing progress during the formative stages or scoring below 80% on the summative assessments
Evidence of Learning – What Students Should Know
Because the Saxon philosophy is to provide students with sufficient time to learn and practice each concept, a lesson’s topic will not be tested until at least five lessons after the topic is introduced.
On the Cumulative Tests that are given during this section of ten lessons, students should be able to demonstrate the following competencies:
• Solve multi-step compound inequalities, and absolute-value equations and inequalities.• Graph quadratic functions, and identify the vertex, the maximum, and the minimum of a parabola.• Simplify complex fractions.• Multiply binomials with radicals.• Factor polynomials, and divide polynomials by monomials.• Use the Pythagorean Theorem, and find the distance between two points.• Find the probability of mutually exclusive events.
Test and Practice Generator CD using ExamView™
The Test and Practice Generator is an easy-to-use benchmark and assessment tool that creates unlimited practice and tests in multiple formats and allows you to customize questions or create new ones. A variety of reports are available to track student progress toward mastery of the standards throughout the year.
NorthStar Math offers you real-time benchmarking, trackingand student progress monitoring.Visit www.NorthStarMath.com for more information.
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Section Overview 10 602F
SE
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0Le ssons 91–10 0, I nve st igat ion 10
Assessment Resources
Resources for Diagnosing and Assessing
• Student Edition• Warm Up• Lesson Practice
• Teacher’s Edition• Math Conversations with the Practice problems• Check for Understanding (closure)
• Course Assessments• Diagnostic Test• Cumulative Tests• Performance Tasks • Benchmark Tests
Resources for Test Prep
• Student Edition Practice• Multiple-choice problems• Multiple-step and writing problems• Daily cumulative practice
• Standardized Test Practice
• College Entrance Exam Practice
• Test and Practice Generator CD using ExamViewTM
Resources for Intervention
• Student Edition • Skills Bank
• Teacher’s Edition• Additional Examples• Scaffolding questions
• Prerequisite Skills Intervention• Worksheets
• Reteaching Masters• Lesson instruction and practice sheets
• Test and Practice Generator CD using ExamViewTM
• Lesson practice problems• Additional tests
Cumulative TestsThe assessments in Saxon Math are frequent and consistently placed after every fi ve lessons to offer a regular method of ongoing testing. These cumulative assessments check mastery of concepts from previous lessons.
Performance Tasks The Performance Tasks can be used in conjunction with the Cumulative Tests and are scored using a rubric.
After Lesson 95 After Lesson 100 For use with Performance Tasks
Name ___________________________________________ Date_______________ Class _______________
Cumulative Test
© Saxon. All rights reserved. 89 Saxon Algebra 1
18B
1. (82) Of five puppies in a litter, four have weights of 4.8 lb, 4.9 lb, 4.5 lb, and 4.4 lb. What could be the weight of the fifth puppy if the average weight of all five puppies is to be between 4.2 and 4.8 lb?
2. (84) Use a table to graph the
function f x( ) = 3x 2 .
3. (61) Simplify 10,000,000 using powers
of ten.
4. (74) Solve | x 5 |= 2 .
5. (76) Find the product 5 = 5( )2
.
6. (69) A rectangular patio has a length of
363 feet and a width of 243 feet. What is the patio’s perimeter?
7. (73) Write a compound inequality that represents all real numbers that are less than 0 or greater than 2. Graph the solution.
8. (68) Nelida has a bag of bagels. There are 4 plain bagels, 5 garlic bagels, 3 sesame bagels, and 6 onion bagels. If Nelida picks a bagel at random, what is the probability she will pick a plain or an onion bagel?
9. (86) Find the distance between the points (4, –2) and (6, 4).
10. (87) Factor 4x 2 = 8xy = 3x = 6y .
11. (62) The stem-and-leaf plot below shows scores on a history test. Find the median, mode, range, and relative frequency of 89.
Scores on a History Test
Stem Leaves
6 0, 1, 1
7 2, 2, 4, 5, 5, 5, 7
8 1, 3, 5, 9, 9
9 5
Key 7 | 2 = 72
Name ___________________________________________ Date_______________ Class _______________
Cumulative Test
© Saxon. All rights reserved. 87 Saxon Algebra 1
18A
1. (82) A farmer randomly chose 4 pumpkins out of a patch of 15 pumpkins. Three of the pumpkins weigh 15 lb, 18 lb, and 16 lb. What could be the weight of the fourth pumpkin if the average weight of all 4 pumpkins is to be between 15 and 20 lb?
2. (84) Use a table to graph the
function f x( ) = 2x 2 .
3. (61) Simplify 1,000,000 using powers of
ten.
4. (74) Solve | x 2 |= 9 .
5. (76) Find the product 3 2( )2
.
6. (69) A rectangular rug has a length of
98 feet and a width of 50 feet. What is the rug’s perimeter?
7. (73) Write a compound inequality that represents all real numbers that are less than –1 or greater than 4. Graph the solution.
8. (68) Jed opens a drawer containing T-shirts. There are 5 white shirts, 3 blue shirts, 2 red shirts, and 2 gray shirts. If Jed picks a shirt at random, what is the probability he will pick a white or a blue T-shirt?
9. (86) Find the distance between the points (2, –3) and (5, 6).
10. (87) Factor 3x 2 + 9xy + 4x + 12y .
11. (62) The stem-and-leaf plot below shows ages of students in a cooking class. Find the median, mode, range, and relative frequency of age 35.
Ages of Students in a Cooking Class
Stem Leaves
1 6, 8, 8
2 4, 5, 5, 5, 6, 8
3 0, 1, 2, 5, 5, 5, 5, 8
4 0, 2, 9
Key 2 | 4 = 24
Name ___________________________________________ Date_______________ Class _______________
Cumulative Test
© Saxon. All rights reserved. 93 Saxon Algebra 1
19B
1. (89) A rock is thrown from a height of 25 feet above the ground. The rock starts with a vertical speed of 96 feet per second. Ignoring friction, the equation y = 16t 2 + 96t + 25 gives the height y
as a function of time t. Find the highest point the rock reaches and how long it takes to reach this point.
2. (84) Determine whether the equation below represents a quadratic function.
y + 7x 2 = 5x 4
3. (83) Determine whether the trinomial below is a perfect-square trinomial. If it is, factor the trinomial.
32x 2 48x + 18
4. (86) Find the midpoint of the line segment with endpoints (–8, 5) and (2, 2).
5. (76) Find the product 5 + 6( )2.
6. (78) Identify the asymptotes for the rational function below.
y =2
x + 8+ 4
7. (73) Write a compound inequality that describes the graph below.
8. (68) What is the probability of rolling either a sum of 7 or a sum of 9 using two different number cubes, each numbered 1 to 6?
Factor the polynomials in problems 9–10.
9. (79) 4x3 + 8x 2 + 12x
10. (87) 3y 2 + 20y + 12 + 5y 3
Name ___________________________________________ Date_______________ Class _______________
Cumulative Test
© Saxon. All rights reserved. 91 Saxon Algebra 1
19A
1. (89) A ball is thrown from a height of 10 feet above the ground. The ball starts with a vertical speed of 64 feet per second. Ignoring friction, the equation y = 16t 2 + 64t + 10 gives the height y
as a function of time t. Find the highest point the ball reaches and how long it takes to reach this point.
2. (84) Determine whether the equation below represents a quadratic function.
y + 2x = 3x 2 + 6
3. (83) Determine whether the trinomial below is a perfect-square trinomial. If it is, factor the trinomial.
8x 2 8x + 2
4. (86) Find the midpoint of the line segment with endpoints (6, 1) and (3, –5).
5. (76) Find the product 2 + 7( )2
.
6. (78) Identify the asymptotes for the rational function below.
y =3
x + 7+ 2
7. (73) Write a compound inequality that describes the graph below.
8. (68) What is the probability of rolling either a sum of 5 or a sum of 8 using two different number cubes, each numbered 1 to 6?
Factor the polynomials in problems 9–10.
9. (79) 3x3 + 3x 2 18x
10. (87) 2y 2 + 3y 3 + 9y + 6
© Saxon. All rights reserved. xi Saxon Algebra 1
Student Rubric
StudentEvalutaion
Knowledge andSkil lsUnderstanding
CommunicationandRepresentation
Process andStrategies
4
I understand and can
justify my reasoning in
more than one way.
I explained my work in
detail and justified my
solution. I described
my work in great
detail and illustrated
my thinking.
I selected the most
appropriate strategy
and used the process
to solve the problem.
3I understand the task
and can show that I
understand.
I can explain my
thinking. I can show
my work. .
I selected a strategy
and followed the
process.
2
I have some
understanding of the
task.
I can explain some of
my thinking. I can
show some of my
work.
I selected a strategy
but became confused
on the process.
1
I need help in
understanding the
task.
I need help in
explaining my thinking.
I need help in showing
my work.
I need help in choosing
a strategy.
© Saxon. All rights reserved. xx SSaxon Algebra 1
TTeacher Rubric
CriteriaPerformance
Knowledge andSkil lsUnderstanding
CommunicationandRepresentation
Process andStrategies
4
The student got it!
The student did it in
new ways and
showed how it
worked. The student
knew and understood
what math concepts
to use.
The student clearly
detailed how he/she
solved the problem.
The student included
all the steps to show
his/her thinking. The
student used math
language, symbols,
numbers, graphs
and/or models to
represent his/her
solution.
The student had an
effective and inventive
solution. The student
used big math ideas
to solve the problem.
The student
addressed the
important details. The
student showed other
ways to solve the
problem. The student
checked his/her
answer to make sure
it was correct
3
The student
understood the
problem and had an
appropriate solution.
All parts of the
problem are
addressed.
The student clearly
explained how he/she
solved the problem.
The student used
math language,
symbols, tables,
graphs, and numbers
to explain how he/she
did the problem.
The student had a
correct solution. The
student used a plan
to solve the problem
and selected an
appropriate strategy.
2
The student
understood parts of
the problem. The
student started, but
he/she couldn’t finish.
The student explained
some of what he/she
did. The student tried
to use words,
symbols, tables,
graphs and numbers
to explain how he/she
did the problem.
The student had part
of the solution, but
did not know how to
finish. The student
was not sure if he/she
had the correct
answer. The student
needed help.
1
The student did not
understand the
problem.
The student did not
explain how he/she
solved the problem.
He/she did not use
words, symbols,
tables or graphs to
show how he/she
The student couldn’t
get started. The
student did not know
how to begin.
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Saxon Algebra 1602
Solving Absolute-Value Inequalities
Warm Up
91LESSON
1. Vocabulary An equation with one or more absolute-value expressions is called an . absolute-value equation
Simplify.
2. �8 - 15� 7 3. �-3 + 9� 6
Solve.
4. �x - 4� = 7 x = 11, -3 5. �x + 7� = 2 x = -5, -9
An absolute-value inequality is an inequality with at least one absolute-value expression. The solution to an absolute-value inequality can be written as a compound inequality.
The inequality �x� < 6 describes all real numbers whose distance from 0 is less than 6 units. The solutions are all real numbers between -6 and 6. The solution can be written -6 < x < 6 or as the compound inequality x > -6 AND x < 6.
640 2-2-4-6
The inequality �x� > 6 describes all real numbers whose distance from 0 is greater than 6 units. The solutions are all real numbers less than -6 or greater than 6. The solution can be written as the compound inequality x < -6 OR x > 6.
640 2-2-4-6
Example 1 Solving Absolute-Value Inequalities by Graphing
Solve each inequality by graphing.
a. �x� < 4 b. �x� > 7
SOLUTION
If the absolute value of x is less than 4, then x is less than 4 units from zero on a number line.
40 2-2-4
The graph shows x < 4 AND x > -4. This can also be written -4 < x < 4.
SOLUTION
If the absolute value of x is greater than 7, then x is more than 7 units from zero on a number line.
6 840 2-2-4-6-8
The graph shows x > 7 OR x < -7.
(74)(74)
(5)(5) (5)(5)
(74)(74) (74)(74)
New ConceptsNew Concepts
Math Language
The absolute value of a number is its distance from zero on the number line.
Reading Math
For the inequality -4 < x < 4, you can say, “x is between -4 and 4.”
Math Reasoning
Analyze Why is the word “OR” used here to describe the solution?
Sample: A number cannot be both greater than 7 and less than -7.
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MATH BACKGROUNDLESSON RESOURCES
Student Edition Practice Workbook 91
Reteaching Master 91Adaptations Master 91Challenge and Enrichment
Master C91
In this lesson, the absolute-value inequalities have the form
⎢x - a� > b,
where a and b are constants. When graphed on the number line, the graph is always symmetric about the center point a.
In this case, the inequality can be understood as “the distance from x to a is greater than b.” Students might have an easier time understanding why such graphs on the number line are symmetrical; one can take b steps from a in two directions.
Warm Up1
602 Saxon Algebra 1
91LESSON
Problems 2–5
Remind students that the expression inside the absolute- value bars can be negative, but the absolute value of any expression cannot be negative.
2 New Concepts
In this lesson, students learn to solve absolute-value inequalities.
Discuss the defi nition of an absolute-value inequality. Explain that the solution to an absolute-value inequality can be written as a compound inequality.
Example 1
Remind students that the absolute value of a number is its distance from zero on the number line.
Additional Example 1
Solve each inequality by graphing.
a. ⎢x� < 2.5 -2.5 < x < 2.5;
20 1-1-2
b. ⎢x� > 10 x > 10 OR x < -10;
100 5-5-10
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Lesson 91 603
Example 2 Isolating the Absolute Value to Solve
Solve and graph each inequality.
a. �x� + 7.4 ≤ 9.8
SOLUTION
Begin by isolating the absolute value.
�x� + 7.4 ≤ 9.8
___-7.4 ___-7.4 Subtraction Property of Inequality
�x� ≤ 2.4 Simplify.
Since the absolute value of x is less than or equal to 2.4, it is 2.4 units or less from zero.
0 2 31-2-3 -1
The solution can be written x ≥ -2.4 AND x ≤ 2.4 or -2.4 ≤ x ≤ 2.4.
b. �x�
_ 4 > 2
SOLUTION
Begin by isolating the absolute value.
�x�
_ 4 > 2
4 · �x�
_ 4 > 2 · 4 Multiplication Property of Inequality
�x� > 8 Simplify.
The absolute value of x is greater than 8, so it is more than 8 units from zero.
1280 4-4-8-12
The solution is x > 8 OR x < -8.
c. -2�x� < -6
SOLUTION
Begin by isolating the absolute value.
-2�x� < -6
-2�x�
_ -2
> -6 _ -2
Division Property of Inequality
�x� > 3 Simplify.
Since the absolute value of x is greater than 3, it is more than 3 units from zero.
40 2-2-4
The solution is x > 3 OR x < -3. Online Connection
www.SaxonMathResources.com
Caution
Be sure to reverse the direction of the inequality sign if you multiply or divide by a negative number when solving the inequality.
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For this lesson, explain the meaning of the word absolute. Say:
“The word absolute means defi nite or fi nal. It can describe something that exists independently and not in relation to other things.”
Connect the meaning of absolute with absolute value by pointing to the defi nition of the absolute value and emphasizing that the distance from n to 0 is not relative to
another value—it is absolute.
Discuss examples of other uses of the word absolute: the absolute-temperature scale Kelvin, absolute power, etc.
ENGLISH LEARNERS
Lesson 91 603
Example 2
Point out that in solving absolute-value inequalities, it is often easier to isolate the absolute value before solving for the variable.
Error Alert When graphing inequalities on the number line, students often will use open and closed circles incorrectly. Remind them to pay attention to what symbol is used in the expression.
Additional Example 2
Solve and graph each inequality.
a. ⎢x� - 2.2 ≤ 7.8 -10 ≤ x ≤ 10;
100 5-5-10
b. 3⎢x� > 2.7 x > 0.9 OR x < -0.9;
0.5 10-0.5-1
c. -3⎢x⎢ ≤ -9 x ≥ 3 OR x ≤ -3;
0 3-3
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Saxon Algebra 1604
Some absolute-value inequalities have variable expressions inside the absolute-value symbols. The expression inside the absolute-value symbols can be positive or negative.
The inequality �x + 1� < 3 represents all numbers whose distance from -1 is less than 3.
0 2-2-4
3 units3 units
The inequality �x + 1� > 3 represents all numbers whose distance from -1 is greater than 3.
0 2-2-4
3 units3 units
Rules for Solving Absolute-Value InequalitiesFor an inequality in the form �K� < a, where K represents a variable expression and a > 0, solve -a < K < a or K > -a AND K < a.
For an inequality in the form �K� > a, where K represents a variable expression and a > 0, solve K < -a OR K > a.
Similar rules are true for �K� ≤ a or �K� ≥ a.
Example 3 Solving Inequalities with Operations Inside
Absolute-Value Symbols
Solve each inequality. Then graph the solution.
a. �x - 5� ≤ 3
SOLUTION
Use the rules for solving absolute-value inequalities to write a compound inequality.
�x - 5� ≤ 3
x - 5 ≥ -3 AND x - 5 ≤ 3 Write the compound inequality.
__ +5 __ +5 __ +5 __ +5 Addition Property of Inequality
x ≥ 2 AND x ≤ 8 Simplify.
Now graph the inequality.
b. �x + 7� > 3
SOLUTION
Use the rules for solving absolute-value inequalities to write a compound inequality.
�x + 7� > 3
x + 7 < -3 OR x + 7 > 3 Write the compound inequality.
__ -7 __ -7 __ -7 __ -7 Subtraction Property of Inequality
x < -10 OR x > -4 Simplify.
Now graph the inequality.
80 4-4-8-12
6 840 2 6 840 2
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INCLUSION
Have students draw a number line from -5 to 5 with every whole number labeled and 1 inch apart. Have them locate the point 2, and with a ruler, have them then mark all the points that are less than 3 inches away from 2.
Lastly, have them show that the absolute-value inequality that represents the graph is ⎢x - 2� < 3 and the solution is -1 < x < 5.
604 Saxon Algebra 1
Example 3
Point out that the variable expression inside the absolute- value bars “shifts” the center of the graph. The expression x + ashifts the center of the graph to -a and the expression x - b shifts the center of the graph to b when a and b are non-negative constants.
Additional Example 3
Solve each inequality. Then graph the solution.
a. ⎢x + 4� ≤ 5 x ≤ 1 AND x ≥ -9
0 2-2-4-6-8-10
b. ⎢x - 2� > 3 x < -1 OR x > 5
4 620-2
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Lesson 91 605
Example 4 Solving Special Cases
Solve each inequality.
a. �x� + 6 ≤ 4
SOLUTION
�x� + 6 ≤ 4
�x� ≤ -2 Subtract 6 from both sides.
This inequality states that a number’s distance from 0 is less than or equal to -2. No distance can be negative. Therefore, there are no solutions to this inequality. The solution is identified as { } or ∅, the empty set.
b. �x� + 6 > 1
SOLUTION
�x� + 6 > 1
�x� > -5 Subtract 6 from both sides.
This inequality states that a number’s distance from 0 is greater than -5. Since all distances (and absolute values) are positive, all numbers on the number line are solutions. The solution is identified as �, the set of all real numbers. It means that the inequality is an identity; it works for all real numbers.
Example 5 Application: Polling
A poll finds that candidate Garcia is favored by 46% of the voters surveyed and Jackson is favored by 44%. The poll has an accuracy of plus or minus 3%.
a. Write an absolute-value inequality to show the true percentage of voters for Garcia.
SOLUTION
Let the true percentage of voters for Garcia be g.
For g to be within 3%, the distance from g to 46 must be less than or equal to 3. The distance is represented by the absolute value of their difference.
�g - 46� ≤ 3
b. Solve the inequality to find the range for the true percentage of voters who support Garcia.
SOLUTION
�g - 46� ≤ 3
-3 ≤ g - 46 ≤ 3 Write the inequality without an absolute value.
43 ≤ g ≤ 49 Add 46 to all 3 parts of the inequality.
The range is between 43% and 49%.
Math Reasoning
Justify According to the poll, Garcia and Jackson are in a “statistical dead heat”— a tie. Explain why the poll would not say that Garcia (46%) is leading Jackson (44%).
Sample: Because the accuracy of the poll is plus or minus 3%, either could actually be ahead. Garcia’s percentage could actually be 3% less, at 43%. Jackson’s percentage could actually be 3% more, at 47%, and Jackson would be leading Garcia.
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Instead of writing an absolute value inequality, have students think about the defi nition of accuracy. Since the problem states that the poll has an “accuracy of plus or minus 3%,” have students write the range for the true percentage of voters that support Garcia.
The limits of the range will be 46 ± 3%, thus the range is between 43% and 49%.
ALTERNATE METHOD FOR EXAMPLE 5
Lesson 91 605
Example 4
Remind students that they should always isolate the absolute value fi rst to identify special cases.
Additional Example 4
Solve each inequality.
a. ⎢x� + 3 < 3{ } or ∅; There is no graph.
b. ⎢x� + 3 > 2 �;
0
Example 5
Explain to students that a measurement, such as a poll, has an uncertainty or a range of values within which the true value lies.
Extend the Example
Let the true percentage of voters for Jackson be j. Write and solve an absolute-value inequality to show that j is within 3% of 44.⎪j - 44⎥ ≤ 341 ≤ j ≤ 47
Additional Example 5
Jim and Tom went fi shing and wanted to see who caught the biggest fi sh. Unfortunately, the scale they used was not very accurate and was only good to ±2 ounces. Jim’s fi sh weighed in at 5 pounds 5 ounces and Tom’s was 5 pounds 10 ounces.
a. Let the true weight of Jim’s fi sh be j and Tom’s fi sh be t. Write the absolute-value inequalities.⎢j - 85 oz� ≤ 2 oz⎢t - 90 oz� ≤ 2 oz
b. Can one confi dently say that Tom’s fi sh is bigger than Jim’s? yes
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Practice Distributed and Integrated
Lesson Practice
a. Solve and graph the inequality � x � < 12. -12 < x < 12
b. Solve and graph the inequality � x � > 19. x < -19 OR x > 19
c. Solve and graph the inequality � x � + 2.8 ≤ 10.4. -7.6 ≤ x ≤ 7.6
d. Solve and graph the inequality �x�
_ -5 < -1. x < -5 OR x > 5
e. Solve and graph the inequality ⎪x - 10⎥ ≤ 12. -2 ≤ x ≤ 22
f. Solve and graph the inequality ⎪x + 12⎥ > 18. x < -30 OR x > 6
g. Solve the inequality � x � + 21 ≤ 14. ∅
h. Solve the inequality � x � + 33 > 24. �
Industry A machine part must be 15 ± 0.2 cm in diameter.
i. Write an inequality to show the range of acceptable diameters.
j. Solve the inequality to find the actual range for the diameters.14.8 ≤ m ≤ 15.2
a.–f. See Additional Answers.a.–f. See Additional Answers. (Ex 1)(Ex 1)
(Ex 1)(Ex 1)
(Ex 2)(Ex 2)
(Ex 2)(Ex 2)
(Ex 3)(Ex 3)
(Ex 3)(Ex 3)
(Ex 4)(Ex 4)
(Ex 4)(Ex 4)
(Ex 5)(Ex 5)
i. � m - 15 � ≤ 0.2i. � m - 15 � ≤ 0.2
1. Find the axis of symmetry for the graph of the equation y = - 1 _ 2 x2 + x - 3.
x = 1
Add or subtract.
2. 6rs
_ r2s2 +
18r _ r2s2
6(s + 3)
_rs2
3. b _
2b + 1 -
6 _ b - 4
b2 - 16b - 6__
(2b + 1)(b - 4)
Factor.
4. -4y4 + 8y3 + 5y2 - 10y 5. 3a2 - 27 3(a + 3)(a - 3)-y(4y2 - 5)(y - 2)
Evaluate.
6. 4x2 + 6x - 4 (2x - 1)(2x + 4) 7. 9x2 - 2x + 32 (9x + 16)(x - 2)
*8. Solve and graph the inequality � x � < 96. -96 < x < 96;
*9. Write Explain what � x � ≥ 54 means on a number line. Sample: x can be any value that is 54 or more units from 0.
*10. Justify When solving an absolute-value inequality, a student gets �x� ≥ -5. Justify that any value for x makes this inequality true.
*11. Multiple Choice Which inequality is represented by the graph? B
80 4 12-4-8-12
A �x� < 9 B �x� > 9 C �x� ≤ 9 D �x� < -9
*12. Track A runner finishes a sprint in 8.54 seconds. The timer’s accuracy is plus or minus 0.3 seconds. Solve and graph the inequality �t - 8.54� ≤ 0.3. 8.24 ≤ t ≤ 8.84
8.848.24 8.44 8.64
(89)(89)
(90)(90) (90)(90)
(87)(87) (83)(83)
(75)(75) (75)(75)
(91)(91) 960 48-48-96 960 48-48-96
(91)(91)
(91)(91)10. Sample: No matter what I substitute for x, its absolute value is going to be greater than -5 because absolute value is always positive.
10. Sample: No matter what I substitute for x, its absolute value is going to be greater than -5 because absolute value is always positive.
(91)(91)
(91)(91)
Saxon Algebra 1606
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606 Saxon Algebra 1
Lesson Practice
Problem c
Scaff olding Before graphing the inequality, have students fi rst isolate the absolute-value term and solve it completely before attempting to graph the solution.
Problem d
Error Alert Students may assume that there is no solution to an absolute value inequality when the expression containing the absolute value is related to a negative number. Remind them to isolate the absolute value, then consider whether there is a solution or not.
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“Explain how to solve an absolute-value inequality.” Sample: If the absolute-value term has a coeffi cient, isolate it fi rst. Check for special cases. The expression inside the absolute-value bars can be positive and negative.
“Explain what absolute value means on the graph of an equation on the number line.” Sample: The absolute value of a number n is the distance from n to 0.
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*13. Error Analysis Two students simplified 2c _
c - 6 + 12 _
6 - c . Which student is correct?
Explain the error.
Student A Student B
2c - 12
_ c - 6
2(c - 6)
_ c - 6
= 2
2c + 12 _ c - 6
2(c + 6)
_ c - 6
= -2
*14. Multi-Step A farmer has a rectangular plot of land with an area of x2 + 22x + 72 square meters. He sets aside x2 square meters for grazing and 2x - 8 square meters for a chicken coop. a. Write a simplified expression for the total fraction of the field the farmer has
set aside. x - 2_x + 18
b. Estimate About what percent of the field has the farmer set aside if x = 30? about 60%
15. Geometry Write a simplified expression for the total fraction of the larger rectangle that the triangle and smaller rectangle cover.
2(x - 1) _
5(x + 9)
16. Find the product of ( √ � 4 - 6) 2 . 16
17. Analyze Why is it necessary to understand factoring when dealing with rational expressions? Sample: Factoring makes it easier to simplify complicated expressions.
18. Multi-Step The base of triangle ABC is x2 + y. The height is 4x + 2xy
_ x3 + xy . What is the
area of triangle ABC ? a. Multiply the base of the triangle by its height. 2(2 + y)
b. Multiply the product from part a by 1 _ 2 . 2 + y
*19. Error Analysis Two students tried to find the axis of symmetry for the equation y = 8x + 2x2. Which student is correct? Explain the error.
Student A Student B
x = -b _ 2a
= -2 _
2(8) =
-2 _ 16
= - 1 _ 8
x = -8 _
2(2) =
-8 _ 4 = -2
*20. Space If it were possible to play ball on Jupiter, the function y = -13x2 + 39x would approximate the height of a ball kicked straight up at a velocity of 39 meters per second, where x is time in seconds. Find the maximum height the ball reaches and the time it takes the ball to reach that height. (Hint: Find the time the ball reaches its maximum height first.) 29.25 feet in 1.5 seconds
21. Measurement The coordinates of two landmarks on a city map are A(5, 3) and B(7, 10). Each grid line represents 0.05 miles. Find the distance between landmarks A and B. 0.364 mile
(90)(90)
13. Student A; Sample: Student B multiplied the denominator of one of the expressions by -1, but forgot to multiply the numerator of that expression by -1also.
13. Student A; Sample: Student B multiplied the denominator of one of the expressions by -1, but forgot to multiply the numerator of that expression by -1also.
(90)(90)
2x x - 2
2x 2x + 18
5x
2x
2x x - 2
2x 2x + 18
5x
2x
(90)(90)
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(88)(88)
(88)(88)
(89)(89)19. Student B; Sample: Student A used the wrong values for a and b.The equation in standard form is y = 2 x2 + 8x, so a = 2 and b = 8.
19. Student B; Sample: Student A used the wrong values for a and b.The equation in standard form is y = 2 x2 + 8x, so a = 2 and b = 8.
(89)(89)
3 43 4
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Lesson 91 607
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Lesson 91 607
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 16
Error AlertA common mistake students make when multiplying expressions is not simplifying each expression fi rst. Write the statement:
( √ � 4 - 6) 2 = ( √ � 4 - 6) ( √ � 4 - 6)
= 4 - 12 √ � 4 + 36
= 40 - 12 √ � 4
= 40 - 12(2)
= 16
Compare with the statement:
( √ � 4 - 6) 2 = (2 - 6)2 = (-4)2
= 16
Remind students that it is often benefi cial to simplify the expressions before multiplying them.
Problem 18
Guide the students by asking them the following questions.
“How can x3 + xy be factored?” Sample: Put x outside of the parentheses: x(x2 + y)
“How can the expression
(x2 + y) · 4x + 2xy
________ x(x2 + y) be
simplifi ed?” Sample: Divide out x and x2 + y from both the numerator and denominator: 4 + 2y.
“What is the formula for the area of a triangle?” Sample: one half times the base times the height
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Saxon Algebra 1608
22. Archeology Archeologists use coordinate grids to record locations of artifacts. Jonah recorded that he found one old coin at (41, 37), and a second old coin at (5, 2). Each unit on his grid represents 0.25 feet. How far apart were the coins? Round your answer to the nearest tenth of a foot. 12.6 feet
23. Find the length t to the nearest tenth. 5.3
5
t√3
24. House Painting A house painter leans a 34-foot ladder against a house with the bottom of the ladder 7 feet from the base of the house. Will the top of the ladder touch the house above or below a windowsill that is 33 feet off the ground? above; 342 - 72 > 3 32
25. Graph the function y = 4x2.
26. Generalize What is the factored form of a 2m + 2 a m b n + b 2n ? ( am + bn) 2
27. Shopping Roger has $40 to buy CDs. The CDs cost $5 each. He will definitely buy at least 3 CDs. How many CDs can Roger buy? Use inequalities to solve the problems. x ≥ 3 and 5x ≤ 40; 3 ≤ x ≤ 8
28. Multi-Step A summer school program has a budget of $1000 to buy T-shirts. Twenty free T-shirts will be received when they place their order. The number of T-shirts y that the program can get is given by y = 1000
_ x + 20, where x is the price per T-shirt. a. What is the horizontal asymptote of this rational function? y = 20
b. What is the vertical asymptote? x = 0
c. If the price per T-shirt is $10, how many T-shirts can the program receive? 120
*29. Suppose the area of a rectangle is represented by the expression 4x2 + 9x + 2. Find possible expressions for the length and width of the rectangle. width =
(4x + 1), length = (x + 2) or width = (x + 2) and length = (4x + 1)
30. Simplify the expression 6 √ � 8 · √ � 5 . 12√ �� 10
(86)(86)
(85)(85)
(85)(85)
(84)(84)25.
x
y
8
12
16
4
2 4-2-4
25.
x
y
8
12
16
4
2 4-2-4
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(Inv 9)(Inv 9)
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LOOKING FORWARD
Solving absolute-value inequalities prepares students for
• Lesson 94 Solving Multi-Step Absolute-Value Equations
• Lesson 101 Solving Multi-Step Absolute-Value Inequalities
• Lesson 107 Graphing Absolute-Value Functions
There are also compound absolute-value inequalities. Consider the inequalities:
⎢x� < 6 AND ⎢x - 2� > 6
Have students solve the compound absolute-value inequalities. Sample: The solution will be the overlap of the two graphs on a number line, -6 < x < -4
CHALLENGE
608 Saxon Algebra 1
Problem 27
Extend the Problem
“If the CDs cost $3 each at the used CD store, how many CDs can Roger buy?” 13 or fewer
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Lesson 92 609
Warm Up
92LESSON
1. Vocabulary For any nonzero real number n, the of the number is 1
_ n . reciprocal
Identify the LCM.
2. 4x - 16 and x - 4 4(x - 4) 3. 18x2 and 9x 18x2
Factor.
4. x2 - 4x - 77 (x + 7)(x - 11) 5. 18x2 + 12x + 2 2(3x + 1)(3x + 1)
A complex fraction is a fraction that contains one or more fractions in the numerator or the denominator.
Complex FractionsThere are two ways to write a fraction divided by a fraction.
a _
b _
c _
d =
a _ b ÷
c _ d , when b ≠ 0, c ≠ 0, and d ≠ 0.
A complex fraction can be written as a fraction divided by a fraction. The rules for dividing fractions can be applied to simplify complex fractions.
Example 1 Simplifying by Dividing
Simplify a _ x _
b _ a + x
.
SOLUTION
a _ x _
b _ a + x
= a _ x ÷
b _ a + x Write using a division symbol.
= a _ x ·
a + x _ b Multiply by the reciprocal.
= a(a + x)
_ xb
Multiply.
(11)(11)
(57)(57) (57)(57)
(72)(72) (83)(83)
New ConceptsNew Concepts
Simplifying Complex Fractions
Math Reasoning
Write Explain why the complex fraction
a _ x _ b
_ a + x can be
written as a_x ÷ b_a + x .
Sample: The fraction bar means divide, so the two expressions are equal.
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LESSON RESOURCES
Student Edition Practice Workbook 92
Reteaching Master 92Adaptations Master 92Challenge and Enrichment
Master C92
The key to simplifying complex fractions is to remember that a fraction is just a way to write a series of division expressions. Take the complex fraction:
a __ x ____
b _____ a + y
It is the same as the statement: “a divided by x is divided by b divided by a + y.”
MATH BACKGROUND
Warm Up1
92LESSON
Lesson 92 609
Problems 2 and 3
Point out to students that factoring the expressions should always precede identifying the LCM.
2 New Concepts
In this lesson, students learn to simplify complex fractions.
Discuss the defi nition of a complex fraction.
Example 1
Remind students that dividing by a fraction is the same as multiplying by its reciprocal.
Additional Example 1
Simplify b __ y ____
y _____ a + y
. b(a + y) _______
y2
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Saxon Algebra 1610
The product of a number and its reciprocal is 1. To eliminate the fraction in the denominator of a complex fraction, multiply the numerator and the denominator by the reciprocal of the denominator.
Example 2 Simplifying Using the Reciprocal of the Denominator
Simplify am
_ n _ x _ mn
.
SOLUTION
am
_ n _ x _ mn
= am _ n ·
mn _ x _
x _ mn · mn _ x
Multiply by the reciprocal of the denominator.
= am _ n ·
mn _ x _
x _ mn · mn _ x
Divide out common factors.
= am2
_ x _1
Multiply.
= am2_
x Simplify.
Example 3 Factoring to Simplify
Simplify
3x _
6x + 12 _
9 _
x + 2 .
SOLUTION
3x _
6x + 12 _
9 _
x + 2
= 3x _
6x + 12 ÷
9 _ x + 2
Write using a division symbol.
= 3x _
6(x + 2) ·
x + 2 _ 9 Factor out the GCF and multiply by the reciprocal.
= 3
1 x _
6(x + 2) ·
(x + 2) _
9 3 Divide out common factors.
= x_18
Simplify. Online Connection
www.SaxonMathResources.com
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For this lesson, explain the meaning of the word reciprocal. Say:
“The reciprocal of something is done for another in return for something else.”
Connect the meaning of reciprocal with the fraction 1 __ x and its reciprocal x by emphasizing that the denominator and numerator swapped places.
Discuss examples of other uses of the word reciprocal. For example, when we treat people nicely, we can expect reciprocal comments or treatment.
Also mention that reciprocate is the verb form of the word. Have students come up with things that they do now and would like their friends to reciprocate, e.g., share their snacks.
ENGLISH LEARNERS
610 Saxon Algebra 1
Example 2
Students learn to use the reciprocal of the denominator to simplify complex fractions.
Additional Example 2
Simplify bt __ as ___ r __ st
. bt2 ___ ar
Error Alert A common mistake that students make when simplifying complex fractions is to divide out factors that are on the very top and very bottom of the fraction. Write the statements:
ab ___ x
___ y ___
bx ≠
a __ x __
y __ x
Remind students to write the fractions using a division symbol.
Example 3
Emphasize to students that factoring all the expressions before multiplying and dividing will make the problem easier to solve.
Additional Example 3
Simplify x
2 - x _____ 9 _____
x - 1 _____ 3x . x
2 __
3
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Lesson 92 611
Example 4 Combining Fractions to Simplify
Simplify 1 _ x _
1 - 1 _ x
.
SOLUTION
1 _ x _
1 - 1 _ x
= 1 _ x _
x - 1
_ x Subtract in the denominator.
= 1 _ x ÷
x - 1 _ x Write using a division symbol.
= 1 _ x ·
x _ x - 1
Multiply by the reciprocal.
= 1 _ x ·
x _ x - 1
Divide out common factors.
= 1 _
x - 1 Simplify.
Example 5 Application: Speed Walking
It took Max 3x2 - 12x _
3x minutes to speed walk to the gym that was
5x - 20 _
x3 miles away. Find his rate in miles per minute.
SOLUTION
r = d _ t Solve for r.
r =
5x - 20 _
x3 __
3x2 - 12x
_ 3x
Evaluate for d and t.
= 5x - 20 _
x3 ÷
3x2 - 12x _ 3x
Write using a division symbol.
= 5x - 20 _
x3 ·
3x _ 3x2 - 12x
Multiply by the reciprocal.
= 5(x - 4)
_ x3
· 3x _
3x(x - 4) Factor out any GCFs.
= 5 _ x3
miles per minute Divide out common factors and simplify.
The expression 5 _ x3 represents Max’s speed-walking rate in miles per minute.
Hint
Find the LCD of 1 and 1 _ x to subtract in the denominator.
Hint
Use the formula d = rt.
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Use the following strategy with students who have diffi culty with complex operations. Have students write the expression below.
1 __ 2
____ 3 __ 4
Then have them write the same expression replacing the fractions with their decimal equivalents.
Students should immediately see the order of operations:
0.5 ____ 0.75
= 0.50
____ 0.75
= 2 __ 3
Remind students that the method to simplify a complex fraction is the same.
INCLUSION
Lesson 92 611
Example 4
Students learn to combine fractions to simplify complex fractions.
Additional Example 4
Simplify 1- 1 ______
x + 1 ________
x ___ 2x . 2x ______
(x + 1)
TEACHER TIPEmphasize that the long fraction bar between the numerator and denominator of a complex fraction indicates division and should be the last operation. The expressions in the numerator and denominator need to be simplifi ed or evaluated separately fi rst.
Example 5
Students learn to fi nd a complex fraction that describes the rate of a person’s walking speed.
Extend the Example
If it takes Max 25 minutes to walk to the grocery store that is 1 mile away, what is x ? x = 5
Additional Example 5
On her bike, Emily can make it to her school that is x miles away in 3 + 4x minutes. If Jason’s house is 1 __ x miles away from school, how long will it take Emily to get there from her school? 3 + 4x
______ x 2 minutes
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Practice Distributed and Integrated
Saxon Algebra 1612
Lesson Practice
Simplify.
a. x
_ 4 _
3(x - 3)
_ x
x2_12(x - 3)
b. b _
cd _
2b
_ c 1_
2d
c.
4x2
_ x - 3
_
x _
3x - 9 12x d.
1 _ m + 5 _
2 _ m - x _ m
1 + 5m_2 - x
e. It took Ariel 5x2 - 45x _ 5x
minutes to walk to school that was 3x - 27
_ x3
miles away. Find her rate in miles per minute. 3_x3 miles per minute
(Ex 1)(Ex 1) (Ex 2)(Ex 2)
(Ex 3)(Ex 3) (Ex 4)(Ex 4)
(Ex 5)(Ex 5)
Find the product or quotient.
1. 15x4
_ x - 4
· x2 - 10x + 24 __
3x3 + 12x2
5x2(x - 6) _
(x + 4)
2. x2 + 12x + 36 __
x2 - 36 ÷
1 _ x - 6
x + 6
Solve.
3. -3(r - 2) > -2(-6) r < -2
4. y _
4 +
1 _ 2
< 2 _ 3 y < 2_
3
Simplify.
*5.
5x _
10x + 20 _
15 _
x + 2 x_
30 6. 8 √ � 9 • 2 √ � 5 48 √ � 5
*7. Write Under what conditions is a rational expression undefined? Sample: when the denominator equals zero
*8. Justify Give an example to show a _ b _
c _
d =
a _ b ·
d _ c = ad _ bc
. See Additional Answers.
*9. Skating It took Jim 15 __
x2 + 2x - 3 minutes to skate to the park that was
2x _
8x - 8 + x
_ 4x + 12
miles away. Find his rate in miles per minute. x2 + x_
30 miles
per minute
(88)(88)
(88)(88)
(77)(77)
(77)(77)
(92)(92) (76)(76)
(92)(92)
(92)(92)
(92)(92)
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612 Saxon Algebra 1
Lesson Practice
Problem d
Scaff olding Before writing the quotient using a division symbol, remind students to combine the fractions fi rst.
Problem e
Error Alert Students will commonly calculate the rate incorrectly by dividing the time by the distance. Remind them that since speed has units of miles per minutes, it is “miles divided by minutes.”
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“What is another way to write out
the fraction a _ x
_ b _ y ?” Sample: a __ x ÷ b __ y
“What can be done to a complex fraction to make the denominator equal to 1?” Sample: Multiply the numerator and denominators by the reciprocal of the denominator.
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 6
Before multiplying the expressions together, remind students to check for perfect squares.
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Lesson 92 613
*10. Multiple Choice Two fractions have a denominator of x2 + 6x + 9 and x2 - 9. What is the least common denominator? DA x2 + 9 B x2 + 6x + 9
C 2x2 + 9 D (x + 3)2(x - 3)
11. Solve and graph the inequality ⎢x� < 65.
*12. Error Analysis Two students solve the inequality ⎢x - 15⎥ < -4. Which student is correct? Explain the error.
Student A Student B
⎢x - 15⎢ < -4-4 < x - 15 < 4
11 < x < 19
⎢x - 15⎢ < -4no solution
*13. Geometry A triangle has sides measuring 16 inches and 23 inches. The triangle inequality states that the length of the third side must be greater than 7 inches and less than 39 inches. Write this inequality and graph it.
*14. Multi-Step The grades on a math test were all within the range of 80 points plus or minus 15 points.
a. Write an absolute -value inequality to show the range of the grades. ⎪x - 80⎥ ≤ 15
b. Solve the inequality to find the actual range of the grades. 65 ≤ x ≤ 95
*15. Error Analysis Two students simplified x + 2 _
2x - 5 - 3 - x
_ 2x - 5
. Which student is correct? Explain the error.
Student A Student B
x + 2 - 3 + x __
2x - 5 =
2x - 1 _ 2x - 5
x + 2 - 3 - x __
2x - 5 =
-1 _ 2x - 5
*16. Canoeing Vanya paddled a canoe upstream for 4 miles. He then turned the canoe around and paddled downstream for 3 miles. The current flowed at a rate of c miles per hour. Write a simplified expression to represent his total canoeing time if he kept a constant paddling rate of 6 miles per hour. c + 42__
(6 - c)(6 + c)
17. Multiple Choice Which equation’s graph has a maximum? BA y = -5 + x2 B y = -x2 + 5x
C y = x2 + 5 D y = 5x2 - 1
18. Write Describe 2 ways to find the axis of symmetry for a parabola.
19. Justify Give an example of a quadratic function and an example of a function that is not quadratic. Explain why each function is or is not quadratic. See Additional Answers.
(92)(92)
(91)(91)-65 < x < 65;
0 32.5 65-32.5-65-65 < x < 65;
0 32.5 65-32.5-65
(91)(91)
Student B; Sample: Student A did not realize that an absolute value can never be less than -4because absolute value is always positive.
Student B; Sample: Student A did not realize that an absolute value can never be less than -4because absolute value is always positive.
(91)(91)
7 < s < 39; 393123157
7 < s < 39; 393123157
(91)(91)
(90)(90)
15. Student A; Sample: Student B did not fully distribute the negative sign through the numerator of the second expression.
15. Student A; Sample: Student B did not fully distribute the negative sign through the numerator of the second expression.
(90)(90)
(89)(89)
(89)(89)
18. Sample: One way is to use the zeros of the function. The axis of symmetry goes through the zero when there is 1 zero because the zero is contained in the vertex. It goes through the average of the 2 zeros when there are 2 zeros. The second way is to use the formula x = - b_
2a. This is the
only way to find the axis of symmetry when the function has no zeros.
18. Sample: One way is to use the zeros of the function. The axis of symmetry goes through the zero when there is 1 zero because the zero is contained in the vertex. It goes through the average of the 2 zeros when there are 2 zeros. The second way is to use the formula x = - b_
2a. This is the
only way to find the axis of symmetry when the function has no zeros.
(84)(84)
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Lesson 92 613
Problem 14
Guide students by asking them the following questions.
“What math operation could the phrase plus or minus indicate?” Sample: absolute value
“What operation could the word range indicate?” Sample: distance, difference
Extend the Problem
Translate the inequality into words. The distance between a number and 80 is less than or equal to 15.
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Saxon Algebra 1614
20. Find the length a to the nearest tenth. 10.5
a17
20
21. Subtract 5f + 6
_ f 2 + 7f - 8
- f + 10
_ f 2 + 7f - 8
. 4_f + 8
22. Let A = (-5, 3), B = (0, 7), C = (12, 7), and D = (7, 3). Use the distance formula to determine whether ABCD is a parallelogram. yes
23. Geography Ithaca, New York is almost directly west of Oneonta, New York and directly north of Athens, Pennsylvania. The three cities form a triangle that is nearly a right triangle. Use the distance formula to estimate the distance from Athens to Oneonta. Each unit on the grid represents 5 miles. about 83 miles
x
y
5
5
Ithaca, NY
New YorkPennsylvania
Oneonta, NY
Athens, PA
24. Error Analysis Two students factor the polynomial (3x2 + 6) - (4x3 + 8x). Which student is correct? Explain the error.
Student A Student B
3(x2 + 2) - 4x(x2 - 2) 3(x2 + 2) - 4x(x2 + 2)(3 - 4x)(x2 - 4) (3 - 4x)(x2 + 2)
25. Probability A bag has 2x + 1 red marbles, 3x blue marbles, and x + 2 green marbles. What is the probability of picking a red marble? 1_3
26. Multi-Step A toolbox is 2 feet high. Its volume is represented by the expression 2x2 - 8x + 6.
a. Factor the expression completely. 2(x - 3)(x - 1)
b. Identify the expressions that represent the length and width of the toolbox. Sample: The length is (x - 1) and the width is (x - 3).
27. Hardiness Zones The state of Kansas falls into USDA hardiness zones 5b and 6a. This means that plants in these zones must be able to tolerate an average minimum temperature range greater than or equal to -15°F, and less than -5°F. Write an inequality to represent the temperature range of the hardiness zones in Kansas. -15 ≤ t < -5
(85)(85)
(90)(90)
(86)(86)
(86)(86)
(87)(87)
Student B; Sample: Student A did not distribute the negative sign correctly.
Student B; Sample: Student A did not distribute the negative sign correctly.
(38)(38)
(79)(79)
(82)(82)
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Simplify:
x
2 - 9 ______ x __________
x2 + 5x + 6
___________ x2 x(x - 3)
_______ x + 2
CHALLENGE
614 Saxon Algebra 1
Problem 25
Extend the Problem
What is the probability of picking a red marble if there are 30 blue marbles in the bag? 1 __
3
Does the probability of picking a red marble depend on the value of x ? No, the probability is independent of x.
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Lesson 92 615
28. Multi-Step A rectangular playing field has a perimeter of 8x feet. Its length is 2 feet greater than 2x.a. Write expressions for the length and width of the rectangle.
b. Write an expression for the area of the field. 4x2 - 4
c. The playing field is in a city park. The park is a large square with a side length of x2. Write and factor an expression for the area of the park that does not include the playing field. x4 - 4x2 + 4 = (x2 - 2)2
29. Vertical Motion The height h of an object t seconds after it begins to fall is given by the equation h = -16t2 + vt + s, where v is the initial velocity and s is the initial height. When an object falls, its initial velocity is zero. Write an equation for the height of an object t seconds after it begins to fall from 14,400 feet. Then factor the expression representing the height. h = -16t2 + 14,400; 16(30 + t)(30 - t)
30. Given the equation r = kst _ p , use the terms “jointly proportional to” and “inversely
proportional to” to describe the relation in the equation such that k is the constant of variation. In the equation, r is jointly proportional to s and t, and inversely proportional to p.
(83)(83)
length: 2x + 2, width: 2x - 2length: 2x + 2, width: 2x - 2
(83)(83)
(Inv 8)(Inv 8)
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Solving complex fractions prepares students for
• Lesson 95 Combining Rational Expressions with Unlike Denominators
• Lesson 99 Solving Rational Equations
LOOKING FORWARD
Lesson 92 615
Problem 30
Error Alert Students can sometimes confuse the terms jointly and inversely. Remind them that if the value of a variable increases as the value of a related variable decreases, then the variables are inversely proportional.
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Saxon Algebra 1616
Warm Up
93LESSON
1. Vocabulary A is a monomial or the sum or difference of monomials. polynomial
Divide.
2. 72x2 - 8x _
8x 9x - 1 3. 15x2 - 3x
_ 5x - 1
3x
Factor.
4. 2x2 + x - 3 (2x + 3)(x - 1) 5. 25x2 - 9 (5x + 3)(5x - 3)
A polynomial division problem can be written as a rational expression. To divide a polynomial by a monomial, divide each term in the numerator by the denominator.
Example 1 Dividing a Polynomial by a Monomial
Divide (8x3 + 12x2 + 4x) ÷ 4x.
SOLUTION
(8x3 + 12x2 + 4x) ÷ 4x
= (8x3 + 12x2 + 4x)
__ 4x
Write as a rational expression.
= 8x3
_ 4x
+ 12x2
_ 4x
+ 4x _ 4x
Divide each term by the denominator.
= 2 8x3 2
_ 14x
+ 312x2 1
_ 1 4x
+ 4x _ 4x
Divide out common factors.
= 2x2 + 3x + 1 Simplify.
Division of a polynomial by a binomial is similar to division of whole numbers.
Words Numbers Polynomials
Step 1
Factor the numerator and denominator, if possible.
128 _ 4 =
32 · 4 _ 4
x2 + 4x + 3 __ x + 1
= (x + 3)(x + 1)
__ x + 1
Step 2 Divide out any common factors.
32 · 4 _
4
(x + 3)(x + 1) __
x + 1
Step 3 Simplify. 32 x + 3
(53)(53)
(38)(38) (43)(43)
(75)(75) (83)(83)
New ConceptsNew Concepts
Dividing Polynomials
Online Connection
www.SaxonMathResources.com
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LESSON RESOURCESLESSON RESOURCES
Student Edition Practice Workbook 93
Reteaching Master 93Adaptations Master 93Challenge and Enrichment
Master C93, E93
The equation
(ax2 + bx + c) ÷ (x + d ) = x + e can be written as a rational equation in the form:
a x 2 + bx + c ___________
x + d = x + e
It can also be written in long-division form:
x + d
x + e
� ��������������������������������������������������� a x 2 + bx + c .
Regardless of the form, the quotient times the divisor is always equal to the dividend. This is a good way for students to learn to check their work.
MATH BACKGROUND
Warm Up1
616 Saxon Algebra 1
93LESSON
Problem 5
Remind students that the difference of squares a2x2 - b2 = (ax + b)(ax - b).
2 New Concepts
In this lesson, students learn to divide polynomials.
Discuss the defi nition of a polynomial. Explain that division of polynomials is similar to division of whole numbers.
Example 1
Point out to students that a polynomial is the sum or difference of monomials, and that dividing a polynomial by a monomial can be done by dividing each term by the monomial.
Additional Example 1
Divide (6x4 + 4x3 - 2x2) ÷ 2x. 3x3 + 2x2 - x
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Lesson 93 617
Example 2 Dividing a Polynomial by a Binomial
Divide each expression.
a. (x2 - 6x + 9) ÷ (x - 3)
SOLUTION
(x2 - 6x + 9) ÷ (x - 3)
= x2 - 6x + 9__
x - 3Write as a rational expression.
= (x - 3)(x - 3)__
x - 3Factor the numerator.
= (x - 3)(x - 3)__
(x - 3)Divide out common factors.
= x - 3 Simplify.
b. (x2 - 5x + 6) ÷ (2 - x)
SOLUTION
(x2 - 5x + 6) ÷ (2 - x)
= x2 - 5x + 6__
2 - xWrite as a rational expression.
= (x - 3)(x - 2)
__ 2 - x
Factor the numerator.
= (x - 3)(x - 2)__
-x + 2Write the denominator in descending order.
= (x - 3)(x - 2)
__ -1(x - 2)
Factor out a -1 in the denominator.
= (x - 3)(x - 2)
__ -1(x - 2)
Divide out common factors.
= -x + 3 Simplify.
Just as with whole numbers, long division can also be used to divide polynomials.
27
21 � ����������������� 567
x + 2
x + 1 � ����������������������������������������������� x2 + 3x + 2
__ -42 _____-(x2 + x)
147 2x + 2
___ -147 _____ -(2x + 2)
0 0
Hint
When the signs of the binomials, one in the numerator and one in the denominator, are opposites, factor out -1.
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ENGLISH LEARNERS
For this lesson, explain the meanings of the prefi xes mono-, bi-, and poly-. Say:
“Mono- means one, alone, or single, bi- means two, and poly- means many or much.”
Connect the meaning of monomials and polynomials by the defi nition of polynomials. Emphasize that a polynomial is “an expression with one or multiple monomials.”
Discuss examples of words that use mono-, bi-, and poly- and relate the use of the prefi xes to their defi nitions, e.g. polygon, monopoly, bicycle, polymer, and monorail.
Lesson 93 617
Example 2
Remind students to factor the numerator and denominator, if possible.
Additional Example 2
Divide each expression.
a. (x2 + 5x - 14) ÷ (x + 7) x - 2
b. (x2 + 5x -14) ÷ (2 - x) -(x + 7)
Error Alert A common mistake when dividing by a polynomial is to divide the numerator by each term in the denominator. Write the statement:
x ___________ x 2 + 3x - 4
≠ x ___ x 2
+ x ___ 3x
- x __ 4
Remind students that only terms in the numerator can be divided by the denominator.
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Saxon Algebra 1618
Example 3 Dividing a Polynomial Using Long Division
Divide using long division.
(-25x + 3x2 + 8) ÷ (x - 8)
SOLUTION
(-25x + 3x2 + 8) ÷ (x - 8)
x - 8 � �������������������������������������������������������������� (3x2 - 25x + 8) Write in long-division form with expressions in standard form.
3x
x - 8 � ������������������������������������������������������������ 3x2 - 25x + 8x Divide the first term of the dividend by the
first term of the divisor to find the first term of the quotient.
3x
x - 8 � �������������������������������������������������������� 3x2 - 25x + 8
Multiply the first term of the quotient by the binomial divisor. Write the product under the dividend. Align like terms. _____ 3x2 - 24x
3x
x - 8 � �������������������������������������������������������� 3x2 - 25x + 8
Subtract the product from the dividend. Then bring down the next term in the dividend.
______ -(3x2 - 24x)
-x + 8
3x - 1
x - 8 � �������������������������������������������������������� 3x2 - 25x + 8
_____ -3x2 + 24x
-x + 8 Repeat the steps to find each term of the quotient. _____ -(-x + 8)
0 The remainder is 0.
The quotient is (3x - 1) remainder 0.
Check Multiply the quotient and the divisor.
(3x - 1)(x - 8)
= 3x2 - 24x - x + 8
= 3x2 - 25x + 8
The divisor is not always a factor of the dividend. When it is not, the remainder will not be 0. The remainder can be written as a rational expression using the divisor as the denominator.
Example 4 Long Division with a Remainder
Divide using long division.
(2x2 - 9 - 7x) ÷ (-4 + x)
Caution
Be sure to put the divisor and dividend in descending order before dividing.
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618 Saxon Algebra 1
Example 3
Emphasize to students that long division can be used to divide polynomials.
Additional Example 3
Divide using long division.
(13x2 + 5x - 42) ÷ (x + 2) 13x - 21
Example 4
Polynomial long division may have remainders. Students can multiply the divisor by the quotient and then add the remainder to that product to check their work.
Additional Example 4
Divide using long division.
(4x2 + 7x + 17) ÷ (x - 3) 4x + 19 + 74 _____
x - 3
TEACHER TIPEmphasize that when using long division to divide polynomials, the remainder is a rational expression using the divisor as the denominator.
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Lesson 93 619
SOLUTION
(2x2 - 9 - 7x) ÷ (-4 + x)
x - 4 � ��������������������������������������������������� 2x2 - 7x - 9 Write in long-division form with expressions in standard form.
2x
x - 4 � ��������������������������������������������������� 2x2 - 7x - 9
Divide the first term of the dividend by the first term of the divisor to find the first term of the quotient.
2x
x - 4 � ��������������������������������������������������� 2x2 - 7x - 9
Multiply the first term of the quotient by the binomial divisor. Write the product under the dividend. Align like terms. ____ 2x2 - 8x
2x
x - 4 � ��������������������������������������������������� 2x2 - 7x - 9
Subtract the product from the dividend. Then bring down the next term in the dividend.
______ -(2x2 - 8x)
x - 9
2x + 1
x - 4 � ��������������������������������������������������� 2x2 - 7x - 9
Repeat the steps to find each term of the quotient.
______ -(2x2 - 8x)
x - 9
____ -(x - 4)
-5
The quotient is 2x + 1 - 5 _
x - 4 .
Example 5 Dividing a Polynomial with a Zero Coefficient
Divide (-2x + 5 + 3x3) ÷ (-3 + x).
SOLUTION
(-2x + 5 + 3x3) ÷ (-3 + x)
(3x3 - 2x + 5) ÷ (x - 3) Write each polynomial in standard form.
x - 3 � ������������������������������������������������������������������������� 3x3 + 0x2 - 2x + 5 Write in long division form. Use 0x2 as a placeholder for the x2-term.
3x2 + 9x + 25
x - 3 � �������������������������������������������������������������������������� 3x3 + 0x2 - 2x + 5
3x3 ÷ x = 3x2
______ -(3x3 - 9x2) Mulitply 3x2(x - 3). Then subtract.
9x2 - 2x Bring down -2x. 9x2 ÷ x = 9x
______ -(9x2 - 27x) Multiply 9x(x - 3). Then subtract.
25x + 5 Bring down 5. 25x ÷ x = 25
______ -(25x - 75) Multiply 25(x - 3). Then subtract.
80 The remainder is 80.
The quotient is 3x2 + 9x + 25 + 80 _
x - 3 .
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INCLUSION
Using long division, have students divide the trinomial x2 - 3x - 18 by the binomial x - 6 on grid paper to help them keep their work properly aligned.
Once the students have the answer, have them multiply the divisor by their answer to check that the product of the two expressons is equal to the dividend.
Next, have the students write a rational expression using the same polynomials.
x2 - 3x - 18 ____________ x - 6
Have them factor and divide out like factors to fi nd the quotient.
Point out that either method, fi nding the quotient using long division or using a rational expression and factoring, should result in the same quotient.
Lesson 93 619
Example 5
Point out to students that when using long division to divide polynomials, it is very important to write the polynomial in standard form and to use 0 as the placeholder for any missing terms.
Additional Example 5
Divide (6x2 - 7 + 3x3) ÷ (4 + x). 3x2 - 6x + 24 - 103 ______
x + 4
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Practice Distributed and Integrated
Saxon Algebra 1620
Example 6 Application: Length of a Garden
Jim wants to find the length of the rectangular garden outside his office. The area is (x2 - 11x + 30) square feet. The width is (x - 6) feet. What is the length of the garden?
SOLUTION
l = A _ w Solve for l.
= x2 - 11x + 30 __
(x - 6) Evaluate for A and w.
= (x - 6)(x - 5)
__ (x - 6)
Factor the numerator.
= (x - 6) (x - 5)
__ (x - 6)
Divide out common factors.
= (x - 5) Simplify.
The length of the garden is (x - 5) feet.
Lesson Practice
Divide each expression.
a. (7x4 + 7x3 - 84x2) ÷ 7x2 x2 + x - 12
b. (x2 - 10x + 25) ÷ (x - 5) x - 5
c. (3x2 - 14x - 5) ÷ (5 - x) -3x - 1
Divide using long division.
d. (8x2 + x3 - 20x) ÷ (x - 2) x2 + 10x
e. (-3x2 + 6x3 + x - 33) ÷ (-2 + x) 6x2 + 9x + 19 + 5_x - 2
f. (6x + 5x3 - 8) ÷ (x - 4). 5x2 + 20x + 86 + 336_x - 4
g. Carlos wants to find the width of his rectangular deck. The area is (x2 - 10x + 24) square feet and the length is (x - 4) feet. What is the width? (x-6) feet
(Ex 1)(Ex 1)
(Ex 2)(Ex 2)
(Ex 2)(Ex 2)
(Ex 3)(Ex 3)
(Ex 4)(Ex 4)
(Ex 5)(Ex 5)
(Ex 6)(Ex 6)
1. Find the distance between (-3, 2) and (9, -3). Give the answer in simplest radical form. 13
2. Solve 5 _
16 y +
3 _ 8 ≥
1 _ 2
, and graph the solution. y ≥ 2_5
; 10-1
Factor.
3. 2x2 + 12x + 16 2(x + 4)(x + 2) 4. 3x3 - 5x2 - 9x + 15 (3x - 5)(x2 - 3)
(86)(86)
(77)(77)
(79)(79) (87)(87)
Hint
To find the length, solve the formula for the area of a rectangle, A = lw, for the length.
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620 Saxon Algebra 1
Example 6
In this example, division of polynomials is used to solve a word problem.
Extend the Example
“What is the domain of x ? How do you know ?” x > 6; Since x - 6 is the width and x - 5 is the length of the garden, both have to be greater than zero for there to be a garden.
“If Jim added a fence around his garden and that added another foot to both the width and length, what is the new area?” (x - 5)(x - 4) = x2 - 9x + 20 square feet
Additional Example 6
Jillian wants to fi nd the radius of the circular fountain in the park. The area is 16πx2 + 24πx + 9π square feet. What is the radius of the fountain? 4x + 3 feet
Lesson Practice
Problem b
Scaff olding Have students write a rational expression, and then have them factor each term completely and divide out common factors.
Problem f
Error Alert Students may forget to include the divisor as the denominator of the remainder. Remind them that the remainder is a fractional part of the divisor.
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Lesson 93 621
*5. Find the quotient: 4x3 + 42x2 - 2x __ 2x
. (2x2 + 21x - 1)
6. Find the axis of symmetry for the graph of the equation y = x2 - 2x. x = 1
Simplify.
7.
7x4
_ 4x + 18
_
3x2
_ 6x + 27
7x2_2
*8. 1 _ x3 _
1 _ x3 +
1 _ x3
1_2
*9. Write Explain how to check that (5x + 6) is the correct quotient of (15x2 + 13x - 6) ÷ (3x - 1). Sample: Multiply the divisor by the quotient. The product should equal the dividend.
*10. Justify Show that the quotient of (x2 - 4) ÷ (x + 2) can be found using two different methods.
*11. Swimming The city has decided to open a new public pool. The area of the new rectangular pool is (x2 - 16x + 63) square feet and the width is (x - 7) feet. What is the length? (x - 9) feet
*12. Multiple Choice Simplify x3 - 7x + 3x2 - 21 __
x + 3 . A
A x2 - 7 B x3 - 7
C -3 D x2 - 7
_ 2x
*13. Error Analysis Students were asked to simplify 6x2 - 6x _ 8x2 + 8x
_
3x - 3
_ 4x2 + 4x
. Which student is correct?
Explain the error.
Student A
6x2 - 6x _ 8x2 + 8x
· 4x2 + 4x _ 3x - 3
= 6x(x - 1)
_ 8x(x + 1)
· 4x(x + 1)
_ 3(x - 1)
= x
Student B
6x2 - 6 _
8x2 + 8x ·
4x2 + 4x _ 3x - 3
= 6x(x - 1)
_ 8x(x + 1)
· 4x(x + 1)
_ 3(x - 1)
= 24x2
_ 24x
14. Multi-Step Brent rode his scooter 8x2 - 48x _
24x5 minutes to get to baseball practice that was 7x - 42
_ 4x2 miles away.
a. Find his rate in miles per minute. 21x2_
4 miles per minute
b. If the rate is divided by 1 _ x , what is the new rate? 21x3_
4 miles per minute
*15. Geometry The area of a parallelogram is m + n _
5 square inches and the height is
m2 + n2
_ 15
inches. What is the length of the base? 3(m + n)
_m2 + n2 inches
16. Solve and graph the inequality ⎢x� > 84. x < -84 OR x > 84
(93)(93)
(89)(89)
(92)(92) (92)(92)
(93)(93)
(93)(93)Method 1:
x2 - 4_
x + 2 =
(x - 2)(x + 2) _
(x + 2)
= (x - 2)
Method 2: x - 2 x + 2� ���������������������������������������������� x2 + 0x - 4
____-x2 - 2x
-2x - 4
____ +2x + 4
0
10.10. Method 1:
x2 - 4_
x + 2 =
(x - 2)(x + 2) _
(x + 2)
= (x - 2)
Method 2: x - 2 x + 2� ���������������������������������������������� x2 + 0x - 4
____-x2 - 2x
-2x - 4
____ +2x + 4
0
10.10.
(93)(93)
(93)(93)
(92)(92) Student A; Sample: Student B did not write solution in simplest form.Student A; Sample: Student B did not write solution in simplest form.
(92)(92)
(92)(92)
(91)(91) 0 42 84-42-84 0 42 84-42-84
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Lesson 93 621
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“Explain some of the fi rst steps that should be taken before dividing a polynomial.” Sample: Check if the polynomial can be factored, and write the polynomial in standard form.
“When doing long division, why are placeholders necessary for the missing terms in the polynomial ?” Sample: Different places in the polynomial represent different powers of the variable.
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 7
Before students simplify the complex fractions, encourage them to factor the polynomials.
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Saxon Algebra 1622
*17. Census The census of England and Wales has a margin of error of ±104,000people. The 2001 census found the population to be 52,041,916. Write an absolute-value inequality to express the possible range of the population. Then solve the inequality to find the actual range for the population. ⎢x - 52,041,916� ≤ 104,000; 51,937,916 ≤ x ≤ 52,145,916
18. Error Analysis Two students solve the inequality ⎢x + 11� > -15. Which student is correct? Explain the error.
Student A
⎢x + 11� > -15x can be any real number.
Student B
⎢x + 11� > -15x = Ø
19. Multiple Choice When simplifying 1 _ x2 - 5x - 50
+ 1 _ 2x - 20
, what is the numerator? D
A 1 B 2
C x + 5 D x + 7
*20. Analyze Explain what can be done so that 1 _ 3 - r
and 5r _
r - 3 have like denominators.
21. Probability Mr. Brunetti writes quadratic equations on pieces of paper and puts them in a hat, and then tells his students each to choose two at random. After each student picks, his or her two papers go back in the hat. The functions are y = 4x2 - 3x + 7, y = -7 + x2, y = -2x + 6x2, y = 0.5x2 + 1.1, and y = - 1 _
2 x2 + 7x + 5. What is the probability of a student choosing two functions
that have a minimum? 3_5
22. Error Analysis Two students divide the following rational expression. Which student is correct? Explain the error.
Student A
m2
_ 6m2
÷ (m2 + 2)
m2
_ 6m2
· 1 _
m2 + 2 =
1 _ 6(m2 + 2)
Student B
m2
_ 6m2
÷ (m2 + 2)
m2
_ 6m2
· m2 + 2 _
1 =
m2 + 2 _ 6
23. Cost A bakery sells specialty rolls by the dozen. The first dozen costs 6b + b2 dollars. Each dozen thereafter costs 4b + b2 dollars. If Marcello buys 4 dozen rolls, how much does he pay? 2b(9 + 2b) dollars
24. Find the vertical and horizontal asymptotes and graph y = 4 _ x + 2
. x = -2; y = 0
25. Flooring Theo is installing new kitchen tiles. The design on the tile includes a square within a square. The smaller square has a side length of s centimeters. The expression 4s2 + 12s + 9 describes the area of the entire tile. What is the difference between the length of the tile and the length of the square within the tile? s + 3 centimeters
(91)(91)
(91)(91) Student A; Sample: Student B did not realize that all absolute values are greater than -15.Student A; Sample: Student B did not realize that all absolute values are greater than -15.
(90)(90)
(90)(90)20. Sample: Multiply either of the expressions by -1_
-1 because
-1(3 - r) =
-3 + r = r - 3, or -1(r - 3) =
-r + 3 = 3 - r.
20. Sample: Multiply either of the expressions by -1_
-1 because
-1(3 - r) =
-3 + r = r - 3, or -1(r - 3) =
-r + 3 = 3 - r.
(89)(89)
(88)(88) Student A; Sample: Student B did not multiply by the reciprocal of the rational expression.Student A; Sample: Student B did not multiply by the reciprocal of the rational expression.
(87)(87)
(78)(78)
24.
x
y
O4
4
-4
-8
8
824.
x
y
O4
4
-4
-8
8
8
(83)(83)
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CHALLENGE
Polynomial division can include multiple variables. Divide:
2 x 2 + 3xy + y 2
_____________ x + y
Have students check their answers by multiplying the divisor and quotient. (2x + y)
622 Saxon Algebra 1
Problem 19
Guide the students by asking them the following questions.
“How can the denominator of the fi rst term be factored ?” (x - 10)(x + 5)
“How can the denominator of the second term be factored ?” 2(x - 10)
Problem 23
Extend the Problem
“If Marcello had 6b2 + 28b dollars, how many dozen rolls could he buy?” 6 dozen
“If Marcello bought 6 dozen rolls, how much money would he have left?” 2b dollars
Problem 25
Error Alert A common mistake with word problems is to solve the wrong problem. Remind students that the question asks for the difference between the lengths and not the lengths of the two tiles.
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Lesson 93 623
26. Multi-Step Tony is sketching the view from the top of a 256-foot-tall observation tower and accidentally drops his pencil. a. Use the formula h = -16t2 + 256 to make a table of values showing the height h
of the pencil, 1, 2, and 3 seconds after it is dropped. (1, 240), (2, 192), (3, 112)
b. Graph the function.
c. About how long does it take for the pencil to hit the ground? about 4 seconds
27. Pendulums The time it takes a pendulum to swing back and forth depends on its length. The formula l = 2.45 t
2
_ π
2 approximates this relationship. Graph the function using 3.14 for π. Use the graph to estimate the time it takes a pendulum that is 1 meter long to swing back and forth. about 2 sec
28. Write Explain how to determine whether a triangle with side lengths 5, 7, and 10 is a right triangle.
29. Multi-Step A bag of marbles contains 3 red, 5 blue, 2 purple, and 4 clear marbles.a. Make a graph that represents the frequency distribution.
b. What is the probability of drawing a red or a clear marble? 7_14
= 1_2
30. Given the following table, find the value of the constant of variation and complete the missing values in the table given that y varies directly with x and inversely with z. k = 3
y x z
1 1 3
3 2 2
6 4 2
9 6 2
2 8 12
(84)(84)
26 b.
x
y
180
240
120
60
1 2 3 4O
26 b.
x
y
180
240
120
60
1 2 3 4O
(84)(84)
27.
x
y
3
4
2
1
1 2 3 4O
27.
x
y
3
4
2
1
1 2 3 4O
(85)(85)28. Sample: Substitute 5 for a, 7 for b, and 10 for c in the equation a2 + b2 = c2
and simplify the equation. If the equation is true, then the triangle is a right triangle. If the equation is false, then the triangle is not a right triangle.
28. Sample: Substitute 5 for a, 7 for b, and 10 for c in the equation a2 + b2 = c2
and simplify the equation. If the equation is true, then the triangle is a right triangle. If the equation is false, then the triangle is not a right triangle.
(80)(80)
29.
543210 Red Blue Purple Clear
Marbles
Colors
29.
543210 Red Blue Purple Clear
Marbles
Colors
(Inv 8)(Inv 8)
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LOOKING FORWARD
Dividing polynomials prepares students for
• Lesson 95 Combining Rational Expressions with Unlike Denominators
• Lesson 99 Solving Rational Equations
Lesson 93 623
Problem 27
If students graph this function without a graphing calculator, tell them to use multiples of π for t to make the calculations easier. For example, let t = 3 · 3.14. Then (3 · 3.14)2 = 32 · (3.14)2 and the (3.14)2 will divide out.
Problem 30
A review of the defi nitions of direct and inverse variation may be helpful.
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Saxon Algebra 1624
Warm Up
94LESSON
1. Vocabulary The of a number n is the distance from n to 0 on a number line. absolute value
Simplify.
2. ⎪-9⎥ - 5 4 3. ⎪12 - 23⎥ 11
Solve.
4. 6x - 7 = 11 x = 3 5. 11x + 8 = 41 x = 3
To solve an absolute-value equation, begin by isolating the absolute value. Then use the definition of absolute value to write the absolute-value equation as two equations. Solve each equation, and write the solution set. There are often two answers to an absolute-value equation. The solutions can be graphed on a number line by placing a closed circle at each value in the solution set.
Example 1 Solving Equations with Two Operations
Solve each equation. Then graph the solution.
a. ⎪x⎥
_ 5 + 3 = 18
SOLUTION
First isolate the absolute value. Write the equation so that the absolute value is on one side of the equation by itself.
⎪x⎥
_ 5 + 3 = 18
__ -3 = __ -3 Subtraction Property of Equality
⎪x⎥
_ 5 = 15 Simplify.
5 · ⎪x⎥
_ 5 = 5 · 15 Multiplication Property of Equality
⎪x⎥ = 75 Simplify.
x = 75 or x = -75 Write as two equations without an absolute value.
The solution set is {-75, 75}.
Graph the solution on a number line.
0 25 50 75-25-50-75
(5)(5)
(5)(5) (5)(5)
(26)(26) (26)(26)
New ConceptsNew Concepts
Solving Multi-Step Absolute-Value Equations
Online Connection
www.SaxonMathResources.com
Math Language
The absolute value of a number n is the distance from 0 to n on a number line. The absolute value of 0 is 0.
Hint
Isolate the absolute value using inverse operations.
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MATH BACKGROUNDLESSON RESOURCES
Student Edition Practice Workbook 94
Reteaching Master 94Adaptations Master 94Challenge and Enrichment
Master C94
Solving absolute-value equations involves many of the same steps as solving multi-step equations in order to isolate the absolute value. Then, the defi nition of absolute value is applied to fi nd the solution set.
It is critical to understand that the absolute value of a number is its distance from zero. This distance can be to the left or right of 0. Even though the numbers are negative in one direction, the distance is still positive.
Beware of equations which simplify to an absolute value equal to a negative number. For these equations, there are no solutions because absolute value is never negative.
Emphasize to students that the absolute value expression must be isolated. The expression’s distance from zero is given on the opposite side of the equal sign. If the absolute value has other operators on the same side with it, then the distance of the quantity in the absolute value symbol from zero is not clear.
Warm Up1
624 Saxon Algebra 1
94LESSON
Problem 2
Remind students that the absolute value of a number is its distance from zero. Distance is always positive.
2 New Concepts
In this lesson, students learn to solve multi-step absolute-value equations and to graph the solutions.
Example 1
Students solve equations involving two operations.
Additional Example 1
Solve each equation. Then graph the solution.
a. ⎢x�
____ 2 - 1 = 14 {-30, 30};
3020100-30 -20 -10
b. -3⎢x� - 10 = -5 Ø; There is no graph.
TEACHER TIPExplain to students that the absolute-value symbol is a grouping symbol. Like parentheses, it must be isolated before it can be “undone” in an equation.
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Lesson 94 625
b. 4⎪x⎥ - 9 = 15
SOLUTION
First isolate the absolute value.
4⎪x⎥ - 9 = 15
4⎪x⎥ = 24 Add 9 to both sides.
⎪x⎥ = 6 Divide both sides by 4.
x = 6 and x = -6 Write as two equations without an absolute value.
The solution set is {-6, 6}.
Graph the solution on a number line.
640 2-2-4-6
Example 2 Solving Equations with More than Two Operations
Solve each equation.
a. 5⎪x⎥
_ 2 + 4 = 4
SOLUTION
5⎪x⎥ _
2 + 4 = 4
5⎪x⎥
_ 2 = 0 Subtract 4 from both sides.
5⎪x⎥ = 0 Multiply both sides by 2.
⎪x⎥ = 0 Divide both sides by 5.
Since the absolute value is equal to zero, there is only one solution. The solution set is {0}.
b. 2⎪x⎥
_ 6 + 3 = 1
SOLUTION
2⎪x⎥ _
6 + 3 = 1
2⎪x⎥
_ 6 = -2 Subtract 3 from both sides.
2⎪x⎥ = -12 Multiply both sides by 6.
⎪x⎥ = -6 Divide both sides by 2.
By the definition of absolute value, we know that there are no solutions to this equation. The absolute value is never negative. The solution set is empty. You can write this as {} or Ø.
Math Reasoning
Analyze Why can the absolute value never be negative?
Sample: A distance can only be represented by 0 or a positive number.
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INCLUSION
Use the following strategy with students who have diffi culty following multiple steps. The steps to solving these equations should be clearly identifi ed.
1. Isolate the absolute value.
2. Set up two equations to show the defi nition of absolute value.
3. Solve the two equations.
Having students label their work with the steps will lead them through the steps in the appropriate order. Have a bulletin board or poster modeling these steps with an example problem for students to see as they work.
Lesson 94 625
Example 2
Students learn to solve equations with more than two operations.
Additional Example 2
Solve each equation.
a. 5⎢x�
____ 2 + 4 = 19 { -6, 6}
b. 2⎢x�
____ 3 + 9 = 11 { -3, 3}
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Saxon Algebra 1626
Example 3 Solving Equations with Operations Inside the
Absolute-Value Symbols
Solve each equation.
a. ⎪2x⎥ + 9 = 15
SOLUTION
⎪2x⎥ + 9 = 15
⎪2x⎥ = 6 Subtract 9 from both sides.
Write the equation as two equations without the absolute value. Then solve both equations.
2x = 6 or 2x = -6
x = 3 Divide both sides by 2. x = -3
The solution set is {3, -3}.
b. 6⎪x + 3⎥ - 8 = 10
SOLUTION
6⎪x + 3⎥ - 8 = 10
6⎪x + 3⎥ = 18 Add 8 to both sides.
⎪x + 3⎥ = 3 Divide both sides by 6.
Write the equation as two equations without the absolute value and solve.
x + 3 = 3 or x + 3 = -3
x = 0 Subtract 3 from both sides. x = -6
The solution set is {0, -6}.
c. 5 ⎪ x
_ 3 - 2⎥ = 15
SOLUTION
5 ⎪ x
_ 3 - 2⎥ = 15
⎪ x
_ 3 - 2⎥ = 3 Divide both sides by 5.
Write the equation as two equations without the absolute value and solve.
x
_ 3 - 2 = 3 or
x _
3 - 2 = -3
x
_ 3 = 5 Add 2 to both sides.
x _
3 = -1
x = 15 Multiply both sides by 3. x = -3
The solution set is {15, -3}.
Math Reasoning
Write Why do you often have to solve two equations without absolute value to find the solution to one equation with absolute value?
Sample: The defi nition of absolute value creates two possibilities for each absolute-value sign.
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626 Saxon Algebra 1
Example 3
Students learn to solve equations with operations inside the absolute-value symbols.
Additional Example 3
Solve each equation.
a. ⎢5x� -10 = 25 {-7, 7}
b. 2 ⎪x - 4⎥ + 10 = 30 {-6, 14}
c. 3 ⎪ x __ 2 - 1⎥ = 18 {-10, 14}
Error Alert Students may try to isolate the variable without isolating the absolute value fi rst. Be sure to emphasize that once the absolute value is isolated, the defi nition of absolute value must be applied.
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Lesson 94 627
Example 4 Application: Archery
A 60-centimeter indoor archery target has several rings around a circular center. If the average diameter of a ring is d, and an arrow landing in that ring scores p points, the inner and outer diameters of the ring are given by the absolute-value equation ⎪d + 6p - 63⎥ = 3. Find the inner and outer diameters of the 8-point ring.
SOLUTION
⎪d + 6p - 63⎥ = 3
⎪d + 6 · 8 - 63⎥ = 3 Substitute 8 for p.
⎪d + 48 - 63⎥ = 3 Multiply.
⎪d - 15⎥ = 3 Subtract.
Write the absolute-value equation as two equations and solve.
d - 15 = 3 or d - 15 = -3
d = 18 Add 15 to both sides. d = 12
The inner diameter of the ring is 12 centimeters and the outer diameter of the ring is 18 centimeters.
Lesson Practice
Solve each equation. Then graph the solution.
a. ⎪x⎥_7 + 10 = 18 {56, -56}; 0 28 56-28-56
b. 3⎪x⎥ - 11 = 10 {7, -7}; 0 3.5 7-3.5-7
Solve each equation.
c.4⎪x⎥_
9+ 23 = 11 Ø
d.⎪x⎥ + 3_
2- 2 = 1 {3, -3}
e. ⎪7x⎥ + 2 = 37 {5, -5}
f. 5 ⎪x + 1⎥ - 2 = 23 {4, -6}
g. 9 ⎪ x_2
- 1⎥ = 45 {12, -8}
h. Investments A factory produces items that cost $5 to make. The factory would like to invest $100 plus or minus $10 in the first batch. Use the equation ⎪5x - 100⎥ = 10 to find the least and greatest number of items the factory can produce. 18 items, 22 items
8 pt.
inner diameterouter diameter
8 pt.
inner diameterouter diameter
(Ex 1)(Ex 1)
(Ex 2)(Ex 2)
(Ex 2)(Ex 2)
(Ex 3)(Ex 3)
(Ex 3)(Ex 3)
(Ex 3)(Ex 3)
(Ex 4)(Ex 4)
Caution
The absolute-value bars act as grouping symbols. Be sure to use the order of operations to simplify any expression within them.
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For Example 4, explain the meaning of the word ring. Tape a picture on the chalkboard and draw a circle around it. Say:
“This picture has a ring drawn around it. A ring is a circular line that surrounds an object. Circus acts are performed inside a large ring.”
Ask volunteers to give examples of other types of rings. Sample: rings around a planet; a wedding ring
ENGLISH LEARNERS
Lesson 94 627
Example 4
Students use absolute-value equations to solve an archery problem.
Extend the Example
Find the inner and outer diameters of the 6-point ring. 24 cm and 30 cm
Additional Example 4
A family wants to spend $200 plus or minus $50 on a campsite for their entire trip. The campsite costs $25 per night. Use the equation ⎪200 - 25x⎥ = 50 to fi nd the greatest and least number of nights the family can camp. 6 and 10 nights
Lesson Practice
Problem c
Error Alert Students need to stop when they isolate the absolute value and fi nd that it equals a negative number. There is no need to go any further. There is no solution.
Problem f
Scaff olding Have students isolate the absolute value. Then apply the defi nition of absolute value.
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“What are the steps for solving a multi-step absolute-value equation?” Sample: Isolate the absolute value. Apply the defi nition of absolute value. Set up two equations, and then solve them to fi nd two solutions.
“What happens if the absolute value is equal to a negative number?” Sample: There is no solution.
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Practice Distributed and Integrated
Saxon Algebra 1628
Add or subtract.
1. m2 _
m - 4 - 16 _
m - 4 m + 4 2.
-66 __ w2 - w - 30
+ w _ w - 6
w + 11_w + 5
*3. Write Explain why some absolute-value equations have no solutions.
*4. Multiple Choice The solution set {-12, 60} correctly solves which absolute-value equation? D
A 6 ⎪ x
_ 4 - 1⎥ = 42 B -2 ⎪
x _
4 - 1⎥ = 16
C 8 ⎪ x
_ 3 - 2⎥ = 48 D -5 ⎪
x _
6 - 4⎥ = -30
5. Refurbishing Rudy has x + y junk cars in his lot. He fixes them up and sells
each car for $400 + $100x
_ y . If he sells 30% of them, how much profit will he make?
6. Physics The function y = -16x2 + 80x models the height of a droplet of water from an in-ground sprinkler x seconds after it shoots straight up from ground level. Explain how you know when the droplet will hit the ground.
Factor.
7. 2a2 + 8ab + 6a + 24b 2(a + 3)(a + 4b)
8. zx10 - 4zx9 - 21zx8 zx8(x - 7)(x + 3)
9. Find the product of b - 4
_ b + 9 · (b2 + 11b + 18). (b - 4)(b + 2)
*10. Error Analysis Students were asked to simplify (15x4 + 4x2 + 3x3) ÷ (x - 6). Which student set their problem up correctly? Explain the error. Student B; Sample: Student A did not put the dividend in descending order or insert a placeholder.
Student A
(x - 6) � ������������������������������������������������������������������������ (15x4 + 4x2 + 3x3)
Student B
(x - 6) � ������������������������������������������������������������������������������������������������������������ (15x4 + 3x3 + 4x2 + 0x + 0)
Simplify.
*11. -1 _ 10x - 10
_
x5
_ 10x2 - 10
1 - x_
x5
12.
2x _
3x + 12 __
6x2
__ x2 + 8x + 16
x + 4_9x
(90)(90) (90)(90)
(94)(94)
3. Sample: An absolute value cannot be negative, so any absolute-valueequation that sets an absolute value equal to a negative number has no solution.
3. Sample: An absolute value cannot be negative, so any absolute-valueequation that sets an absolute value equal to a negative number has no solution.
(94)(94)
(88)(88)
5.30(x2 + 4x + 4y + xy)
___y dollars5.
30(x2 + 4x + 4y + xy) ___
y dollars
(89)(89)6. Sample: The droplet is at ground level when y = 0. This first occurs at 0 seconds, before the water has shot out from the sprinkler. The maximum value occurs at x = - b_
2a= - 80_
-32= 2.5.
Because of symmetry, the droplet is at ground level 2.5 seconds before and after its maximum point, so it hits the ground 5 seconds after it shoots up.
6. Sample: The droplet is at ground level when y = 0. This first occurs at 0 seconds, before the water has shot out from the sprinkler. The maximum value occurs at x = - b_
2a= - 80_
-32= 2.5.
Because of symmetry, the droplet is at ground level 2.5 seconds before and after its maximum point, so it hits the ground 5 seconds after it shoots up.
(87)(87)
(79)(79)
(88)(88)
(93)(93)
(92)(92)
(92)(92)
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628 Saxon Algebra 1
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 1
Combine the numerators, factor, and then cancel common factors in the numerator and denominator.
Problem 6
Remind students that when the droplet hits the ground, its height is 0.
Problem 7
Since the expression has 4 unlike terms, the students should factor by grouping. Factor pairs of terms rather than all four terms.
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Lesson 94 629
*13. Canoeing A canoe rental company charges $10 for the canoe and an additional charge per person. There are 4 people going on the trip and they have planned on spending a total of $50. They hope that the total cost is within $20 of the planned spending total. What is the minimum and maximum they can be charged per person? $5, $15
Find the quotient.
14. (1 + 4x4 - 10x2) ÷ (x + 2) 4x3 - 8x2 + 6x - 12 + 25_x + 2
15. 25x3 + 20x2 - 5x
__ 5x (5x - 1)(x + 1)
*16. Solve the equation ⎪x⎥
_ 11
+ 9 = 15 and graph the solution.
*17. Justify Show that the solution set to the equation ⎪x⎥
_ -3 + 1 = 5 is Ø.
*18. Multi-Step Marty measured the area of his rectangular classroom. He determined that the area is (-2x2 + x3 - 98 - 49x) square feet. The length is (x + 6) feet. a. What is the width? (x2 - 8x - 1 - 92_
x + 6 ) feet
b. If the area is (x2 - 36) square feet, what is the width? (x - 6) feet
19. Geometry The area of a triangle is (10y2 + 6y) square centimeters. The base is (5y + 3) centimeters. What is the height? 4y centimeters
*20. Error Analysis Students were asked to simplify 4x
_ 8x + 16
_ 12
_ x + 2
. Which student is correct? Explain the error. Student B; Sample: Student A did not multiply by the reciprocal.
Student A
4x _ 8x + 16
· 12 _
x + 2
= 4x _
8(x + 2) ·
12 _ x + 2
= 6x __
(x + 2)(x + 2)
Student B
4x _ 8x + 16
· x + 2 _
12
= 4x _
8(x + 2) ·
x + 2 _ 12
= x _ 24
*21. Commuting It took Taylor 1 __ x2 + 3x - 40
minutes to get to work, which was
x2
_ 6x + 48
miles away. Find his rate in miles per minute. x3 - 5x2_
6x miles per minute
22. Verify Show that 0 is a solution to the inequality ⎪x - 14⎥ < 30. Sample:⎪0 - 14⎥ = ⎪-14⎥ = 14 and 14 < 30
23. Multiple Choice Which inequality is represented by the graph? C
20100-10-20
A ⎪x⎥ ≤ - 21 B ⎪x⎥ < 21
C ⎪x⎥ ≤ 21 D ⎪x⎥ > 21
24. Analyze When a line segment is horizontal, which expression under the radical in the distance formula is 0: (x2 - x1)
2 or (y2 - y1)2? (y2 - y1)2
(94)(94)
(93)(93)
(93)(93)
(94)(94){-66, 66};
660 33-33-66{-66 366}
660 33-33-66
(94)(94) 17.3 3 33 3 3 3 3g ⎪x⎥
-3=3 3T 33 3 3
-33 3g ⎪x⎥ =
- 3 33v 33 3 g v 3
3 3 3
17.3 3 33 3 3 3 3g ⎪x⎥
-3=3 3T 33 3 3
-33 3g ⎪x⎥ =
- 3 33v 33 3 g v 3
3 3 3
(93)(93)
(93)(93)
(92)(92)
(92)(92)
(91)(91)
(91)(91)
(86)(86)
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Lesson 94 629
Problem 15
Extend the Problem
Multiply the dividend by the divisor to verify that the solution is equal to the quotient.
Problem 17
Isolate the absolute value. Note that it is equal to a negative number.
Problem 19
The formula for the area of a triangle is A = 1 __
2 bh.
Problem 22
Substitute 0 into the equation and see if it makes the inequality true.
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Saxon Algebra 1630
25. Multi-Step Ornella hiked 8 miles on easy trails and 3 miles on difficult trails. Her hiking rate on the easy trails was 2.5 times faster than her rate on the difficult trails. a. Write a simplified rational expression for Ornella’s total hiking time.
b. Find Ornella’s hiking time if her hiking rate on the difficult trails was 2 miles per hour. 3.1 hours
26. Find the vertical and horizontal asymptotes and graph y = 1 _ x - 2
- 4.
27. Multi-Step Phone Plan A costs $12 per month for local calls and $0.06 per minute for long-distance calls. Phone Plan B costs $15 per month for local calls plus $0.04 per minute for long-distance calls. How many minutes of long-distance calls would Jenna have to make for Plan B to cost less than Plan A? a. Formulate Write an inequality to answer. 12 + 0.06m > 15 + 0.04m
b. Solve the inequality and answer the question. m > 150; 150 minutes
c. Graph the solution. 20015050 100
28. United States Flag An official American flag should have a length that is 1.9 times its width. The area of an official American flag can be found by the function y = 1.9x2. Graph the function. Then use the graph to approximate the width of a flag that has an area of 47.5 square feet.
29. Multi-Step A rectangular garden that is 25 feet wide has a diagonal length that is 50 feet long. a. Find the length of the garden in simplest radical form. 25√ � 3 ft
b. Find the perimeter of the garden to the nearest tenth of a foot. 136.6 ft
30. The volume of a sphere is V = 4 _ 3 πr3. Describe in words the relationship between
the volume of a sphere and its radius. Identify the constant of variation. Thevolume of a sphere is directly proportional to the cube of its radius. The constant of variation is equal to 4_
3π.
(90)(90)
8_2.5r
+ 3_r = 15.5_
2.5r8_
2.5r+ 3_
r = 15.5_2.5r
(78)(78)
xy
4-2
-2
-6
26. x = 2; y = -4
xy
4-2
-2
-6
26. x = 2; y = -4
(81)(81)
(84)(84)
28.
x
y
30
40
20
10
1 2 3 4O
; about 5 feet28.
x
y
30
40
20
10
1 2 3 4O
; about 5 feet
(85)(85)
(Inv 8)(Inv 8)
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CHALLENGE LOOKING FORWARD
Write a multi-step absolute-value equation that has no solution. Sample: ⎪ x __
4 - 2⎥ + 15 = 12
Solving multi-step absolute-value equations prepares students for
• Lesson 101 Solving Multi-Step Absolute-Value Inequalities
• Lesson 107 Graphing Absolute-Value Functions
630 Saxon Algebra 1
Problem 30
Error Alert When fi nding the constant of variation, students may think that π is a variable. Remind students that π is a constant.
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Lesson 95 631
Combining Rational Expressions with
Unlike Denominators
Warm Up
95LESSON
1. Vocabulary One of two or more numbers or expressions that are multiplied to get a product is called a (n) . factor
Find the LCM.
2. 8x4y and 12x3y2 24x4y2
3. (9x - 27) and (4x - 12) 36(x - 3)
Factor.
4. x2 + 4x - 21 (x + 7)(x - 3)
5. 10x2 + 13x - 3 (5x - 1)(2x + 3)
The steps for adding rational expressions are the same as for adding numerical fractions. Fractions with unlike denominators cannot be added unless you first find their least common denominator.
Example 1 Finding a Common Denominator
Find the least common denominator (LCD) for each expression.
a. 3 _
(x + 3) -
9 __ (x2 - 2x - 15)
SOLUTION
3 _ (x + 3)
- 9 __
(x2 - 2x - 15)
= 3 _
(x + 3) -
9 __ (x + 3)(x - 5)
Factor each denominator, if possible.
To find the LCD of (x + 3) and (x + 3)(x - 5), use every factor of each denominator the greatest number of times it is a factor of either denominator. Each denominator has a factor of (x + 3). One denominator also has a factor of (x - 5). The product of these two factors is the LCD.
LCD = (x + 3)(x - 5)
b. 2x _
4x2 - 196 +
12x __ x2 + x - 56
SOLUTION
2x _ 4x2 - 196
+ 12x __
x2 + x - 56
= 2x __
4(x - 7)(x + 7) +
12x __ (x - 7)(x + 8)
Factor each denominator completely.
LCD = 4(x - 7)(x + 7)(x + 8)
(2)(2)
(57)(57)
(57)(57)
(72)(72)
(75)(75)
New ConceptsNew Concepts
Online Connection
www.SaxonMathResources.com
Math Language
A least common
denominator (LCD) is the least common multiple (LCM) of the denominators.
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Working with rational expressions can test the students’ understanding of fractions. If the algorithm for adding and subtracting fractions is simply memorized, then adding and subtracting rational expressions can be challenging. Rational expressions can be treated as fractions. In order to add or subtract rational expressions, a common denominator must be found. Any common denominator will work, but the LCD is the most effi cient. Keeping the terms simpler will help eliminate error.
Once a common denominator is found, each term must be multiplied by a factor equal to 1 so that equivalent fractions with a common denominator are found. Combine the fractions while keeping the denominator in factored form. After simplifying the numerator, it is important to factor the numerator to divide out factors that form quotients equal to one.
LESSON RESOURCES
Student Edition Practice Workbook 95
Reteaching Master 95Adaptations Master 95Challenge and Enrichment
Master C95
MATH BACKGROUND
Warm Up1
95LESSON
Lesson 95 631
Problem 3
Remind students to factor fi rst and then to fi nd the LCM.
2 New Concepts
In this lesson, students combine rational expressions with unlike denominators.
Example 1
Students fi nd a common denominator for rational expressions.
Additional Example 1
Find the least common denominator (LCD) for each expression.
a. 5 ______ x - 2
+ 6 __________ x2 - x - 2
(x - 2)(x + 1)
b. 3x ____________ x 2 + 8x + 15
+ 4x _________ 5 x 2 - 500
5(x + 3)(x + 5)(x + 10)(x - 10)
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Saxon Algebra 1632
Example 2 Using Equivalent Fractions to Add with Unlike
Denominators
Add 6x2
_ x2 - 16
+ x - 1 _ 2x - 8
.
SOLUTION Factor each denominator.
6x2
_ x2 - 16
+ x - 1 _ 2x - 8
= 6x2
__ (x - 4)(x + 4)
+ x - 1 _
2(x - 4)
LCD = 2(x - 4)(x + 4)
Write an equivalent fraction for each addend with the LCD as a denominator.
6x2
__ (x - 4)(x + 4)
· 2 _ 2 =
2(6x2) __
2(x - 4)(x + 4)
Multiply the numerator and denominator of the first fraction by 2.
x - 1 _ 2(x - 4)
· x + 4 _ x + 4
= (x - 1)(x + 4)
__ 2(x - 4)(x + 4)
Multiply the numerator and denominator of the second fraction by x + 4_
x + 4.
2(6x2) __
2(x - 4)(x + 4) +
(x - 1)(x + 4) __
2(x - 4)(x + 4)
Write the sum of the equivalent fractions.
= 2(6x2) + (x - 1)(x + 4)
___ 2(x - 4)(x + 4)
Add.
= 12x2 + x2 + 3x - 4 __
2(x - 4)(x + 4) Expand the numerator.
= 13x2 + 3x - 4 __
2(x - 4)(x + 4) Combine like terms in the
numerator.
Example 3 Using Equivalent Fractions to Subtract with
Unlike Denominators
Subtract 4x2
_ 9x - 27
- 2x - 5 _ x2 - 9
.
SOLUTION Factor each denominator.
4x2
_ 9x - 27
- 2x - 5 _ x2 - 9
= 4x2
_ 9(x - 3)
- 2x - 5 __
(x - 3)(x + 3)
LCD = 9(x - 3)(x + 3)
4x2(x + 3)
__ 9(x - 3)(x + 3)
- 9(2x - 5)
__ 9(x - 3)(x + 3)
Write the difference of the equivalent fractions.
= 4x2(x + 3) - 9(2x - 5)
___ 9(x - 3)(x + 3)
Subtract.
= 4x3 + 12x2 - 18x + 45 __
9(x - 3)(x + 3) Expand the numerator.
There are no like terms, so the difference is 4x3 + 12x2 - 18x + 45
__ 9(x - 3)(x + 3) .
Math Reasoning
Analyze What does it mean to write an equivalent fraction?
Sample: It means to multiply the numerator and denominatorof a fraction by the same factor resulting in a fraction that is equal to the original fraction.
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To help students combine rational expressions, set up boxes. After students factor the denominators and determine the LCD for Example 3, have them draw unit factor boxes next to each term.
(4 x2)
________ 9(x - 3)
- (2x - 5)
_____________ (x - 3)(x + 3)
Then have the students compare the denominator of the terms to the LCD.
Say, “Since the fi rst term is missing a factor of (x + 3), multiply by the unit
factor (x + 3)
______
(x + 3)
.” Have students determine
what to multiply the second term by. After the expression has been setup with common denominators it will be ready to simplify.
(x + 3)
________ (x + 3)
(4x2)
___________ 3 · 3(x - 3)
- 9 __ 9 (2x - 5)
_____________ (x - 3)(x + 3)
INCLUSION
632 Saxon Algebra 1
Example 2
Students learn to use equivalent fractions to add rational expressions with unlike denominators.
Additional Example 2
Add 3 x 2 _____ x 2 - 9
+ x + 2 ______
4x - 12 .
13 x 2 + 5x + 6 _____________ 4(x - 3)(x + 3)
Example 3
Students learn to use equivalent fractions to subtract rational expressions with unlike denominators.
Additional Example 3
Subtract 2x _______ 4x + 16
- 3x - 4 _______ x 2 - 16
.
x2 - 10x + 8 _____________ 2(x - 4)(x + 4)
Error Alert Students may forget to distribute the negative sign to each term in the numerator of the rational expression being subtracted. Putting parentheses around the terms in the numerator will help students to remember that they must subtract each term.
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Lesson 95 633
Example 4 Adding and Subtracting with Unlike Denominators
a. Add -35 _
56 - 7x +
-6x - 2 __ x2 - 6x - 16
.
SOLUTION
-35 _ 56 - 7x
+ -6x - 2 __
x2 - 6x - 16
= -35 _
-7(x - 8) +
-6x - 2 __ (x - 8)(x + 2)
Factor each denominator.
LCD = -7(x - 8)(x + 2) Find the LCD.
-35(x + 2) __
-7(x - 8)(x + 2) +
-7(-6x - 2) __
-7(x - 8)(x + 2) Write each fraction as an
equivalent fraction.
= -35(x + 2) + (-7)(-6x - 2)
___ -7(x - 8)(x + 2)
Add.
= 7x - 56 __
-7(x - 8)(x + 2) Expand the numerator and
collect like terms.
= 7(x - 8)
__ -7(x - 8)(x + 2)
Factor the numerator.
= - 1 _
x + 2 Divide out common factors.
b. Subtract x - 5 _ 2x - 6
- x - 7 _
4x - 12 .
SOLUTION
x - 5 _ 2x - 6
- x - 7 _
4x - 12
= x - 5 _
2(x - 3) -
x - 7 _ 4(x - 3)
Factor each denominator.
LCD = 4(x - 3) Find the LCD.
2(x - 5) _
4(x - 3) -
(x - 7) _
4(x - 3) Write each fraction as an equivalent fraction.
= 2(x - 5) - 1(x - 7)
__ 4(x - 3)
Subtract.
= 2x - 10 - x + 7 __
4(x - 3) Expand the numerator.
= x - 3_
4(x - 3)Collect like terms.
= 1 _ 4 Divide out common factors.
Caution
Before factoring, write expressions in descending order.
Hint
Always check to see if the numerator of the sum can be factored and if a common factor can be divided out.
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Lesson 95 633
Example 4
Students add and subtract rational expressions with unlike denominators. Then the sum or difference is simplifi ed by dividing out like factors.
Additional Example 4
a. Add -24 _______ 32 - 8x
+ 4x + 1 ___________
x2 - 7x + 12 .
7x - 8 ______________ (x - 4)(x - 3)
b. Subtract x + 4 _______ 3x + 9
- x + 9 ________
5x + 15 .
2x - 7 __________ 15(x + 3)
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Practice Distributed and Integrated
Saxon Algebra 1634
Example 5 Application: Traveling
A pilot’s single-engine aircraft flies at 230 kph if there is no wind. The pilot plans a round trip to a city that is 400 kilometers away. If there is a tailwind
of w kilometers per hour, the time for the outbound flight is 400
_ 230 + w hours. The time for the return flight with a headwind of w kilometers per hour is 400
_ 230 - w hours. What is the total time for the round trip?
SOLUTION
400 _ 230 + w
+ 400 _
230 - w Add to find the total time.
LCD = (230 + w)(230 - w) Find the LCD.
400(230 - w)
__ (230 + w)(230 - w)
+ 400(230 + w)
__ (230 + w)(230 - w)
Write equivalent fractions.
= 184,000
__ (230 + w)(230 - w)
hours Expand the numerator and collect like terms.
The round trip takes 184,000
__ (230 + w)(230 - w)
hours.
Lesson Practice
Find the LCD for each expression.
a. 5x _
5x - 45 -
44 _ x2 - 81
LCD = 5(x - 9)(x + 9)
b. 3x _
x + 4 -
12 __ x2 + 2x - 8
LCD = (x + 4)(x - 2)
c. Add 3x2
_ x2 - 25
+ x - 1 _
4x - 20 .
13x2 + 4x - 5__4(x - 5)(x + 5)
d. Subtract 2x2
_ 6x - 24
- 3x - 4 _ x2 - 16
. x3 + 4x2 - 9x + 12__
3(x - 4)(x + 4)
e. Add x - 1 _ x2 - 1
+ 2 _
5x + 5 . 7_
5(x + 1)
f. Subtract 2 _ x2 - 36
- 1 _
x2 + 6x . 1_
x(x - 6)
g. Trenton hiked 4x
_ x2 - 64 miles on Saturday and 12
_ 7x - 56 miles on Sunday. How many miles did he hike altogether? 40x + 96__
7(x - 8)(x + 8) miles
(Ex 1)(Ex 1)
(Ex 2)(Ex 2)
(Ex 3)(Ex 3)
(Ex 4)(Ex 4)
(Ex 4)(Ex 4)
(Ex 5)(Ex 5)
Factor completely.
1. 3x3 - 9x2 - 30x 3x(x + 2)(x - 5) 2. 8x3y2 + 4x2y - 12xy3
3. 32x3 - 24x4 + 4x5 4x3(x - 2)(x - 4) 4. mn3 - 10mn2 + 24mn
(Inv 9)(Inv 9) (Inv 9)(Inv 9)4xy(2x2y + x - 3y2)4xy(2x2y + x - 3y2)
(79)(79) (79)(79)mn(n - 6)(n - 4)mn(n - 6)(n - 4)
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634 Saxon Algebra 1
Example 5
Students add rational expressions to solve a rate problem.
Extend the Example
About how many hours does the trip take if the tailwind is 10 kilometers per hour? about 3.485 hours
Additional Example 5
Sarah biked 7x ______ x 2 - 49 kilometers
on Saturday. She biked 15 ______ 5x + 35
kilometers on Sunday. How many kilometers did she bike altogether? 10x - 21 ____________
(x + 7)(x - 7) kilometers
Lesson Practice
Problem a
Scaff olding Factor both denominators completely. Then have students fi nd the LCM of the denominator.
Problem f
Error Alert Students may not recognize x as a factor in the LCD. Remind them that every factor must be in the LCD.
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“How do you fi nd the LCD?” Sample: Factor each denominator. Find the LCM of the denominators. To fi nd the LCM of the denominators, use every factor of each denominator the greatest number of times it is a factor of either denominator.
“When adding rational expressions with unlike denominators, what do you do after you fi nd the LCD?” Sample: Find equivalent fractions with their denominators equal to the LCD.
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Lesson 95 635
Find the quotient.
5. 4m
_ 17r
÷ 12m2
_ 5r
5_51m 6. (x2 - 16x + 64) ÷ (x - 8) (x - 8)
7. Find the zeros of the function shown. no zeros
x
y
4
6
2 4-4 -2
*8. Find the LCD of 4 _ x + 4
- 8 __
x2 + 6x + 8 . LCD = (x + 4)(x + 2)
*9. Add 2x _
2x2 - 128 +
5 __ x2 - 7x - 8
. x2 + 6x + 40__
(x - 8)(x + 8)(x + 1)
*10. Solve the equation 9⎢x� - 22 = 14 and graph the solution. {-4, 4};
11. Error Analysis Students were asked to subtract 6x2
_ x2 + 6x - 16
- 3 _
8x - 16 . Which student
is correct? Explain the error. Student A; Sample: Student B didn’t distribute the negative all the way through the second numerator.
Student A
6x2
__ x2 + 6x - 16
- 3 _
8x - 16
6x2
__ (x + 8)(x - 2)
- 3 _
8(x - 2)
6x2(8) __
8(x + 8)(x - 2) -
3(x + 8) __
8(x - 2)(x + 8)
48x2 - 3x - 24
__ 8(x + 8)(x - 2)
Student B
6x2
__ x2 + 6x - 16
- 3 _
8x - 16
6x2
__ (x + 8)(x - 2)
- 3 _
8(x - 2)
6x2(8) __
8(x + 8)(x - 2) -
3(x + 8) __
8(x - 2)(x + 8)
48x2 - 3x + 24
__ 8(x + 8)(x - 2)
*12. Generalize Explain how to find the LCD of two algebraic rational expressions.
*13. Running Michele is training for a marathon. She ran 3x2
_ x2 - 100 miles on Monday and
x - 1
_ 2x - 20 miles on Tuesday. How many miles did she run in all?
*14. Multi-Step The girls’ track team sprinted 2x
_ 4x2 - 196 meters Thursday and 12x
_ x2 + x - 56 meters Friday.
a. What was the total distance that the track team sprinted?
b. If their rate was 2x
_ x + 8 meters per minute, how much time did it take them to sprint on Thursday and Friday? 25x + 176__
4(x - 7)(x + 7) minutes
(88)(88) (93)(93)
(89)(89)
(95)(95)
(95)(95)
(94)(94) 40-4 40-4
(95)(95)
(95)(95)
12. Sample: Factor each denominator. The LCD must contain each factor of each denominator and use each factor the greatest number of times it occurs in either denominator.
12. Sample: Factor each denominator. The LCD must contain each factor of each denominator and use each factor the greatest number of times it occurs in either denominator.
(95)(95)
7x2 + 9x - 10__
2(x + 10)(x - 10) miles
7x2 + 9x - 10__2(x + 10)(x - 10)
miles
(95)(95)
14a. 25x2 + 176x___2(x - 7)(x + 7)(x + 8)
meters14a. 25x2 + 176x___2(x - 7)(x + 7)(x + 8)
meters
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Lesson 95 635
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 5
To divide by a fraction is to multiply by its reciprocal. Before multiplying, divide out factors common to the numerator and denominator, if any.
Problem 9
Guide the students by asking them the following questions.
“What is the LCD of the fractions?” 2(x - 8)(x + 8)(x +1) or (x - 8)(x + 8)(x + 1)
“How do you change the fractions so that they share the LCD?” Sample: Multiply the numerator and denominator of each fraction by its missing factors of the LCD.
“Once the fractions share the LCD, what is the next step?” Sample: Add the fractions.
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Saxon Algebra 1636
*15. Error Analysis Two students solve the equation 6⎪x + 3⎢ - 8 = -2. Which student is correct? Explain the error.
Student A
6⎪x + 3⎢ - 8 = -2no solutionAbsolute values cannot equal a negative number.
Student B
6⎢x + 3� - 8 = -26⎢x + 3� = 6⎢x + 3� = 1x + 3 = 1 and x + 3 = -1 x = -2 and x = -4
16. Geometry The perimeter of a square must be 34 inches plus or minus 2 inches. What is the longest and the shortest length each side can be? 8 inches, 9 inches
*17. Multi-Step A student budgets $35 for lunch and rides each week. He gives his friend $5 for gas and then pays for 5 lunches a week. He has a $2 cushion in his budget, meaning that he can spend $2 more or less than he budgeted. a. Write an absolute-value equation for the minimum and maximum he can spend
on each lunch. ⎢5 + 5x - 35� = 2
b. What is the maximum and the minimum he can spend on each lunch? $5.60,$6.40
18. Error Analysis Students were asked to simplify x2 + 10x + 24
_ x . Which student is correct?
Explain the error. Student B; Sample: Student A canceled terms that were not common factors.
Student A
(x + 6)(x + 4)
__ x = (x + 6)(4)
Student B
(x + 6)(x + 4)
__ x
*19. Carpeting The rectangular public library is getting new carpet. The area of the room is (x3 - 18x2 + 81x) square feet. The length of the room is (x - 9) feet. What is the width? ( x2 - 9x) feet
*20. Justify Give an example to show a _ b · c _
d = ac
_ bd
.
21. Multiple Choice Simplify: x2 - 9
_ x2 - 5x + 6
_ x
2 + 5x + 6
_ x2 - 4 . A
A 1 B (x - 3)2
_ (x - 2)2 C -1 D
(x - 3) _
(x - 2)
22. Solve and graph the inequality ⎢x� > 17. x < -17 or x > 17;
23. Measurement When measuring something in centimeters, the accuracy is within 0.1 centimeter. A board is measured to be 15.6 centimeters. The accuracy of the measurement can be represented by the absolute-value inequality ⎢x - 15.6� ≤ 0.1. Solve the inequality. 15.5 ≤ x ≤ 15.7
(94)(94)15. Student B; Sample:Student A did not isolate the absolute value and assumed that because the equation was equal to a negative number, the absolute value would be equal to a negative number.
15. Student B; Sample:Student A did not isolate the absolute value and assumed that because the equation was equal to a negative number, the absolute value would be equal to a negative number.(94)(94)
(94)(94)
(93)(93)
(93)(93)
(92)(92)Sample: 4_
2 · 3_
9 = 2 · 1_
3 = 2_
3 and 4 · 3_
2 · 9 = 12_
18 = 2_
3Sample: 4_
2 · 3_
9 = 2 · 1_
3 = 2_
3 and 4 · 3_
2 · 9 = 12_
18 = 2_
3
(92)(92)
(91)(91) 200 10-10-20 200 10-10-20
3 43 4
(91)(91)
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In Problem 17, explain the meaning of the word cushion. Say:
“Cushioning provides or gives security or support to ensure safety. For example, the couch has cushions which provide comfort when we sit on it.”
Explain to students that not only pillows can provide cushioning.
Say:
“Buying extra amounts of ingredients to bake a cake provides a cushion in case the cake doesn’t come out correctly the fi rst time.”
Ask volunteers to give examples of other types of cushioning. Sample: bringing extra money to buy something when the price is estimated
ENGLISH LEARNERS
636 Saxon Algebra 1
Problem 16
Extend the Problem
What is the greatest and least area the square can have? 64 square inches and 81 square inches
Problem 18
Remind students that when a term is within parentheses, it cannot be separated. In order to cancel these factors, they must both be identical.
Problem 19
Factor the expression that describes area. Then divide by the length to get the width.
Problem 22
Extend the Problem
Solve and graph ⎢x� ≤ 17. -17 ≤ x ≤ 17
170-17
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Lesson 95 637
24. Multi-Step Reggie has made a stew that is at a temperature of 100°F, and he either wants to heat it up to 165°F or cool it down to 45°F. To heat up the stew will take at most 45 minutes plus 1 minute for each degree the temperature is raised. To cool down the stew will take at least 10 minutes plus 2 minutes for each degree the temperature is lowered. How much time does Reggie need to allow for changing the food temperature either up or down? a. How many degrees up and down does the stew need to go?
b. How much time will it take to cool down or heat up the stew?
c. Is it faster to heat up or cool down the stew? To heat up the stew is faster.
25. Olympic Swimming Pool An Olympic swimming pool must have a width of 25 meters and a length of 50 meters. Find the length of a diagonal of an Olympic swimming pool to the nearest tenth of a meter. 55.9 meters
26. Multi-Step A graphing calculator screen is 128 pixels wide and 240 pixels long. The “pixel coordinates” of three points are shown.
128
0 240P(61, 43)
Q(124, 106)
R(155, 35)
a. Find the distance in pixels between each pair of points.
b. List the line segments in order from shortest to longest. −−−
QR,−−
PQ,−−
PR
27. Write Create a polynomial in which each term has a common factor of 4a, and then factor the expression. Sample: 8 a2b + 4a3 + 12ab + 16ab2; 4a(2ab + a2 + 3b + 4b2)
28. Hiking In celebration of getting to the top of Beacon Rock in southern Washington, Renata throws her hat up and off the top of Beacon Rock. The height of the hat x seconds after the throw (in meters) can be approximated by the function y = -5x2 + 10x + 260. After how many seconds will the hat be at its maximum height? What is this height? 1 second; 265 feet
29. Commuting Mr. Shakour’s round trip commute to and from work totals 30 miles. Because of traffic, his speed on the way home is 5 miles per hour less than what it is on the way to work. Write a simplified expression to represent Mr. Shakour’s total commuting time. 15(2x - 5)
_x(x - 5)
hours
30. A teacher randomly picks a shirt and a skirt from her closet to wear to school. Use the table to find the theoretical probability of the teacher choosing an outfit with a blue shirt and khaki skirt. 1_
10
Ski
rts
ShirtsRed Blue Blue White White
Khaki KR KB KB KW KWNavy NR NB NB NW NWBlack BR BB BB BW BWWhite WR WB WB WW WW
(82)(82)
65 degrees up or 55 degrees down65 degrees up or 55 degrees downheat: t ≤ 110 minutes; cool: t ≥ 120 minutesheat: t ≤ 110 minutes; cool: t ≥ 120 minutes
(85)(85)
(86)(86)
PQ = 89 pixels, QR = 77 pixels, PR = 94 pixelsPQ = 89 pixels, QR = 77 pixels, PR = 94 pixels
(87)(87)
(89)(89)
(90)(90)
(80)(80)
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Find the LCD of 3 _________ mz 2 - mz
- 6 ________ pz 2 - p 2 z
.
mpz(z - 1)(z - p)
Combining rational expressions with unlike denominators prepares students for
Lesson 99 Solving Rational Equations
CHALLENGE LOOKING FORWARD
Lesson 95 637
Problem 25
Use the Pythagorean Theorem.
Problem 28
Error Alert Students may set y equal to 0 to fi nd the highest point. However, this is when the hat hits the ground. Remind students that the hat’s maximum height is represented by the y-value at the vertex.
Problem 30
Remind students that the probability of an event is the number of desired outcomes over the total number of outcomes.
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Saxon Algebra 1638
Warm Up
96LESSON
1. Vocabulary A(n) __________ is a line that divides a figure or graph into two mirror-image halves. axis of symmetry
Evaluate for the given value.
2. y = 4 x 2 - 6x - 4 for x = 1 _ 2 -6 3. y = -x2 + 5x - 6 for x = -2 -20
Find the axis of symmetry using the formula.
4. y = -2x2 + 4x - 5 x = 1 5. y = x2 - 3x - 4 x = 3_2
A quadratic function can be graphed using the axis of symmetry, the vertex, the y-intercept, and pairs of points that are symmetric about the axis of symmetry. The quadratic function in the standard form f(x) = ax2 + bx + c can be used to find these parts of the graph of the parabola.
The equation of the axis of symmetry and the x-coordinate of the vertex of a quadratic function is x = - b
_ 2a
. To find the y-coordinate of the vertex, substitute the x-coordinate of the vertex into the function. The y-intercept of a function is the point on the graph where x = 0. For quadratic functions in standard form, the y-intercept is c.
Example 1 Graphing Quadratics of the Form y = x2 + bx + c
Graph the function.
y = x 2 + 4x + 5
SOLUTION
Step 1: Find the axis of symmetry.
x = - b _
2a Use the formula.
= - 4 _
2 · 1 = -
4 _ 2
= -2 Substitute values for b and a.
The axis of symmetry is x = -2.
Step 2: Find the vertex.
y = x 2 + 4x + 5
= (-2 ) 2 + 4(-2) + 5 = 1 Substitute -2 for x.
The vertex is (-2, 1).
Step 3: Find the y-intercept.
The y-intercept is c, or 5.
(89)(89)
(9)(9) (9)(9)
(89)(89) (89)(89)
New ConceptsNew Concepts
Graphing Quadratic Functions
Online Connection
www.SaxonMathResources.com
Math Reasoning
Analyze Why can a not equal 0 in the standard form of a quadratic equation?
Sample: If a = 0, then there is no x2, or quadratic term. The function would no longer be a quadratic function.
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MATH BACKGROUNDLESSON RESOURCES
Student Edition Practice Workbook 96
Reteaching Master 96Adaptations Master 96Challenge and Enrichment
Master C96
Graphing quadratic functions is a key element of algebra. The graph can help students analyze and solve problems that can be modeled by a quadratic function.
Begin by fi nding the equation of the axis of symmetry, which passes through the vertex. Then plot the y-intercept and one other point found by substituting a value of x into the function. Finally, refl ect these points across the axis of symmetry. Connect all the points with a smooth curve.
Symmetry is an important characteristic of quadratic functions. By fi nding one point on the curve, a second point can be quickly identifi ed by refl ection across the axis of symmetry. Connecting the points with a smooth curve is critical. The graph of a quadratic function is different from a linear function because it is not straight. The vertex is a critical point because the y-value at the vertex represents the minimum or maximum value of the function.
Warm Up1
638 Saxon Algebra 1
96LESSON
Problem 3
Remind students that after substituting, (-2) should be squared and then the product should be multiplied by -1.
2 New Concepts
In this lesson, students graph quadratic functions.
Example 1
Students graph quadratics of the form y = x2 + bx + c.
Additional Example 1
Graph the function y = x2 + 5x + 6.
x
y
O
6
2
2-2-4-6
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Lesson 96 639
Step 4: Find one point not on the axis of symmetry.
y = x 2 + 4x + 5
y = (1 ) 2 + 4(1) + 5 = 10 Substitute 1 for x.
A point on the curve is (1, 10).
Step 5: Graph.
Graph the axis of symmetry x = -2, the vertex (-2, 1), the y-intercept (0, 5). Reflect the point (1, 10) over the axis of symmetry and graph the point (-5, 10). Connect the points with a smooth curve.
Example 2 Graphing Quadratics of the Form y = a x 2 + bx + c
Graph the function.
y = 3 x 2 + 18x + 13
SOLUTION
Step 1: Find the axis of symmetry.
x = - b_2a
Use the formula.
= - 18_
2(3)= -3 Substitute values for b and a.
The axis of symmetry is x = -3.
Step 2: Find the vertex.
y = 3 x 2 + 18x + 13
= 3(-3 )2 + 18(-3) + 13 = -14 Substitute -3 for x.
The vertex is (-3, -14).
Step 3: Find the y-intercept.
The y-intercept is c, or 13.
Step 4: Find one point not on the axis of symmetry.
y = 3 x 2 + 18x + 13
= 3(-1 ) 2 + 18(-1) + 13 = -2 Substitute -1 for x.
A point on the curve is (-1, -2).
Step 5: Graph.
Graph the axis of symmetry x = -3, the vertex (-3, -14), the y-intercept (0, 13). Reflect the point (-1, -2) across the axis of symmetry to get the point (-5, -2). Connect the points with a smooth curve.
x
y
16
24
8
4-4-8
x
y
16
24
8
4-4-8
x
20
-10
-2-4-6
x
20
-10
-2-4-6
Hint
Count how far the plotted point (1, 10) is from the axis of symmetry. Then check that the reflected point is the same distance from the axis of symmetry, but in the opposite direction.
Hint
Identify the values of a, b, and c first.
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Lesson 96 639
Example 2
Students learn to graph quadratic functions of the formy = ax2 + bx + c.
Additional Example 2
Graph the function.
y = 5x2 - 20x - 3
x
y
-10
-20
2-2O
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Saxon Algebra 1640
Example 3 Graphing Quadratics of the Form y = a x 2 + c
Graph the function.
y = 5 x 2 + 4
SOLUTION
Step 1: Find the axis of symmetry.
x = - b_2a
= - 0_
2(5)= 0
The axis of symmetry is x = 0.
Step 2: Find the vertex.
y = 5 x 2 + 4
= 5(0 )2 + 4 = 4 Substitute 0 for x.
The vertex is (0, 4).
Step 3: Find the y-intercept.
The y-intercept is c, or 4.
Step 4: Find one point not on the axis of symmetry.
y = 5x2 + 4.
= 5(-1 )2 + 4 = 9 Substitute -1 for x.
A point on the curve is (-1, 9)
Step 5: Graph.
Graph the axis of symmetry x = 0, the vertex (0, 4), the y-intercept (0, 4). Reflect the point (-1, 9)
across the axis of symmetry to get the point (1, 9). Connect the points with a smooth curve.
A zero of a function is an x-value for a function where f(x) = 0. It is the point where the graph of the function meets or intersects the x-axis. The standard form of a quadratic equation ax2 + bx + c = 0, where a ≠ 0, is the related equation to the quadratic function. The quadratic equation is used to find the zeros of a quadratic function algebraically. Alternatively, a graphing calculator can help find zeros of a quadratic function.
Example 4 Finding the Zeros of a Quadratic Function
Find the zeros of the function.
a. y = x2 - 6x + 9
SOLUTION
Use a graphing calculator to graph y = x2 - 6x + 9.
The zero of the function is 3.
x
y
O
8
16
24
2 4-2-4
x
y
O
8
16
24
2 4-2-4
Math Language
A zero of a function is another name for an x-intercept of the graph.
Graphing
Calculator
For help with finding zeros, see graphing calculator keystrokes in Lab 8 on p. 583.
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A table can be used to graph a quadratic function. Choose values for the x-coordinate and substitute them into the function to fi nd the y-coordinates. Then plot the points and connect them with a smooth curve.
ALTERNATE METHOD
640 Saxon Algebra 1
Example 3
Students learn to graph quadratic functions of the form y = ax2 + c.
Additional Example 3
Graph the function y = 8x2 + 2.
x
y
O
8
2 4
-4
-2-4
-8
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Lesson 96 641
b. y = x2 - 3x - 10
SOLUTION
Use a graphing calculator to graph y = x2 - 3x - 10.
There are two zeros for this function, 5 and -2.
c. y = -2x2 - 3
SOLUTION
Use a graphing calculator to graph y = -2x2 - 3.
There are no real zeros for this function.
When an object is thrown or kicked into the air, it follows a parabolic path.
S
You can calculate its height in feet after t seconds using the formula h = -16t2 + vt + s. The initial vertical velocity in feet per second is v, and s is the starting height of the object in feet.
Example 5 Application: Baseball
A baseball is thrown straight up with an initial velocity of 50 feet per second. The ball leaves the player’s hand when it is 4 feet above the ground. At what time does the ball reach its maximum height?
SOLUTION Substitute the values given for the initial velocity and starting height into the formula h = -16t2 + vt + s. Then find the x-coordinate of the vertex. Use the formula x = - b
_ 2a
.
h = -16t2 + 50t + 4
x = - b _ 2a
= - 50 _
2(-16) = 1.5625 Substitute values for b and a.
The ball reaches its maximum height 1.5625 seconds after it has been thrown.
Math Language
If a quadratic function has no real zeros,
then there are no real numbers that when substituted for x result in y = 0. The graph of such a function does not cross the x-axis.
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Use the following strategy with students who have diffi culty with graphing. To help them connect the algebraic representation with the graphical representation, create a color-coded chart. Color code each part of the graph and, in the same color, show how to fi nd that part algebraically.
For example, show x = -b ___ 2a
in red and then show the axis of symmetry in red on the graph. Do this for the vertex, the y-intercept, and a pair of symmetric points.
INCLUSION
Lesson 96 641
Example 4
Students learn to fi nd the zeros of a quadratic function.
Additional Example 4
Find the zeros of the function.
a. y = x2 - 16x + 64 8
b. y = x2 + 3x - 28 4 and -7
c. y = 5x2 + 9 No real zeros
Error Alert Students may confuse the zeros of a function and the y-intercept. Explain that the zeros are the x-intercepts, and the y-intercept is where x equals zero.
Example 5
Students use quadratic functions to solve vertical motion problems.
Extend the Example
When does the baseball land on the ground? after about 3.2 seconds
Additional Example 5
An egg is thrown from the top of a wall that is 20 feet tall. Its initial velocity is 32 feet per second. How long will it take for the egg to reach its maximum height? 1 second
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Practice Distributed and Integrated
Saxon Algebra 1642
Lesson Practice
Graph each function.
a. y = x2 - 4x + 7 b. y = 2x2 - 16x + 24
c. y = 2x2 - 9
Find the zeros of each function.
d. y = x2 + 10x + 25. -5
e. y = 3x2 - 21x + 30. 2, 5
f. y = - 1 _ 2 x2 - 1 no real zeros
g. Scoccer The height of a soccer ball that is kicked can be modeled by the function f(x) = -8x2 + 24x, where x is the time in seconds after it is kicked. Find the time it takes the ball to reach its maximum height. 1.5 seconds
(Ex 1)(Ex 1)
a.
x
y
O
4
2
4-2-4
6
a.
x
y
O
4
2
4-2-4
6
(Ex 2)(Ex 2)b.
x
y
O
8
8
16
24
4
-8
b.
x
y
O
8
8
16
24
4
-8
(Ex 3)(Ex 3)
c.
x
y
8
4
4
-4
-4
c.
x
y
8
4
4
-4
-4
(Ex 4)(Ex 4)
(Ex 4)(Ex 4)
(Ex 4)(Ex 4)
(Ex 5)(Ex 5)
1. Find the zeros of the function shown. 0 and 4
2. Add 25 _
16x2y +
xy _
32y5 .
50y3 + x3_32x2y4
*3. Solve the equation 10|x|
_ 3 + 18 = 4 and graph the solution.
Ø; no graph 4. Solve -0.3 + 0.14n = 2.78. n = 22 5. Solve
6 _ x - 3
= 3 _ 10
. x = 23
6. Find the LCD of 6 _
x + 6 - 12 _
x2 + 8x + 12 . LCD = (x + 6)(x + 2)
7. The table lists the ordered pairs from a relation. Determine whether they form a function. function
Domain (x) Range (y)
10 15
11 17
8 11
9 13
5 5
8. Simplify
-x5
_ 21x + 3
_
5x9
_ 28x + 4
. -4_
15x4
*9. Graph the function y = x2 - 2x - 8.
*10. Write Explain how to reflect a point across the axis of symmetry to get a second point on the parabola.
(89)(89)x
y
O
2
2 4 6
-4
-6
-2
y = 1_2
x2+ 2x
x
y
O
2
2 4 6
-4
-6
-2
y = 1_2
x2+ 2x
(90)(90)
(94)(94)
(24)(24) (31)(31)
(95)(95)
(25)(25)
(92)(92)
(96)(96)
9. xyO2
-2
-4
-4
-8
9. xyO2
-2
-4
-4
-8
(96)(96)Sample: The second point will have the same y-value and will be the same horizontal distance from the axis of symmetry, but on the other side.Sample: The second point will have the same y-value and will be the same horizontal distance from the axis of symmetry, but on the other side.
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642 Saxon Algebra 1
Lesson Practice
Problem a
Error Alert Students may refl ect points across the y-axis rather than the axis of symmetry. Emphasize that the graph is symmetric about the axis of symmetry.
Problem b
Scaff olding To graph the function, fi rst fi nd the vertex, the y-intercept, and the axis of symmetry. Then fi nd a pair of points that are symmetric about the axis of symmetry.
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“How do you fi nd the axis of symmetry?” Sample: Use the formula x = -b ___
2a .
“What is the signifi cance of the vertex?” Sample: The y-value at the vertex is the maximum or minimum value of the function. The axis of symmetry passes through the vertex.
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 1
Extend the Problem
What is another name for the zeros of the function? x-intercepts
Problem 5
Error AlertStudents may forget to distribute when cross-multiplying with a binomial. Encourage them to put parentheses around the binomial before cross-multiplying.
Problem 7
For each value of the domain of a function, there can be exactly one value in the range.
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Lesson 96 643
*11. Justify Show that the vertex of y = 4x2 - 24x + 9 is (3, -27).
*12. Multiple Choice Which function has the vertex (6, -160)? AA y = 6x2 - 72x + 56 B y = 2x2 - 8x + 48
C y = 3x2 + 42x - 12 D y = 5x2 - 5x + 43
*13. Diving A diver moves upward with an initial velocity of 10 feet per second. How high will he be 0.5 seconds after diving from a 6-foot platform? Use h = -16t2 + vt + s. 7 feet
14. Multiple Choice Find the LCD of 3 _
2x - 10 and 5x
_ 2x2 - 4x - 30
. A
A 2(x - 5)(x + 3) B (x - 5)(x + 3)
C (x + 5)(x - 3) D 2 __ (x - 5)(x + 3)
*15. Geometry One side of a triangle is 2 _ x + 2
yards and two sides are each -5 _
3x + 6 yards.
Find the perimeter of the triangle. -4_3(x + 2)
yards
*16. Measurement Carrie measured a distance of 3x2
_ 9x - 18
yards and Jessie measured a distance of 4x - 5
_ x2 - 4
yards. How much longer is Carrie’s measurement than Jessie’s?
x3 + 2x2 - 12x + 15__3(x - 2)(x + 2)
yards
17. Banking The dollar amount in a student’s banking account is represented by the absolute-value inequality |x - 200| ≤ 110. Solve the inequality and graph the solution.
18. Generalize Why is a place holder needed for missing variables in a polynomial dividend? Sample: It helps line up like terms for the dividend and quotient.
19. Multiple Choice Simplify (-5x + 2x2 - 3) ÷ (x - 3). B
A 2x - 1 B 2x + 1 C x - 2 _
2x D
x2 - 3 _ 5x
*20. Physics A family is going to see friends that live in two different towns. They will have to travel 100 miles plus or minus 10 miles to see either of them. They want to spend 2 hours in the car. What are the minimum and maximum rates that they need to go? 45 miles per hour; 55 miles per hour
*21. Error Analysis Two students graph the solution to the equation |2x + 10| = 8. Which student is correct? Explain the error.
Student A Student B
{-9, -1} {-9, -1}
-2-4-6-8-10 0 -2-4-6-8-10 0
22. Art Jeremy’s picture frame has an area of x2 - 18x + 80 square inches. He has two square pictures in it, one measuring 1 _
4 x inch on each side and the other
measuring 1 _ 2 x inch on each side. Write a simplified expression for the total fraction
of the frame covered by pictures. 0.3125x2__(x - 8)(x - 10)
(96)(96)
11. Sample: x = -b_
2a= 24_
8= 3. Then,
substitute 3 into the equation to get 4(3)2 - 24(3) + 9 = -27.
11. Sample: x = -b_
2a= 24_
8= 3. Then,
substitute 3 into the equation to get 4(3)2 - 24(3) + 9 = -27.
(96)(96)
(96)(96)
(95)(95)
(95)(95)
3 43 4
(95)(95)
(91)(91)
90 ≤ x ≤ 310; 255 31020090 145
90 ≤ x ≤ 310; 255 31020090 145
(93)(93)
(93)(93)
(94)(94)
(94)(94)21. Student A; Sample: Student B graphed all values between -9 and -1 in addition to the solution set.
21. Student A; Sample: Student B graphed all values between -9 and -1 in addition to the solution set.
(90)(90)
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In problem 13, explain the meaning of the word diver. Say:
“A diver is a person who jumps into the water head fi rst with his body straight and his arms out in front.”
Ask volunteers to demonstrate how to position their arms to dive into the water.
ENGLISH LEARNER
Lesson 96 643
Problem 14
Factor fi rst. Then use the factors to fi nd the LCD.
Problem 19
Order the powers in the trinomial in descending order before dividing.
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Saxon Algebra 1644
23. Verify Divide the rational expression 3x2 + 2x _
9y ÷
y + 2 _ y . How can you check your
answer?
24. Multi-Step The volume of a prism is 6x3 + 14x2 + 4x cm3. Find the dimensions of the prism. a. Factor out common terms. 2x(3x2 + 7x + 2)
b. Factor completely to find the dimensions. 2x(3x + 1)(x + 2)
25. Baseball A baseball diamond is a square that is 90 feet long on each side. To use a coordinate grid to model positions of players on the field, place home plate at (0, 0), first base at (90, 0), second base at (90, 90), and third base at (0, 90). An outfielder located at (150, 80) throws to the first-baseman. How long is the throw? Round your answer to the nearest foot. 100 feet
26. Multi-Step A square has a side length of a centimeters. A smaller square has a side length of b centimeters. a. If the difference in the areas of the squares is a2 - 16, what is the value of b? 4 cm
b. If the area of the larger square is 36b2 + 60b + 25, what is the side length in terms of b? 6b + 5
c. Using your answers to parts a and b, find the side length and area of the larger square. 29 cm; 841 c m 2
27. Determine if the inequality 6x > 7x is never, sometimes, or always true. If it is sometimes true, identify the solution set. sometimes true; It is true for all negative values of x.
28. Use the graph to find the theoretical probability of choosing each color.
Favorite Color10
8
6
4
2
0
Red
Green
Yellow
Blue
Orang
e
Purple
Color
29. Suppose d varies inversely with b and jointly with a and c. Find the constant of variation when a = 4, b = 5, c = 2, and d = 8. Express the relationship between these quantities. What is d when a = 9, b = 15, and c = 6? k = 5; d = 5ac_
b; d = 18
30. Is (x + 10)(x - 2) the correct factorization for x2 - 8x - 20? Explain.
(88)(88) x(3x + 2) _
9(y + 2) ; Sample: Substitute real numbers for the variables x and y before and after dividing.
x(3x + 2) _
9(y + 2) ; Sample: Substitute real numbers for the variables x and y before and after dividing.
(87)(87)
(86)(86)
(83)(83)
(81)(81)
(80)(80)
P(red) = 6_24
= 1_4, P(green) = 2_
24= 1_
12,
P(yellow) = 8_24
= 1_3, P(blue) = 5_
24,
P(orange) = 2_24
= 1_12
, P(purple) = 1_24
P(red) = 6_24
= 1_4, P(green) = 2_
24= 1_
12,
P(yellow) = 8_24
= 1_3, P(blue) = 5_
24,
P(orange) = 2_24
= 1_12
, P(purple) = 1_24
(Inv 8)(Inv 8)
(Inv 9)(Inv 9) no; If (x + 10)(x - 2) is multiplied, the result is x2 + 8x - 20. Changing the signs to (x - 10)(x + 2) would produce the correct factorization.no; If (x + 10)(x - 2) is multiplied, the result is x2 + 8x - 20. Changing the signs to (x - 10)(x + 2) would produce the correct factorization.
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Graphing quadratic functions prepares students for
• Lesson 98 Solving Quadratic Equations by Factoring
• Lesson 100 Solving Quadratic Equations by Graphing
• Lesson 102 Solving Quadratic Equations Using Square Roots
• Lesson 110 Using the Quadratic Formula
The height in feet of a dolphin as it jumps out of the water at an aquarium show can be modeled by the function f (x) = -16x2 + 32x, where x is the time in seconds after it exits the water. Find how long the dolphin is in the air. 2 seconds
LOOKING FORWARDCHALLENGE
644 Saxon Algebra 1
Problem 23
Write the division problem as the dividend times the reciprocal of the divisor. Factor each term and simplify.
Problem 25
Use the distance formula.
Problem 27
Test different types of numbers: positive, negative, zero, and fractions.
Problem 30
Remind students to be mindful of the signs when multiplying binomials.
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Lab 9 645
L AB
9Linear inequalities in two variables can be graphed by hand or with a graphing calculator. Begin by following the process used to graph a linear equation. Then change the settings to shade the region of the coordinate system that makes the inequality true.
Graph the solution set of the inequality y > 2x - 7.
1. Enter the equation y = 2x - 7 into the Y = editor.
2. Graph the equation by pressing and selecting 6:ZStandard.
The line represents the boundary of the solution set of the inequality y > 2x - 7. The solution set of the inequality is either the region above or the region below the line y = 2x - 7.
Use a test point to determine which region makes the inequality true.
Choose a test point that is not on the graph of the line y = 2x - 7.
The point (0, 0) is a good test point because it does not fall on the boundary line.
Substitute 0 for both x and y in the inequality. This substitution gives 0 > 2 · 0 - 7, or 0 > -7, which is true.
Since the point (0, 0) satisfies the inequality, the solution set is the region that contains the point (0, 0).
3. Shade the region above the line y = 2x - 7.
To graph this region, press . Then press
the key twice. The cursor moves to the
left of Y1 over an icon that looks like a line segment, \.
Press twice to choose the icon, which resembles a shaded
region above a line.
Graphing Calculator Lab (Use with Lesson 97)
Graphing Linear Inequalities
Online Connection
www.SaxonMathResources.com
Graphing
Calculator Tip
For help with entering an equation into the Y=
editor, see the graphing calculator keystrokes in Lab 3 on page 305.
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Lab 9 645
9LAB
Materials
• graphing calculator
Discuss
The solution set and graph for a linear inequality is a region of the rectangular coordinate system. Remind students that the graph of a linear equation is a straight line. The inequality sign extends the solution set to a region on a side of the line on the graph.
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Saxon Algebra 1646
(Pressing a third time allows you to choose the icon, which
resembles a shaded region below a line.)
4. Press to view the graph of the solution set.
All points in the shaded region are in the solution set of y > 2x - 7.
Note that the solution set does not include points on the line y = 2x - 7. The graph of the solution set of the inequality y ≥ 2x - 7 does include points on the boundary line y = 2x - 7.
Lab Practice
Graph the solution set of each inequality. a. y < 3x + 5; Is the point (1, 1) a solution of the inequality? yes
b. y ≥ 2x - 5; Is the point (7, 2) a solution of the inequality? no
c. y < -2x + 3; Is the point (0, 0) a solution of the inequality? yes
a.a.
b.b.
c.c.
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646 Saxon Algebra 1
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Lesson 97 647
Warm Up
97LESSON
1. Vocabulary The of the equation of a line is y = mx + b, where m is the slope of the line and b is the y-intercept. slope-intercept form
Determine the slope and the y-intercept of each equation.
2. y = - 1 _ 3 x - 5 3. 2x + 2y = 6
slope is -1; y-intercept is 3
Graph each of the following inequalities on a number line.
4. y < 3 0 2 4-2
5. x ≥ -2 0 2-2
A linear inequality is similar to a linear equation, except that a linear inequality has an inequality symbol instead of an equal sign. A solution of a linear inequality is any ordered pair that makes the inequality true.
You can evaluate an inequality with an ordered pair to find out if the ordered pair makes the inequality true and is a solution.
Example 1 Determining Solutions of Inequalities
Determine if each ordered pair is a solution of the given inequality.
a. (0, 4); y > 5x - 1
SOLUTION
y > 5x -1
4 > 5(0) -1 Evaluate the inequality for the point (0, 4).
4 > -1 Simplify.
The inequality is true. The ordered pair (0, 4) is a solution.
b. (3, -3); y < -3x + 6
SOLUTION
y < -3x + 6
-3 < -3(3) + 6 Evaluate the inequality for the point (3, -3).
-3 < -3 Simplify.
The inequality is not true because -3 is not less than -3. The ordered pair (3, -3) is not a solution.
c. (-4, 8); y ≤ 9
SOLUTION
y ≤ 9
8 ≤ 9 Evaluate the inequality for the point (-4, 8).
The inequality is true. The ordered pair (-4, 8) is a solution.
(49)(49)
slope is - 1_3; y-intercept
is -5slope is - 1_
3; y-intercept
is -5 (49)(49) (49)(49)
(50)(50) (50)(50)
New ConceptsNew Concepts
Graphing Linear Inequalities
Online Connection
www.SaxonMathResources.com
Hint
The inequalities y ≤ 9 and y ≤ 0x + 9 are equivalent. If an ordered pair is a solution of the inequality, then the y-coordinate is less than or equal to 9, and the x-coordinate can be any real number.
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LESSON RESOURCES
Student Edition Practice Workbook 97
Reteaching Master 97Adaptations Master 97Challenge and Enrichment
Master C97
The word solution means answer. For solutions of linear inequalities, there will be more than one ordered pair that will give a correct answer, or solution.
The graph of a simple inequality on a number line can be compared to the solution of a linear inequality. Note the differences:
the linear inequality contains two variables while a simple inequality has only one variable. The solution of an inequality with only one variable has a linear solution. The solution of a linear inequality has an area as a solution.
MATH BACKGROUND
Warm Up1
97LESSON
Lesson 97 647
Problem 4
Remind students to represent the inequality using an open or closed circle.
2 New Concepts
In this lesson, students are introduced to graphing linear inequalities on the coordinate plane.
Example 1
This example shows whether a given ordered pair is or is not a solution of a linear inequality.
Additional Example 1
Determine if the ordered pair is a solution of the given inequality.
a. (3, -2); y ≤ 2x - 8 (3, -2) is a solution.
b. (2, -2); y > x - 3 (2, -2) is not a solution.
c. (-2, 8); x > -4 (-2, 8) is a solution
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Saxon Algebra 1648
Exploration Exploration Graphing Inequalities
a. Graph the equation y = x + 2 on a coordinate plane.
b. Test three points that lie above the graph of y = x + 2. Substitute the coordinates of each point for the x- and y-values in the inequalityy < x + 2. If the statement is true, mark the point of the graph.
c. Test three points that lie below the graph of y = x + 2. Substitute the coordinates of each point for the x- and y-values in the inequalityy < x + 2. If the statement is true, mark the point of the graph.
d. To graph the inequality y < x + 2, would you choose points above or below y = x + 2? below
e. Would you graph points above or below the graph of y = x - 3 to graph the inequality y > x - 3? above
A linear inequality describes a region of a coordinate plane called a half-plane. The boundary line for the region is the related equation.
To graph an inequality, begin by graphing the boundary line. Test points are helpful in deciding which half-plane makes the inequality true.
The boundary line is a dashed line when the inequality contains the symbol < or >. The boundary line is a solid line when the inequality contains the symbol ≤ or ≥.
Example 2 Graphing Linear Inequalities without Technology
Graph each inequality.
a. y ≥ - 3 _ 4
x - 3
SOLUTION
Graph the boundary line y = - 3 _ 4 x - 3 using a solid line because the
inequality contains the symbol ≥.
Next use an ordered pair as a test point to find which half-plane should be shaded on the coordinate plane.
Test (0, 0).
y ≥ - 3_4
x - 3
0 ≥ - 3 _ 4 (0) - 3 Evaluate for (0, 0).
0 ≥ -3 Simplify.
The point (0, 0) satisfies the inequality, so it is a solution. The half-plane that contains the point should be shaded.
a–c. Sample:
x
y
O
4
2 4
-2
-4
-4
a–c. Sample:
x
y
O
4
2 4
-2
-4
-4
x
y
O
8
8
-4
4
-4 4-8
-8
x
y
O
8
8
-4
4
-4 4-8
-8
Hint
The point (0, 0) is a good test point if it is not on the boundary line.
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Explain the meaning of a shaded area.Say:
“An area that has been shaded is an area that has been darkened by coloring. To shade, artists may color more heavily in places.”
Have students discuss other situations where artists use shading in their drawings.
ENGLISH LEARNERS
648 Saxon Algebra 1
ExplorationExploration
This exploration will connect the shaded area of the graph with the solution of the inequality.
Extend the Exploration
Graph the equation x = 3. Where would you graph the inequality x ≤ 3? Graph the points of the inequality to the left of the line for x = 3.
2
x
y
O
4
2 4
-2
-2-4
-4
Example 2
This example introduces students to graphing a linear inequality on the coordinate plane.
Additional Example 2
Graph each inequality.
a. y ≤ 1 __ 2 x + 2
x
y
O
4
2 4
-2
-2-4
-4
b. y > 3
2
x
y
O
4
2 4
-2
-2-4
-4
Error Alert Watch for students who identify the area to shade without checking. Have students verify their solutions using substitution.
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Lesson 97 649
b. x < 4
SOLUTION
Graph the boundary line x = 4 using a dashed line because the inequality contains the symbol <.
Test (0, 0).
x < 4
0 < 4 Evaluate for (0, 0).
The point (0, 0) satisfies the inequality, so it is a solution. The half-plane that contains the point should be shaded.
To graph an inequality using a graphing calculator, the inequality must first be solved for y.
Example 3 Graphing Linear Inequalities with Technology
Graph each inequality using a graphing calculator.
a. 16x + 4y ≤ 8
SOLUTION
Solve for y.
16x + 4y ≤ 8
4y ≤ -16x + 8 Subtract 16x from both sides.
4y
_ 4 ≤
-16x _ 4 +
8 _ 4 Divide all three terms by 4.
y ≤ -4x + 2 Simplify.
Enter the inequality into your graphing calculator to view the graph.
b. 3x - y < -4
SOLUTION
Solve for y.
3x - y < -4
-y < -3x - 4 Subtract 3x from both sides.
-y
_ -1
> -3x _ -1
- 4 _
-1 Divide all three terms by -1.
y > 3x + 4 Simplify.
Enter the inequality into your graphing calculator to view the graph.
x
y
O
8
4
8
-4
-4-8
-8
x
y
O
8
4
8
-4
-4-8
-8
Caution
Remember to reverse the inequality when multiplying or dividing both sides by a negative number.
Graphing
Calculator Tip
For help with graphing inequalities, see graphing calculator Lab 9 on p. 645.
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Students may have diffi culty deciding which area to shade to represent the inequality. Once the equation of the related line is graphed, have them draw a dotted or solid line. For Example 3, copy and use the charts.
y ≤ -4x + 2
output ≤ line
Say, “Since the output is less than or equal to the y-values on the line, shade below the line to represent the output.”
y > 3x + 4
output > line
Say, “Since the output is greater than the y-values on the line, shade above the line to represent the output.”
INCLUSION
Lesson 97 649
Example 3
This example demonstrates graphing a linear inequality using a graphing calculator. Simplifying the inequality fi rst is necessary.
Additional Example 3
Graph each inequality using a graphing calculator.
a. 2y - 8x > -6
b. 5 - 2y ≤ 3x
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Saxon Algebra 1650
Example 4 Writing a Linear Inequality Given the Graph
Write an inequality for the region graphed on each coordinate plane.
a.
x
y
O
4
2
2 4
-2
-2-4
-4
SOLUTION
Determine the equation of the boundary line. It is a horizontal line that cuts through the y-axis at -1.
The equation of the boundary line is y = -1.
Then decide which inequality symbol to use for this graph.
The graph is shaded below the solid boundary line, so the inequality contains the ≤ symbol.
The inequality shown on the graph is y ≤ -1.
b.
x
y
O
8
4
4 8-4-8
-8
SOLUTION
Determine the equation of the boundary line. It has a y-intercept at -4 and a slope of 1.
The equation of the boundary line is y = x - 4.
Then decide which inequality symbol to use for this graph.
The graph is shaded above the dashed boundary line, so the inequality contains the > symbol.
The inequality shown on the graph is y > x - 4.
Example 5 Application: Carnival
Kia will attend a school carnival and she plans to spend no more than $12. Each game costs $2 and each item of food costs $3. Write and graph an inequality to describe the total cost of the carnival.
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650 Saxon Algebra 1
Example 4
Students learn to formulate an inequality from a graph.
Additional Example 4
Write an inequality for the region graphed on each coordinate plane.
a.
x
y
O
4
2
2 4
-2
-2-4
-4
x > 3
b.
2
-4
x
y
O
4
2 4
-2
-2
-4
y ≤ - 3 __ 2
Example 5
This example helps students formulate inequalities that describe maximum cost.
Extend the Example
Kia’s father gave her an extra $5 for the carnival and her mother gave her an extra $8. Kia also learned that the cost of food items is $3.50 this year. Write a new inequality and graph the solution. 2x + 3.5y ≤ 25;
20
x
y
0
30
10 30
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Lesson 97 651
SOLUTION
Write the inequality that models the situation.
cost of games plus cost of food is no more than 12
2x + 3y ≤ 12
Solve the inequality for y.
2x + 3y ≤ 12
3y ≤ -2x + 12 Subtract 2x from both sides.
y ≤ - 2 _ 3 x + 4 Divide all three terms by 3 and simplify.
Graph the solutions.
Since Kia cannot buy a negative amount of games and food, use only Quadrant I. Graph the boundary line y = - 2 _
3 x + 4. Use a solid
line for ≤.
Shade below the line. Kia must buy whole numbers of games and food items. All the points on or below the line with whole-number coordinates are the different combinations of games and food Kia can buy.
Lesson Practice
Determine if each ordered pair is a solution of the given inequality.
a. (2, 6); y > 3x - 2 yes
b. (4, 1); y < -4x + 1 no
c. (-6, 2); y ≤ 5 yes
Graph each inequality.
d. 4x + 5y ≥ -10
e. x < 6
Graph each inequality using a graphing calculator.
f. 4x + 2y ≤ 6
g. y > 2x + 6
Write an inequality for the region graphed on each coordinate plane.
h.
x
y
O
8
4 8
-4
-4-8
-8
i.
x
y
O
8
8
-4
4
-4-8
-8
y ≤ 4 y > x - 5
x
y
2 4 6 8
2
4
6
8
0x
y
2 4 6 8
2
4
6
8
0
(Ex 1)(Ex 1)
(Ex 2)(Ex 2)
d.
x
y
O
4
2
2 4-4
-4
d.
x
y
O
4
2
2 4-4
-4
e.
x
y
O
8
4
4 8
-4
-4-8
-8
e.
x
y
O
8
4
4 8
-4
-4-8
-8
(Ex 3)(Ex 3)f.f.
g.g.
(Ex 4)(Ex 4)
Caution
The graph of an inequality will represent all of the solutions of the inequality, but only selected points on the graph may represent solutions of the word problem.
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Lesson 97 651
Additional Example 5
Student tickets for a play cost $5 and adult tickets cost $10. Write an inequality to describe the minimum number of tickets that need to be sold to earn $525 in sales. 5x + 10y ≥ 525
x
y
60
80
40
20
30 60 90O
Lesson Practice
Problem c
Error Alert Make sure that students select the correct coordinate of the ordered pair to use in the inequality.
Problem f
Scaff olding Have students solve for y. Then enter the inequality into the graphing calculator.
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“Explain how to graph an inequality on the coordinate plane.” Graph the boundary line. Then shade the area of the graph that contains the solution set of the inequality.
“How can you determine which area of the graph to shade to show the solution set?” Substitute an ordered pair into the inequality. If the point is a solution, then the points on that side of the line make up the solution set. If the point is not a solution, then the points on the other side make up the solution set.
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Practice Distributed and Integrated
Saxon Algebra 1652
j. Nila has plans to attend the school bookfair and she wants to spend no more than $25. Each book series costs $15 and each book costs $5. Write an inequality to describe the total cost of the books Nila can buy and graph the inequality. 15x + 5y ≤ 25; See Additional Answers.
(Ex 5)(Ex 5)
Simplify.
1. 30x-2y12
_ 6y-5
5y17_x2
2. √ ��� 0.09q2r + q √ �� 0.04r 0.5q√ � r
3. 16g4
_ 2g + 3
- 81 _
2g + 3 (4g2 + 9)(2g - 3)
4. Find the range of the data set that includes the ages of 9 members of a chess club:23, 7, 44, 31, 18, 27, 35, 39, 66. 59
5. Find the product (4x2 + 8)(2x - 7) using the FOIL method.
*6. Add 9 _ 9x - 36
+ -24 _
3x2 - 48 . 1_
x + 4
*7. Jim ran a total of x _
x2 + 2x + 1 miles in the gym and x + 2
_ x + 1
miles outside. How
many more miles did he run inside? -x2 - 2x - 2__(x + 1)(x + 1)
miles
8. Find the quotient of (x2 - 14x + 49) ÷ (x - 7). (x - 7)
Solve and graph the inequality.
9. 13 ≤ 2x + 7 < 15 3 ≤ x < 4;
10. ⎢x�
_ 6 > 8 x < -48 or x > 48;
11. Determine if the inequality 3x - 4x ≥ 6 - x + 8 is never, sometimes, or always true. If it is sometimes true, identify the solution set. never true
*12. Determine if the ordered pair (2, 6) is a solution of the inequality y > 3x - 2.Yes, it satisfies the inequality.
*13. Graph the function. y = x2 + 2x - 24
*14. Write What points on a graph of an inequality satisfy the inequality? Explain.
*15. Generalize How do you know which half-plane to shade for the graph of a linear inequality?
(32)(32)
(69)(69)
(90)(90)
(48)(48)
(58)(58)8x3 - 28x2 + 16x - 56.8x3 - 28x2 + 16x - 56.
(95)(95)
(95)(95)
(93)(93)
(82)(82) 4 5 62 31 4 5 62 31
(91)(91) 400 20-20-40 400 20-20-40
(81)(81)
(97)(97)
(96)(96)
13.
x
y
-4-8
-30
-10
10
-20
13.
x
y
-4-8
-30
-10
10
-20
(97)(97)
14. Sample: All the points that are on a solid boundary line and all the points that fall in the shaded half-plane satisfy the inequality.
14. Sample: All the points that are on a solid boundary line and all the points that fall in the shaded half-plane satisfy the inequality.
(97)(97)
15. Sample: Choose a test point and evaluate the inequality for that point. If the point satisfi es the inequality, shade the half-plane that contains that point. If it does not satisfy the inequality, shade the remaining half-plane.
15. Sample: Choose a test point and evaluate the inequality for that point. If the point satisfi es the inequality, shade the half-plane that contains that point. If it does not satisfy the inequality, shade the remaining half-plane.
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652 Saxon Algebra 1
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 4
Extend the Problem
Find the median and mean of the data set. median: 31; mean: 32.2
Problem 5
Error AlertStudents who do not read the problem carefully may expect a product with three terms. Caution students to be sure that they have copied the factors carefully before multiplying.
Problem 7
Guide the students by asking them the following questions.
“What is the fi rst step needed to solve this problem?” Factor the denominator of the fi rst expression.
“What are the factors of the denominator in the fi rst expression?” The factors are (x + 1) and (x + 1).
“What is the LCD of the two expressions?” (x + 1)(x + 1)
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Lesson 97 653
*16. Multiple Choice Which inequality represents the graph on the coordinate plane? B
x
y
O
4
2
-2
-4
-4 -2
A y ≥ - 2 _ 3 x + 2 B y ≤ -
2 _ 3 x + 2
C y ≤ 2 _ 3
x + 2 D y < - 2 _ 3 x + 2
17. Football Tickets for the school football game cost five dollars for adults and three dollars for students. In order to buy new helmets, at least $9000 worth of tickets must be sold. Write an inequality that describes the total number of tickets that must be sold in order to buy new helmets. 5x + 3y ≥ 9000
*18. Error Analysis Two students find the vertex for y = x2 - 6x + 19. Which student is correct? Explain the error. Student B; Sample: Student A did not substitute the x-value into the original equation to find the y-value.
Student A
-b _ 2a
= 6 _ 2
= 3
The vertex is (3, 0).
Student B
-b _ 2a
= 6 _ 2 = 3
32 - 6(3) + 19 = 10 The vertex is (3, 10).
*19. Geometry The area of a rectangle is 48 square inches. The length is three times the width. Find the width of the rectangle by finding the positive zero of the function y = 3w2 - 48. 4 inches
*20. Multi-Step The height y of a golf ball in feet is given by the function y = -16x2 + 49x. a. What is the y-intercept? 0
b. What does this y-intercept represent? Sample: The ball starts on the ground.
c. What answer does the equation give for the height of the ball after 5 seconds?
d. What does that height mean? Sample: After 5 seconds, the ball has already landed. It cannot have a negative height.
*21. Commuting Jeff traveled 1 _ 2x2 - 4x
miles for his job on Monday and 1 _ x3 - 2x2 miles for
his job on Tuesday. How many more miles did he travel on Tuesday? - 1_
2x2 miles
22. Multiple Choice Add 3y + 2
_ y + z + 4 _
2y + 2z . A
A 3y + 4
_ y + z B y + 4
_ y + z
C 6y
_ 4mn
D 3y - 4
_ y - z
(97)(97)
(97)(97)
(96)(96)
(96)(96)
(96)(96)
-155 feet-155 feet
(95)(95)
(95)(95)
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Lesson 97 653
Problem 16
Guide the students by asking them the following questions.
“Which choices can you immediately discard? Why?” Choice D because the line of the graph is solid; choice A because the shading of the graph is below the line.
“What is the y-intercept of the graph?” 2
“What is the slope of the graph?” - 2 __
3
“What choice can you eliminate now? Why?” choice C because the slope of the graph is negative
Problem 17
Extend the Problem
Adult tickets to the game cost $15. Student tickets cost $6. If 432 adult tickets are sold, how many student tickets must be sold to reach the goal of $9000? 420
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Saxon Algebra 1654
23. Verify Show that -11 is a solution to the inequality ⎢3x� - 2 = 31. Sample: ⎢3 (-11)� - 2 = ⎢-33� - 2 = 33 - 2 = 31
24. Multiple Choice Solve 4⎢x - 8� = 12. CA {-5, 5} B {11, -11}
C {5, 11} D {-20, 20}
25. Bike Riding Ron rode his bike for 10 _
45x2 + 4x - 1 minutes to get to his grandmother’s
house that was 1 _ 45x - 5
+ 2x _
25x + 5 miles away. Find his rate in miles per minute.
26. Carpentry A carpenter uses a measuring tape with an accuracy of ± 1 _ 32
inches. He measures the height of a bookshelf to be 95 5
_ 8 inches. Solve the inequality
⎢ ⎢ x - 95 5 _ 8 � � ≤ 1
_ 32 to find the range of the height of the bookshelf.
27. Generalize How are the number of zeros of a function related to the location of the vertex of the function’s parabola? See Additional Answers.
28. Probability The probability of winning a certain game is 2x4y2
_ 15xy3 . The probability
of winning a different game is 5x2y
_ 8x3y2
. What is the probability of winning both games? x2_
12y2
29. Rate An orange juice machine squeezes juice out of x2 + 30x oranges every hour. How much time in days will it take to squeeze 3000 oranges? 125_
x(x + 30) days
30. Multi-Step A circle has a radius of x. Another circle has a radius of 3x. a. Write equations for the areas of both circles. A = πx2; A = 9πx2
b. Graph both functions in the same coordinate plane.
c. Compare the graphs. Sample: The graph of the area of the larger circle is much narrower.
(94)(94)
(94)(94)
(92)(92) 18x2 + 3x + 1__50
miles
per minute
18x2 + 3x + 1__50
miles
per minute
(91)(91)
95 19_32
≤ x ≤ 95 21_32
95 19_32
≤ x ≤ 95 21_32
(89)(89)
(88)(88)
(88)(88)
(84)(84)
30b.
x
y
6
8
4
2
1 2 3 4O
30b.
x
y
6
8
4
2
1 2 3 4O
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Find the solution set to the system of inequalities by graphing. The solution set is the region where the areas overlap.
y < 3x - 5
2x - 3y ≤ 4
2
x
y
O2
-2
-2
Graphing linear inequalities prepares students for
• Lesson 109 Graphing Systems of Linear Inequalities
CHALLENGE LOOKING FORWARD
654 Saxon Algebra 1
Problem 26
Extend the Problem
Using the same tape measure, the carpenter measures the door opening at 31 1 __ 4 inches. Find the range of possible measures by solving the inequality ⎢x - 31 1 __ 4 � ≤ 1 ___
32 . 31 7 ___
32 ≤ x ≤ 31 9 ___
32
Problem 27
Guide the students by asking them the following questions.
“Think of the graph of a quadratic function. Under what circumstances would you say there is a zero of the function?” If the parabola touches or crosses the x-axis, there would be a zero in the function.
“Under what circumstances would there be one zero?” There would be one zero if the vertex was on the x-axis.
“When would there be two zeros?” There would be two zeros if the vertex was below the x-axis opening upward, or if the vertex was above the x-axis opening downward.
“When would there be no zeros?” There would be no zeros when the vertex is above the x-axis opening upward, or when the vertex is below the x-axis opening downward.
Problem 28
Error AlertWatch for students who add instead of multiplying the given probabilities. Review how to fi nd the probability of two independent outcomes.
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Lesson 98 655
Warm Up
98LESSON
1. Vocabulary A is an x-value for the function where f (x) = 0.zero of a function
Factor.
2. x2 + 3x - 88 (x + 11)(x - 8) 3. 6x2 - 7x - 5 (3x - 5)(2x + 1)
4. 4x2 + 28x + 49 (2x + 7)(2x + 7) 5. 12x2 - 27 3(2x + 3)(2x - 3)
A root of an equation is the solution to an equation. A quadratic equation can have zero, one, or two roots. The roots of a quadratic equation are the x-intercepts, or zeros, of the related quadratic function.
To find the roots of a quadratic equation, set the equation equal to 0. If the quadratic expression can be factored, the equation can be solved using the Zero Product Property.
Zero Product Property
If the product of two quantities equals zero, at least one of the quantities equals zero.
Example 1 Using the Zero Product Property
Solve.
(x - 4)(x + 5) = 0
SOLUTION
By the Zero Product Property, one or both of these factors must be equal to 0. To find the solutions, set each factor equal to zero and solve.
x - 4 = 0 x + 5 = 0 Set each factor equal to zero.
x = 4 x = -5 Solve each equation for x.
Check Substitute each solution into the original equation to show it is true.
(x - 4)(x + 5) = 0 (x - 4)(x + 5) = 0
(4 - 4)(4 + 5) � 0 (-5 - 4)(-5 + 5) � 0
0 · 9 � 0 -9 · 0 � 0
0 = 0 ✓ 0 = 0 ✓
The solution set is {-5, 4}.
(89)(89)
(72)(72) (75)(75)
(83)(83) (83)(83)
New ConceptsNew Concepts
Solving Quadratic Equations by Factoring
Online Connection
www.SaxonMathResources.com
Math Language
The roots of a quadratic equation are the values of x that makeax2 + bx + c = 0.
Math Reasoning
Analyze What is the difference between a quadratic function and a quadratic equation?
Sample: A quadratic equation has one variable, and a quadratic function has two variables.
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LESSON RESOURCES
Student Edition Practice Workbook 98
Reteaching Master 98Adaptations Master 98Challenge and Enrichment
Master C98
The graph of a quadratic function is a parabola. The function is equal to 0 at the x-intercept(s).
Use this principle to set a quadratic equation equal to 0 and then to solve it. Factoring the equation produces two factors. By the Zero Product Property (or Principle of Zero Products), one of the factors must be equal to zero.
By setting each factor equal to zero and solving, the x-intercepts, or roots of the equation are determined.
Solutions can be checked by substituting one solution at a time into the original equation.
As a further check, have the students confi rm their solutions using a graphing calculator. Have them graph the original function on the calculator and then use the calculator to fi nd the x-intercepts. The x-intercepts and their solutions should be the same. Checking by graphing will also emphasize that the roots, solutions, and x-intercepts have the same meaning.
MATH BACKGROUND
Warm Up1
98LESSON
Lesson 98 655
Problems 2–5
These problems review factoring quadratic expressions in preparation for solving quadratic equations by factoring.
2 New Concepts
This lesson introduces the Zero Product Property. Students are informally aware of this principle, but have not applied it.
This property shows that each factor is equal to zero.
Discuss the term root of an equation. Have students note that on the graph of the equation, any x-intercept, where y is equal to zero, is a solution of the equation.
Example 1
This example uses the Zero Product Property to solve a quadratic equation.
Additional Example 1
Solve. (x - 3)(x - 8) = 0 {3, 8}
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Saxon Algebra 1656
Example 2 Solving Quadratic Equations by Factoring
Find the roots.
a. x2 + 2x = 8
SOLUTION
x2 + 2x = 8
x2 + 2x - 8 = 0 Set the equation equal to 0.
(x + 4)(x - 2) = 0 Factor.
Use the Zero Product Property to solve the equation.
x + 4 = 0 x - 2 = 0 Set each factor equal to zero.
x = -4 x = 2 Solve each equation for x.
Check
x2 + 2x = 8
(-4)2 + 2(-4) � 8
16 - 8 � 8
8 = 8 ✓
x2 + 2x = 8
(2)2 + 2(2) � 8
4 + 4 � 8
8 = 8 ✓
The roots are -4 and 2.
b. 7x2 - 6 = 19x
SOLUTION
7x2 - 6 = 19x
7x2 - 19x - 6 = 0 Set the equation equal to 0.
(7x + 2)(x - 3) = 0 Factor.
7x + 2 = 0 x - 3 = 0 Set each factor equal to zero.
7x = -2 x = 3 Solve each equation for x.
x = - 2 _ 7
Check
7x2 - 6 = 19x
7 (- 2 _ 7 )
2
- 6 � 19 (- 2 _ 7 )
( 4 _ 7 ) - 6 � - 38 _ 7
- 38 _ 7 = -
38 _ 7 ✓
7x2 - 6 = 19x
7(3)2 - 6 � 19(3)
63 - 6 � 57
57 = 57 ✓
The roots are - 2 _ 7 and 3.
Hint
Standard form for the equation isax2 + bx + c = 0.
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For this lesson, say and explain the meaning of root. Say:
“The word root means source or foundation. For example, the root of a tooth is under the gums.”
Also relate the word root to the root of a tree, explaining that the roots are the foundation of a tree.
ENGLISH LEARNERS
656 Saxon Algebra 1
TEACHER TIPEncourage students to try to factor the equations mentally rather than listing all the possible combinations fi rst. They may need to adjust their fi rst guesses as they go along, but this will give them practice in mental math as well as help them gain familiarity with the factoring process.
Example 2
These examples demonstrate how to solve quadratic equations by factoring.
Additional Example 2
Find the roots.
a. x2 + 6 = -5xThe roots are -3 and -2.
b. 3x2 = 4x + 7The roots are 7 __
3 and -1.
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Lesson 98 657
Example 3 Finding the Roots by Factoring Out the GCF
Find the roots.
20 - 2x2 = 70 - 20x
SOLUTION
2x2 - 20x + 50 = 0 Set the equation equal to zero.
2(x2 - 10x + 25) = 0 Factor out the GCF.
2(x - 5)(x - 5) = 0 Factor the trinomial expression.
Disregard the factor of 2, since it can never equal 0.
The factor (x - 5) appears twice, but it only needs to be set to equal zero once.
x - 5 = 0
x = 5
Check
20 - 2x2 = 70 - 20x
20 - 2(5)2 � 70 - 20(5)
20 - 2(25) � 70 - 100
20 - 50 � -30
-30 = -30 ✓
The root is 5.
Finding the values of x that satisfy the quadratic equation is another way of finding the roots of a quadratic equation.
Example 4 Application: Gardening
The area of a rectangular garden is 51 square yards. The length is 14 yards more than the width. What are the length and width of the garden?
SOLUTION
Let w be the width and w + 14 be the length.
A = lw Area formula
51 = (w + 14)w Substitute known values into the equation.
51 = w2 + 14w Distribute.
0 = w2 + 14w - 51 Write the equation into standard form.
0 = (w + 17)(w - 3) Factor.
w + 17 = 0 w - 3 = 0 Use the Zero Product Property.
w = -17 w = 3 Solve.
Because the width must be a positive number, the only possible solution is 3 yards. Since the width is 3 yards, the length is w + 14. So the length is 3 + 14, which is 17 yards.
Caution
When checking your answers, use the original equation, not the one that has been rearranged. Also, use the order of operations when simplifying each side of the equation.
Math Reasoning
Justify How could you check your answers?
Sample: Multiply the length and width to see if the product is 51.
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Have students practice factoring quadratic expressions, including expressions for which the GCF can be factored out:
x2 - x - 12 (x - 4)(x + 3)
6x2 - 24x + 24 6(x - 2)(x - 2)
6x3 - 2x2 - 8x 2x(3x - 4)(x + 1)
Encourage students to identify which factors of their solutions would be used to fi nd solutions for quadratic equations.
Help them understand that each factor containing a variable could be used to determine a root of an equation. For example, have students set each factor of 6x2 - 24x + 24 equal to zero. Point out the 6 ≠ 0. Then have them set each factor of 6x3 - 2x2 - 8x equal to zero. Point out that x = 0 is a solution of 2x = 0.
INCLUSION
Lesson 98 657
Example 3
This example requires that the GCF be fi rst factored out of the quadratic equation. Then the equation is factored to solve.
Additional Example 3
Find the roots. 6x2 - 12 = -21x.The roots are 1 __
2 and -4.
Example 4
This example shows how to solve an area problem using the Zero Product Property.
Extend the Example
The area of the garden is increased by 21 square yards. What are the dimensions of the expanded garden? The width is 4 yards and the length is 18 yards.
Additional Example 4
The area of a rectangular pasture is 58,500 square meters. The width is 320 meters less than the length. What are the length and width of the pasture? length = 450 meters; width = 130 meters
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Saxon Algebra 1658
Example 5 Solving Quadratic Equations with Missing Terms
Solve.
a. 18x2 = 8x
SOLUTION
18x2 - 8x = 0 Set the equation equal to zero.
2x(9x - 4) = 0 Factor out the GCF.
2x = 0
x = 0
9x - 4 = 0 Set each factor equal to zero.
x = 4 _ 9 Solve each equation for x.
Check
18x2 = 8x
18 · (0)2 � 8(0)
0 = 0 ✓
18x2 = 8x
18 ( 4 _ 9 )
2
� 8 ( 4 _ 9 )
18 ( 16
_ 81
) � 8 ( 4 _ 9 )
32 _ 9 =
32 _ 9 ✓
The solution set is ⎧ ⎨
⎩ 0, 4 _
9 ⎫ ⎬
⎭ .
b. 4x2 - 25 = 0
SOLUTION
4x2 - 25 = 0 Set the equation equal to 0.
(2x - 5)(2x + 5) = 0 Factor.
2x - 5 = 0 2x + 5 = 0 Set each factor equal to zero.
2x = 5 2x = -5 Solve each equation for x.
x = 2.5 x = -2.5
Check
4x2 - 25 = 0
4(2.5)2 - 25 � 0
4(6.25) - 25 � 0
25 - 25 � 0
0 = 0 ✓
4x2 - 25 = 0
4 · (-2.5)2 - 25 � 0
4(6.25) - 25 � 0
25 - 25 � 0
0 = 0 ✓
The solution set is {-2.5, 2.5}.
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658 Saxon Algebra 1
Example 5
These examples demonstrate a method of solving that requires students to factor quadratic equations with missing terms.
Error Alert Some students may use incorrect signs in the solution. Encourage them to set each factor equal to zero and to solve the resulting equation. When the variable is isolated, the solution has the opposite sign of the term in the factor.
Additional Example 5
a. Solve 18x2 + 9x = 0. The solution set is {0, - 1 __
2 }.
b. Solve 9x2 - 4 = 0.The solution set is {- 2 __
3 , 2 __
3 }.
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Practice Distributed and Integrated
Lesson 98 659
Solve and graph the inequality.
1. 11 < 2(x + 5) < 20 2. ⎢x� + 1.5 ≤ 7.6
3. Determine whether the polynomial -121 + 9x2 is a perfect-square trinomial or a difference of two squares. Then factor the polynomial. difference of two squares; (3x + 11)(3x - 11)
4. Graph the function y = 2x2 + 8x + 6.
5. The number of Apples A and Oranges O grown in a certain fruit orchard can be modeled by the given expressions where x is the number of years since the trees were planted. Find a model that represents the total number of apples and oranges grown in this orchard. S = 35x3 + 28x - 24
A = 15x3 + 17x - 20
O = 20x3 + 11x - 4
6. Write an equation for a line that passes through (1, 2) and is perpendicular to y = - 3 _
4 x + 2 3
_ 4 . Sample: y = 4_3x + 2_
3
7. Solve the equation 4⎢x�
_ 9
+ 3 = 11 and graph the solution.
Simplify.
8. 3x + 6 _ 7x - 7
_
5x + 10
_ 14x - 14
6_5 9. (5xyz)2(3x-1y)2 225y4z2
*10. Determine if the ordered pair (5, 5) is a solution of the inequality y < -5x + 4. no; It does not satisfy the inequality.
*11. Write Explain the Zero Product Property in your own words. Sample: If two numbers multiplied together equal 0, then at least one of the numbers has to be 0.
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1_2
< x < 5; 40 2-2 1_2
< x < 5; 40 2-2
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Lesson Practice
a. Solve (x - 3)(x + 7) = 0. 3, -7
Find the roots. b. x2 + 3x - 18 = 0 3, -6 c. 2x2 + 13x + 15 = 0 - 3_
2, -5
Solve. d. 5x2 - 20x = 10x - 45 3 e. 45x2 = 27x ⎧
⎨
⎩ 0, 3_
5⎫ ⎬
⎭
f. 25x2 - 16 = 0 ⎧ ⎨
⎩ - 4_
5, 4_
5⎫ ⎬
⎭
g. Architecture A rectangular pool has an area of 360 square feet. The length is 6 feet more than three times the width. Find the dimensions of the pool. The width is 10 feet, and the length is 36 feet.
(Ex 1)(Ex 1)
(Ex 2)(Ex 2) (Ex 2)(Ex 2)
(Ex 3)(Ex 3) (Ex 5)(Ex 5)
(Ex 5)(Ex 5)
(Ex 4)(Ex 4)
-6.1 ≤ x ≤ 6.1;
6.1-6.1 0
4.
x
y
6
2 4
-2
-4
{18, -18};180 9-18 -9
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Lesson 98 659
Lesson Practice
Problem d
Error Alert Watch that students correctly combine the terms of the equation as they move all the terms to one side of the equation.
Problem e
Scaff olding Have students write the equation with all of the terms on one side of the equal sign. Have students explain how they decide on which side to isolate the zero.
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“What are some steps used to solve a quadratic equation by factoring?” Sample: Make sure that the equation is set equal to zero. Some terms may need to be moved from one side of the equation to the other so that the like terms can be combined.
“How is the Zero Product Property used when solving quadratic equations by factoring?” Sample: Since either factor may cause the equation to equal zero, each factor can be set equal to zero to fi nd the solution(s).
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 1
Guide the students by asking them the following questions.
“Is this compound inequality combined with AND or OR?” AND
“How is the solution set found?” Sample: Include all values common to both inequalities.
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Saxon Algebra 1660
*12. Justify What property allows you to use the following step when solving an equation? Sample: The Zero Product Property
(x + 4)(x + 5) = 0
x + 4 = 0 x + 5 = 0
*13. Multiple Choice What is the solution set of 0 = (3x - 5)(x + 2)? A
A ⎧
⎨
⎩ 5 _ 3 , -2
⎫ ⎬
⎭ B
⎧ ⎨
⎩ 5 _ 3
, 2 ⎫ ⎬
⎭ C
⎧ ⎨
⎩ -
5 _ 3 , -2
⎫ ⎬
⎭ D
⎧ ⎨
⎩ -
5 _ 3
, 2 ⎫ ⎬
⎭
*14. Ages A girl is 27 years younger than her mother. Her mother is m years old. The product of their ages is 324. How old is each person? The mother is 36 years old and the girl is 9 years old.
*15. Multi-Step Seve plans to go shopping for new jeans and shorts. She plans to spend no more than $70. Each pair of jeans costs $20 and each pair of shorts costs $10. a. Write an inequality that describes this situation. 20x + 10y ≤ 70
b. Graph the inequality. See Additional Answers.
c. If Seve wants to spend exactly $70, what is a possible number of each she can spend her money on? Sample: 3 pairs of jeans and 1 pair of shorts
*16. Geometry The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. The sides of a triangle are labeled 4x inches, 2y inches, and 8 inches. James wrote an inequality that satisfies the Triangle Inequality Theorem. He wrote the inequality 4x + 2y > 8. Use a graphing calculator to graph the inequality. See Additional Answers.
*17. Solve (x + 4)(x - 9) = 0. -4, 9
18. Error Analysis Students were asked to write an inequality that results in a dashed horizontal boundary line. Which student is correct? Explain the error. Student A; Sample: Student B wrote an inequality with a solid horizontal line.
Student A Student B
y > -3 y ≥ 4
19. Horseback Riding It took Joe 2x - 10 _
2x5 minutes to ride his horse to Darrell’s house that
was 3x2 - 15x _
3x miles away. Find his rate in miles per minute. x5 miles per minute
20. Measurement The area of a triangle can be expressed as 4x2 - 2x - 6 square meters. The height of the triangle is x + 1 meters. Find the length of the base of the triangle. 8x - 12 meters
21. Art Michael bought a rectangular painting from a local artist. The area of the painting was (20x + 5 + x3) square inches. The width was (x - 5) inches. What was the length? (x2 + 5x + 45 + 230_
x - 5) inches
22. Analyze Should you find the LCD when multiplying or dividing rational fractions? Sample: No, the LCD is found in addition and subtraction problems so that parts of equal size can be added or subtracted.
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3 43 4
(92)(92)
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CHALLENGE
Factor and solve each quadratic equation. What do they have in common?
x2 + 12x + 36 = 0
16 = -24x - 9x2
16x2 + 48x = -48x - 144
81x2 - 162x + 324 = 162x
-54x = -9x2 - 81
All these equations are products of squares. They have only one solution.
660 Saxon Algebra 1
Problem 13
Error AlertEncourage students who choose an incorrect solution to set each factor equal to zero and to carefully perform the operations necessary to isolate the variable. Then have them substitute their solutions into the respective factors and see if the result is zero.
Problem 14
Extend the Example
Have students solve the problem letting g represent the girl’s age. Have them give the age of the mother in terms of g and explain how the solution compares to the solution of the original problem. Sample: The mother’s age is g + 27. The solution is the same except that the girl’s age is found fi rst and is then used to fi nd the age of her mother.
Problem 21
Remind students that the long division will be easier if a placeholder is used for the missing quadratic term.
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Lesson 98 661
23. Multi-Step Erika and Casey started a new walking program. They walked
4x _
3x + 9 miles Thursday and 16
_ x2 + 12x + 27
miles Friday.
a. What is the total distance that they walked? 4x2 + 36x + 48__3(x + 9)(x + 3)
miles
b. If their rate was 4x
_ x + 3 miles per hour, how much time did it take them to walk on Thursday and Friday?
x2 + 9x + 12__3x(x + 9)
hours
*24. Physics A ball is dropped from 100 feet in the air. What is its height after 2 seconds? Use h = -16t2 + vt + s. (Hint: Its initial velocity is 0 feet per second.) 36 feet
*25. Error Analysis Two students find the zeros of the function y = x2 - 8x - 33. Which student is correct? Explain the error. Student B; Sample: Student A found the y-intercept.
Student A Student B
y = x2 - 8x - 33y = (0)2 - 8(0) - 33y = -33(0, -33)
0 = x2 - 8x - 330 = (x - 11)(x + 3)x - 11 = 0 or x + 3 = 0 x = 11 or x = -311 and -3
26. Multi-Step Hideyo has a picture frame that measures 20 centimeters by 26 centimeters. The frame is 1.5 centimeters wide. a. Find the length and width of the picture area. 23 cm; 17 cm
b. Find the length of the diagonal of the picture area to the nearest tenth of a centimeter. 28.6 cm
27. Profit A business sold x2 + 6x + 5 items. The profit for each item sold
is x2
_ 100x
dollars. a. What is the profit in terms of x?
x(x + 5)(x + 1)__100 dollars
b. What is the profit (in dollars) if x = 50? $1402.50
28. Multi-Step Javed has a garden with an area of 36 square feet. The width of his garden is 9 feet less than the length. What are the dimensions of his garden? a. Write a formula to find the dimensions of the garden and describe how you will
solve it. x(x - 9) = 36; Sample: The formula needs to be set in the formax2 + bx + c = 0 in order to solve for x.
b. What are the dimensions of the garden? 12 feet long and 3 feet wide.
29. Generalize Describe the steps for subtracting x - 6 _
x + 5 from x
2 - x - 30 _
x + 5 .
30. The width of a rectangle is represented by the expression (4x - 6) and the length (8x + 7). Would the area of the rectangle be correctly expressed as32x2 + 20x - 42? If not, what is the correct area? No. A sign error has occurred; The correct expression would be 32x2 - 20x - 42.
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(90)(90)
29. Sample:Distribute the negative sign so that you are subtracting x and adding 6 tox2 - x - 30. Then combine like terms to get x2 - 2x - 24. Finally, try to factor out x + 5 from the numerator. Since you can’t, your answer in simplest
form is x2 - 2x - 24__
x + 5.
29. Sample:Distribute the negative sign so that you are subtracting x and adding 6 tox2 - x - 30. Then combine like terms to get x2 - 2x - 24. Finally, try to factor out x + 5 from the numerator. Since you can’t, your answer in simplest
form is x2 - 2x - 24__
x + 5.
(Inv 9)(Inv 9)
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Solving quadratic equations by factoring prepares students for
• Lesson 100 Solving Quadratic Equations by Graphing
• Lesson 102 Solving Quadratic Equations by Using Square Roots
• Lesson 104 Solving Quadratic Equations by Completing the Square
• Lesson 110 Using the Quadratic Formula
• Lesson 113 Interpreting the Discriminant
LOOKING FORWARD
Lesson 98 661
Problem 26
Guide the students by asking them the following questions.
“How many centimeters does the frame add to the width of the picture?” Sample: 3 cm
“What formula is used to fi nd the length of the diagonal?” the Pythagorean Theorem
Problem 27
Have students check their work by substituting 50 for x in the original expressions for the number of items sold and the profi t for each item. Then have them multiply the results to fi nd the total profi t.
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Saxon Algebra 1662
Warm Up
99LESSON
1. Vocabulary The denominator of a contains a variable. The value of the variable cannot make the denominator equal to zero.
Find the LCM.
2. 7x2y and 3xy3 21x2y3 3. (3x - 6) and (9x2 - 18x) 9x(x - 2)
4. (x + 3) and (2x - 1) 5. (14x - 7y) and (10x - 5y)
A rational equation is an equation containing at least one rational expression. There are two ways to solve a rational equation: using cross products or using the least common denominator.
Either way may lead to an extraneous solution; that is, a solution that does not satisfy the original equation. The solution may satisfy a transformed equation, but make a denominator in the original equation equal 0. If an answer is extraneous, eliminate it from the solution set.
If a rational equation is a proportion, it can be solved using cross products.
Example 1 Solving a Rational Proportion
Solve each equation.
a. 3 _ x
= 5 _
x - 6
SOLUTION
3 _ x
= 5 _
x - 6
3(x - 6) = 5x Use cross products.
3x - 18 = 5x Distribute 3 over (x - 6).
-18 = 2x Subtract 3x from both sides.
-9 = x Divide both sides by 2.
Check Verify that the solution is not extraneous.
3 _ x
= 5 _
x - 6
3 _
-9 �
5 _ -9 - 6
Substitute -9 for x in the original equation.
3 _
-9 �
5 _ -15
Simplify the denominator.
- 1 _ 3
= - 1 _ 3 Simplify each fraction.
The solution is x = -9. ✓
(39)(39)rationalexpressionrationalexpression
(57)(57) (57)(57)
(57)(57)(x + 3)(2x - 1)(x + 3)(2x - 1)
(57)(57)35(2x - y)35(2x - y)
New ConceptsNew Concepts
Solving Rational Equations
Online Connection
www.SaxonMathResources.com
Math Language
A rational expression is a fraction with a variable in the denominator.
Math Reasoning
Analyze Why is it necessary to keep the terms in the denominator grouped?
Sample: They are already grouped by the fraction bar. A set of parentheses is a reminder to distribute when cross multiplying.
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MATH BACKGROUNDLESSON RESOURCES
Student Edition Practice Workbook 99
Reteaching Master 99Adaptations Master 99Challenge and Enrichment
Master C99
For this lesson, consider any expression with a variable in the denominator to be a rational expression.
Rational equations can be solved using cross products or by fi nding the least common denominator to simplify all terms on both sides of the equation.
Using cross products to solve proportions can be applied to solving rational equations that have one term on each side.
Rational equations that contain more than two terms must be solved using the LCD.
Regardless of the method used to solve rational equations, the equations are transformed into the form of an equation, linear or quadratic, that the students already know how to solve.
Warm Up1
662 Saxon Algebra 1
99LESSON
Problems 2–5
Review fi nding the LCM in preparation for fi nding the LCD to solve rational proportions.
2 New Concepts
Remind students of the defi nition of a rational number: a number that can be expressed as the quotient of two numbers. Introduce the term rational equation and have students look over the examples. Point out that all the terms have a variable in the denominator.
Explain that in this lesson, students will solve equations that contain rational terms.
Example 1
These examples introduce students to solving rational equations that take the form of a proportion.
Additional Example 1
Solve each equation.
a. x __ 5 =
2x + 4 _______ 2
x = -2.5
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Lesson 99 663
b. x _ 4 =
3 _ x - 1
SOLUTION
x _ 4 =
3 _ x - 1
x(x - 1) = 12 Use cross products.
x2 - x = 12 Distribute x over (x - 1).
x2 - x - 12 = 0 Subtract 12 from both sides.
(x - 4)(x + 3) = 0 Factor.
x = 4 or x = -3 Use the Zero Product Property to solve.
Check Verify that the solution is not extraneous.
x _ 4 =
3 _ x - 1
; x = 4
4 _ 4 �
3 _ 4 - 1
1 = 1 ✓
or
Substitute.
Simplify each fraction.
x _ 4 =
3 _ x - 1
; x = -3
-3 _ 4 �
3 _ -3 - 1
- 3 _ 4 = -
3 _ 4 ✓
The solution set is {4, -3}.
If a rational equation includes a sum or difference, find the LCD of all the terms to solve.
Example 2 Using the LCD to Solve Addition Equations
Solve 3 _ x
+ 16 _ 2x
= 11.
SOLUTION
3 _ x
+ 16 _ 2x
= 11
The LCD of the denominators is 2x.
(2x) 3 _ x
+ (2x) 16 _ 2x
= (2x)11 Multiply each term by the LCD.
6 + 16 = 22x Simplify each term.
22 = 22x Add.
1 = x Divide both sides by 22.
Check Verify that the solution is not extraneous.
3 _ x +
16 _ 2x
= 11
3 _ 1 +
16 _ 2(1)
� 11 Substitute 1 for x in the original equation.
11 = 11 ✓ Simplify.
The solution is x = 1.
Hint
To find the LCD of the terms, consider the denominator of the whole number to be 1.
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INCLUSION
Use the following strategy with students who have diffi culty with written symbols. Help students identify the terms needed to fi nd the LCD of a rational equation. Since sometimes one denominator will be a factor of another, it is not necessary to multiply by that factor a second time.
Present students with these groups of rational terms. Have them identify the denominators used to fi nd the LCD of the terms.
1. 3 ___ ab ; 6 ___ bc ; 12 ___ b ab, bc
2. 9 ______ x + 2 ; 13 ___ 2x ; 4 _____ x - 3 x + 2, 2x, x - 3
3. 18 ____ 5m ; 6 _______ 2m + 1 ; 3 ___ m 5m, 2m + 1
4. 7 ____ x2y3
; 9 ___ xy ; 2 ____ xy4
x2y3, xy4
5. 5 ______ d - 3 ; 1 ______ d - 1 ; 4 ___ d 2 d - 3, d - 1, d2
Lesson 99 663
b. x __ 3 = 5 ______
x - 2
x = -3 and x = 5
Example 2
This example demonstrates how to solve a rational equation involving addition by multiplying by the LCD.
Additional Example 2
Solve 9 ___ 3x
+ 15 ___ x = 6. x = 3
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Saxon Algebra 1664
Example 3 Using the LCD to Solve Subtraction Equations
Solve 3 _ x - 1
- 2 _ x
= 5 _
2x .
SOLUTION
3 _ x - 1
- 2 _ x
= 5 _
2x
The LCD is 2x(x - 1).
2x(x - 1) 3 _
x - 1 -2x(x - 1)
2 _ x = 2x(x - 1) 5 _
2x Multiply each term by
the LCD.
6x - 4(x - 1) = 5(x - 1) Simplify each term.
6x - 4x + 4 = 5x - 5 Use the Distributive Property.
2x + 4 = 5x - 5 Collect like terms.
4 = 3x - 5 Subtract 2x from both sides.
9 = 3x Add 5 to both sides.
3 = x Divide both sides by 3.
Check Verify that the solution is not extraneous.
3 _
x - 1 -
2 _ x
= 5 _
2x
3 _
3 - 1 -
2 _ 3
� 5 _
2(3) Substitute 3 for x in the original equation.
5 _ 6 =
5 _ 6 ✓ Simplify.
The solution is x = 3.
Example 4 Checking for Extraneous Solutions
Solve the equation.
x - 1 _ x - 2
= x + 9 _ 2x - 4
SOLUTION
x - 1 _ x - 2
= x + 9 _ 2x - 4
(x - 1)(2x - 4) = (x - 2)(x + 9) Use cross products.
2x2 - 6x + 4 = x2 + 7x - 18 Multiply.
x2 - 13x + 22 = 0 Subtract x2, 7x, and -18 from both sides.
(x - 11)(x - 2) = 0 Factor.
x = 11 or x = 2 Use the Zero Product Property to solve.
Caution
Remember to distribute the negative over the new numerator in a fraction following a subtraction sign.
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ENGLISH LEARNERS
Explain the meaning of the word extraneous. Write the word on the board and underline “extra”. Say:
“Extraneous means extra. Something that is extra is not needed. For example, the light from the window is extraneous lighting.”
Have students give other examples using the word extraneous. Sample: He made an extraneous remark about the situation.
664 Saxon Algebra 1
Example 3
This example demonstrates how to solve a rational equation involving subtraction by multiplying by the LCD. Point out to students the caution note regarding distributing the negative sign over the new numerator in a subtraction equation.
Additional Example 3
Solve 12 ___ x + 15 ______ x + 1
= 72 ___ 3x
. x = 4
Example 4
This example shows students how an extraneous solution may occur and how such cases should be handled.
Additional Example 4
Solve the equation.
3x - 1 ______ x + 3
= 2x - 4 ______ x - 2
x = 7 and x = 2;
2 is an extraneous solution, so x = 7.
Error Alert Check that students fully understand that it is possible to get a solution that makes the equation false. Some students will assume they have made an error and continue trying to fi nd a second valid solution.
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Lesson 99 665
Check Verify that the solutions are not extraneous.
x - 1 _ x- 2
= x + 9 _ 2x - 4
; x = 11
11 - 1 _ 11 - 2
� 11 + 9 _
2(11) - 4
10 _ 9
= 10 _ 9 ✓
or
Substitute.
Simplify.
x - 1 _ x - 2
= x + 9 _ 2x - 4
; x = 2
2 - 1 _ 2 - 2
� 2 + 9 _
2(2) - 4
1 _ 0 =
11 _ 0 ✗
2 is an extraneous solution.
The solution is x = 11.
Example 5 Application: Painting
It takes Samuel 7 hours to paint a house. It takes Jake 5 hours to paint the same house. How long will it take them if they work together?
SOLUTION
Understand The answer will be the number of hours h it takes for Samuel and Jake to paint the house.
Samuel can paint the house in 7 hours, so he can paint 1 _ 7 of the house
per hour.
Jake can paint the house in 5 hours, so he can paint 1 _ 5 of the house per hour.
Plan The part of the house Samuel paints plus the part of the house Jake paints equals the complete job. Samuel’s rate times the number of hours worked plus Jake’s rate times the number of hours worked will give the complete time it will take them to paint the house. Let h represent the number of hours worked.
(Samuel’s rate)h + (Jake’s rate)h = complete job
1 _ 7
h + 1 _ 5
h = 1
Solve
1 _ 7 h +
1 _ 5 h = 1
(35) 1 _ 7
h + (35) 1 _ 5 h = (35)1 Multiply each term by the LCD, 35.
5h + 7h = 35 Simplify each term.
h = 35 _ 12
Combine like terms; and divide both sides by 12.
Together, they can paint the house in 2 11
_ 12 hours.
Math Reasoning
Justify Why do you have to check both solutions to the equation?
Sample: Either or both solutions could be extraneous.
Math Reasoning
Verify Show that the solution 35_
12 hours satisfies the original equation.
Sample:1_7h + 1_
5h =1
1_7 ( 35_
12 ) + 1_5 ( 35_
12 ) = 1
5_12
+ 7_12
= 1
1 = 1
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Lesson 99 665
Example 5
This example solves a rate problem using rational expressions.
Extend the Example
Suppose that Samuel takes 1 hour less and Jake 1 hour more to paint a house. Will they still fi nish in the same amount of time as when they worked at their original rates? Explain. no; Sample: Both now have a rate of 6 hours to paint a house. Working together, it will take them 3 hours to paint the house.
Additional Example 5
It takes Bill 8 hours to mow the pasture. It takes Frank 10 hours to mow the same pasture. How long will it take them if they work together? Together, they can mow the pasture in 4 4 __
9 hours.
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Practice Distributed and Integrated
Saxon Algebra 1666
Lesson Practice
Solve each equation.
a. 6 _ x
= 7 _
x - 1 x = -6
b. 2 _
x + 4 =
x _ 6 x = 2, -6
c. 12 _ 2x
+ 16 _ 4x
= 5 x = 2
d. 4 _ x - 2
- 2 _ x
= 1 _
3x x = - 14_
5
e. x + 5 _ x + 4
= x - 2 _ 2x + 8
x = -12
f. Lawn Care It takes John 2 hours to mow the yard. Sarah can do it in 3 hours. How long will it take them if they mow the yard together? 1 1_
5 hours
(Ex 1)(Ex 1)
(Ex 1)(Ex 1)
(Ex 2)(Ex 2)
(Ex 3)(Ex 3)
(Ex 4)(Ex 4)
(Ex 5)(Ex 5)
Solve.
1. 4 _ x
= 8 _
x + 4 x = 4 2. (x - 13)(x + 22) = 0 {13, -22}
3. Physics A student is biking to a friend’s house. He bikes at 10 miles per hour. The friend lives 20 miles away, give or take 2 miles. What are the minimum and maximum times it will take him to get there? 1.8 hours, 2.2 hours
*4. Verify Without graphing, show that the point (2, -6) lies on the graph of y = x2 + x - 12. Sample: y = (2)2 + (2) - 12 = -6
5. Simplify: 4x
_ 12x - 60
+ 1 _ 4x - 16
__
-2
_ 9x - 20 - x2 . 6. Factor x + 4x2 - 5. (4x + 5)(x - 1)
7. Larry weighed 180 pounds. He has lost 2 pounds a month for x months. Write a linear equation to model his weight after 8 months. y = 180 - 2x; 164 pounds
*8. Write What is an extraneous solution? Sample: It is an answer that solves the transformed equation, but not the original one.
9. Find the LCD of 2x _ 2x2 - 72
- 12 __
x2 + 13x + 42 . LCD = 2(x - 6)(x + 6)(x + 7)
10. Population The function y = -0.0003x2 + 0.03x + 1.3 shows the population of Philadelphia County between the years 1900 and 1990, where x is the number of years after 1900 and y is the population for that year in millions of people. Find the vertex of the parabola that represents the function. What does it represent in terms of the scenario? (50, 2.05); The population reached its maximum of about 2,050,000 people in 1950.
(99)(99) (98)(98)
(94)(94)
(96)(96)
(92)(92)
5. 4x2 - 13x - 15__24
5. 4x2 - 13x - 15__24
(75)(75)
(52)(52)
(99)(99)
(95)(95)
(89)(89)
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666 Saxon Algebra 1
Lesson Practice
Problem d
Error Alert Some students may use x and 3x in the LCD. Point out that since the LCD uses each factor to its highest degree, the LCD is 3x(x - 2) rather than 3x(x)(x - 2).
Problem f
Scaff olding Have students fi rst fi nd John’s rate of mowing, and then have them fi nd Sarah’s rate of mowing. Ask them what the variable represents in this problem. Sample: the number of hours worked together
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“What are two methods used to solve rational equations?” Sample: Use cross products or use the LCD.
“Why are the solutions of a rational equation checked?” Sample: because one of the solutions might be extraneous, and that would make the equation false
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 9
Error Alert Make sure that students don’t try to use the denominators as written for the LCD. Remind them to factor polynomials when possible and to delete common factors when fi nding the LCD.
Problem 10
Extend the Problem
Predict the population of Philadelphia County in the year 2000, assuming that the trend shown by the function continues. 1.3 million
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Lesson 99 667
11. Multi-Step The coordinates of three friends’ houses on a city map are P(3, 3), Q(5, 9), and R(11, 3). The friends plan to meet at the point that is half-way between Q and the midpoint of
−− PR .
a. Find the midpoint of −−
PR . (7, 3)
b. Find the coordinates of the point where the friends plan to meet. (6, 6)
12. Find the quotient of (18x2 - 120 + 6x5) ÷ (x - 2).
13. What is the ratio of the volume of a cube with side lengths of 5 to the volume of a cube with side lengths of 3? 125_
27
*14. Formulate How can you quickly tell if a possible solution is extraneous using the denominators in the original equation? Sample: If it would cause one of the denominators to equal 0, the solution is extraneous.
*15. Multiple Choice Which of the following is an extraneous solution to the equation x2
_ x - 1
+ 4x2 - 20x _
(x - 1)(x - 5) = 10
_ 2x - 2 ? C
A x = -1 B x = 0
C x = 1 D x = 5
*16. Housekeeping It takes a man 8 hours to clean the house. His friend can clean it in 6 hours. How long will it take them if they clean together? 24_
7 hours
*17. Error Analysis Two students solve the equation 0 = (x - 5)(x + 11). Which student is correct? Explain the error.
Student A Student B
x - 5 = 0 x + 11 = 0 x = 5 x = -11
x - 5 = 0 x + 11 = 0 x = -5 x = 11
18. Geometry The area of a triangle is 24 square centimeters. The height is four more than two times the base. Find the base and height of the triangle. The base is 4 centimeters and the height is 12 centimeters.
*19. Multi-Step The length of a lawn is twice the width. The lawnmower cuts 2-foot strips. One strip along the length and width has already been cut.
2x feet
x feet
2 feet2 feet
a. Write expressions for the length and width of the area left to be cut.
b. The area left to be cut is 144 square feet. Find the width of the yard. 10 feet
c. What is the length of the yard? 20 feet
20. Generalize How do you know when there is no solution to an absolute-value inequality? Sample: There is no solution when the absolute value is less than 0 because it would be a negative number.
21. Architecture Mia is designing a rectangular city hall. She accidentally spills water on her newly revised sketches. She is only able to determine the area, which is (x2 - 15x + 56) square feet, and the length, which is (x - 7) feet. What is the width? (x - 8) feet
(86)(86)
(93)(93)6x4 + 12x3 + 24x2 + 66x + 132 + 144_
x - 2 6x4 + 12x3 + 24x2 + 66x + 132 + 144_
x - 2
(36)(36)
(99)(99)
(99)(99)
(99)(99)
(98)(98)17. Student A; Sample: Student B has the incorrect signs on both solutions.
17. Student A; Sample: Student B has the incorrect signs on both solutions.
(98)(98)
(98)(98)
2x - 2 and x - 22x - 2 and x - 2
(91)(91)
(93)(93)
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Lesson 99 667
Problem 14
Guide the students by asking them the following questions.
“What kind of denominator would not be allowed in a rational equation?” A denominator equal to 0.
“How could the denominator equal zero?” The value substituted for the variable causes the denominator to equal zero.
“What would this tell about the solution?” It would be an extraneous solution.
Problem 19
Error Alert Make sure that students write the correct expressions for the length and width of the lawn yet to be cut. If students begin with erroneous expressions, they cannot arrive at the correct solution.
Problem 21
Extend the Problem
The height of the rectangular-prism-shaped city hall is (3x + 5). What is the volume of the building? 3x3 - 40x2 + 93x + 280
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Saxon Algebra 1668
*22. Multiple Choice What are the zeros for the function y = 4x2 + 28x - 72? CA 0, 4 B 0, -4
C 2, -9 D 2, -76
*23. Band The school band is performing a music concert. Tickets cost $3 for adults and $2 for students. In order to cover expenses, at least $200 worth of tickets must be sold. Write an inequality that describes the graph of this situation. 3x + 2y ≥ 200
x
y
O40-40
-40
2 units
3 units
24. Measurement Jesse bought a new glass window to go in his front room. The area of the window is (2 + 3x3 - 8x) square inches. The width is (x + 4) inches. What is the length? (3x2 - 12x + 40 - 158_
x + 4 ) inches
*25. Error Analysis Students were asked to write an inequality that results in a solid, vertical boundary line. Which student is correct? Explain the error.
Student A Student B
x ≤ 6 x < -2
26. Graph the inequality 4x + 5y ≥ -7.
27. Multi-Step Pedro biked 6 miles on dirt trails and 12 miles on the street. His biking rate on the dirt trails was 25% of what it was on the street. a. Write a simplified expression for Pedro’s total biking time. 9_
0.25x = 36_x
b. Analyze Explain how finding the simplified expression would change if Pedro’s biking rate on the trails was 50% of what it was on the streets.
28. Graph the function y - 3 = -x2 + 3.
29. A square has a side length s. The square grows larger and has a new area of s2 + 16s + 64. What is the new side length? s + 8
30. Write an equation where j is inversely proportional to m and n and directly proportional to p and q. j =
kpq_mn
(96)(96)
(97)(97)
3 43 4
(93)(93)
(97)(97)25. Student A; Sample: Student B wrote an inequality with a dashed vertical boundary line.
25. Student A; Sample: Student B wrote an inequality with a dashed vertical boundary line.
(97)(97)26.
2
x
y
O
4
2 4
-2
-2-4
-4
26.
2
x
y
O
4
2 4
-2
-2-4
-4
(90)(90)
27b. Sample: The expression for the street time, 12_
x , would be
multiplied by .5_.5
insteadof .25_
.25, making the
simplified expression
12_.5x
= 24_x
.
27b. Sample: The expression for the street time, 12_
x , would be
multiplied by .5_.5
insteadof .25_
.25, making the
simplified expression
12_.5x
= 24_x
.
(84)(84)
28.
x
y
O
4
2
2 4
-2
-4
28.
x
y
O
4
2
2 4
-2
-4
(83)(83)
(Inv 8)(Inv 8)
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LOOKING FORWARD
Solving rational equations will be further developed in other Saxon High School Math courses.
CHALLENGE
Have students apply the skills learned in this lesson to simplify each expression.
1. 15 ___ m2 - 2 __ m - 1
___________ 4 ___ m2 - 5 __ m + 4
-(m - 3)(m + 5)
______________
4m2 - 5m + 4
2. 1 - 12 ________
3b + 10 ______________
b - 8 ________ 3b + 10
1
_____ b + 4
668 Saxon Algebra 1
Problem 24
Guide the students by asking them the following questions.
“What is the fi rst step to perform in solving this problem?” Sample: Rewrite the terms in descending order of the exponents.
“What else must be done before dividing?” Sample: Write a 0x2 term as a placeholder so you can divide.
Problem 29
Extend the Problem
The square grows larger still and has a new area of s4 + 18s2 + 81. What is the new side length?(s2 + 9)
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Lesson 100 669
Solving Quadratic Equations by Graphing
Warm Up
100LESSON
1. Vocabulary The U-shaped curve that results from graphing a quadratic function is called a(n) . parabola
Evaluate each expression for the given values.
2. 3(x - y)2 - 4y2 for x = -5 and y = -2 11
3. -x2 - 3xy + y for x = 3 and y = -1 -1
Determine the direction that the parabola opens.
4. f (x) = 3x2 + x - 4 upward 5. f (x) = -2x2 + x + 1 downward
The solution(s) of a quadratic equation, 0 = ax2 + bx + c, can be found by graphing the related function, f(x) = ax2 + bx + c. The U-shaped graph of a quadratic function is called a parabola. The solutions of the equation are called roots and can be found by determining the x-intercepts or zeros of the quadratic function. These zeros can be found by graphing the related function to see where the parabola intersects the x-axis.
Graphical Solutions
One Real Solution
The graph intersects the x-axis at the vertex. x
y
O
4
2
2 4 6-2
Two Real Solutions
The graph intersects the x-axis at two distinct points.
x
y
O
4
2
4
-2
-4
No Real Solutions
The graph does not intersect the x-axis.
xy
O4 62
-2
-4
-6
(84)(84)
(9)(9)
(9)(9)
(84)(84) (84)(84)
New ConceptsNew Concepts
Online Connection
www.SaxonMathResources.com
Math Language
The same function is described by y = 3 x 2 - 5and f(x) = 3 x 2 - 5.
The function notation
for y is f(x). It is read, “f of x.”
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LESSON RESOURCES
Student Edition Practice Workbook 100
Reteaching Master 100Adaptations Master 100Challenge and Enrichment
Master C100Technology Lab Master 100
MATH BACKGROUND
To solve a quadratic equation, it is written in standard form, where the exponents are in descending order and the polynomial is set equal to zero: 0 = a x 2 + bx + c.
Quadratic equations may have one real solution (when the vertex of the parabola is on the x-axis), two real solutions (when the parabola crosses the x-axis), or no real solutions (when the parabola does not interesect the x-axis). The x-intercepts shown by the graph are the solutions.
Take the opportunity with this lesson, to tell or remind students that when any type of equation is solved, the solution is the x-coordinate of the point where the graph of the equation crosses the x-axis.
Students have already solved linear, quadratic, and absolute-value equations in their study of algebra. In Section 10, they will solve rational and radical equations. In all cases, the graphical interpretation of the solution, or roots, will be the x-intercept(s).
Warm Up1
100LESSON
Lesson 100 669
Problems 4 and 5
Briefl y review graphing quadratic functions in preparation for solving quadratic equations by graphing.
2 New Concepts
Students are already familiar with graphing and solving quadratic functions. Now they will solve quadratic equations by graphing.
Discuss the graphs in the table, and note the number of real solutions indicated by each type of graph. Have students identify the vertex and x-intercept(s) of each graph.
TEACHER TIPHelp students understand that the most important points of the quadratic function to graph are the x-intercepts and the vertex. The axis of symmetry is helpful to use in graphing other points on the parabola. The infi nite number of points remaining to be graphed can be approximated with a smooth curve.
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Saxon Algebra 1670
Example 1 Solving Quadratic Equations by Graphing
Solve each equation by graphing the related function.
a. x2 - 36 = 0
SOLUTION
Step 1: Find the axis of symmetry.
x = - b _ 2a
Use the formula.
x = - 0_
2(1) = 0 Substitute values for a and b.
The axis of symmetry is x = 0.
Step 2: Find the vertex.
f (x) = x2 - 36
f (0) = (0)2 - 36 Evaluate the function for x = 0 to find the vertex.
The vertex is (0, -36).
Step 3: Find the y-intercept.
The y-intercept is c, or -36.
Step 4: Find two more points that are not on the axis of symmetry.
f (5) = 52 - 36 f (7) = 72 - 36
(5, -11) (7, 13)
Step 5: Graph.
Graph the axis of symmetry x = 0, the vertex and the y-intercept, both at coordinate (0, -36). Reflect the points (5, -11) and (7, 13) over the axis of symmetry and graph the points (-5, -11) and (-7, 13). Connect the points with a smooth curve.
From the graph, the x-intercepts appear to be 6 and -6.
Check Substitute the values for x in the original equation.
x2 - 36 = 0; x = 6 x2 - 36 = 0; x = -6
(6)2 - 36 � 0 (-6)2 - 36 � 0
36 - 36 � 0 36 - 36 � 0
0 = 0 ✓ 0 = 0 ✓
The solutions are 6 and -6.
x
y
O4 8
-20
20
40
-40
-8
x
y
O4 8
-20
20
40
-40
-8
Hint
When the coefficient of the x 2 -term is positive, the parabola will open upward.
When the coefficient of the x 2 -term is negative, the parabola will open downward.
Math Reasoning
Write Why are the x-intercepts substituted into the original equation?
The x-intercepts of the function are roots, or solutions, of the equation.
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ENGLISH LEARNERS
For Example 1, explain the meaning of the word related. Say:
“Related means connected with. For example, we are related to the members of our family.”
Have students give other examples using the word related in a sentence. Sample: The notes are related by the key signature of the music.
670 Saxon Algebra 1
Example 1
These examples introduce the concept of solving a simple quadratic equation through graphing the related function.
Error Alert Some students may locate points on only one side of the parabola. Encourage them to use the axis of symmetry to plot points on both sides of the parabola.
Additional Example 1
Solve each equation by graphing the related function.
a. x 2 - 25 = 0 The solutions are -5 and 5.
10
x
y
O8-8
-10
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Lesson 100 671
b. -x2 - 2 = 0
SOLUTION
Graph the related function f(x) = -x2 - 2.
axis of symmetry: x = 0
vertex: (0, -2)
y-intercept: (0, -2)
two additional points: (1, -3) and (3, -11)
Reflect these two points across the axis of symmetry and connect them with a smooth curve.
From the graph, it can be seen that there is no x-intercept because the graph does not intersect the x-axis.
There is no real-number solution.
c. x2 + 16 = 8x
SOLUTION
Write the equation in standard form.
x2 - 8x + 16 = 0
Graph the related function f(x) = x2 - 8x + 16.
axis of symmetry: x = 4
vertex: (4, 0)
y-intercept: (0, 16)
two additional points: (2, 4) and (3, 1)
Reflect these two points across the axis of symmetry and connect them with a smooth curve.
From the graph, the x-intercept appears to be 4.
Check Substitute 4 for x in the original equation.
x2 - 8x + 16 = 0; x = 4
(4)2 - 8(4) + 16 � 0
16 - 32 + 16 � 0
0 = 0 ✓
The solution is 4.
xy
O2 4
-4
-8
-12
-4 -2
xy
O2 4
-4
-8
-12
-4 -2
x
y
O
8
12
16
4
2 6 8
x
y
O
8
12
16
4
2 6 8
Caution
When a parabola does not cross the x-axis, there is no real-number solution to the quadratic equation.
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Lesson 100 671
b. 0 = - x 2 - 3 There are no real-number solutions.
x
y
O
2
2 4
-2
-2-4
-6
c. x 2 -3x -18 = 0 The solutions are 6 and -3.
x
y
O
-20
20
10
4 8-4-8
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Saxon Algebra 1672
Example 2 Solving Quadratic Equations Using a Graphing
Calculator
Solve each equation by graphing the related function on a graphing calculator.
a. -6x - 9 = x2
SOLUTION
Write the equation in standard form.
-x2 - 6x - 9 = 0
Graph the related function
f (x) = -x2 - 6x - 9.
The graph appears to have an x-intercept at -3.
Use the Table function to determine the zeros of this function.
The solution is -3.
b. -6x = -x2 - 13
SOLUTION
Write the equation in standard form.
x2 - 6x + 13 = 0.
Graph the related function f (x) = x2 - 6x + 13.
The graph opens upward and does not intersect the x-axis.
There is no solution.
c. -3x2 + 5x = -7
Round to the nearest tenth.
SOLUTION
Write the equation in standard form.
-3x2 + 5x + 7 = 0
Graph the related function f (x) = -3x2 + 5x + 7.
The graph appears to have x-intercepts at 3 and -1.
Use the Zero function to determine the zeros of this function. Round to the nearest tenth.
The solutions are x = 2.6 and -0.9.
Hint
For help with graphing quadratic functions, see Graphing Calculator Lab 8: Characteristics of Parabolas on p. 583.
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INCLUSION
Use the following strategy with students who have diffi culty with abstract concept processing. Have students practice making tables in preparation for drawing graphs of quadratic equations. Help them decide which numbers to include in their tables. Have them fi nd the vertex fi rst using -b
___ 2a . They should understand that since a parabola is symmetrical, their list of numbers should also be symmetrical about the axis of symmetry.
Students should be able to fi nd the x-intercepts from the table. Any value of x for which the corresponding value of y is 0, is an x-intercept and thus a solution.
Once students have mastered making the table for the graph, help them plot the points and draw a smooth curve.
672 Saxon Algebra 1
Example 2
Once students have graphed quadratic functions by hand, the graphing calculator is a useful tool for quickly graphing and fi nding the solution(s) of quadratic equations.
Additional Example 2
Solve each equation by graphing the related function on a graphing calculator.
a. x 2 + 28 = 11xThe solutions are 4 and 7. See Additional Answers.
b. x 2 + 2x + 7 = 0There is no solution. See Additional Answers.
c. 2 x 2 - 7x + 3 = 0The solutions are 3 and 0.5. See Additional Answers.
Example 3
The example provides a framework for a real-world application of solving quadratic equations by graphing.
Extend the Example
Suppose that Gill drops the baseball off of a platform twice the height of the original platform. How much longer does it take the baseball to hit the ground? about 0.8 second
Additional Example 3
Wendy dropped a golf ball from the top of a platform 102 feet off the ground. The height of the golf ball is described by the quadratic equation h = -16 t 2 + 102, where h is the height in feet and t is the time in seconds. Find the time t when the golf ball hits the ground. t ≈ 2.52 seconds
t
h
O
40
60
80
20
2 4-4
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Practice Distributed and Integrated
Lesson 100 673
Example 3 Application: Physics
Gill drops a baseball from the top of a platform 64 feet off the ground. The height of the baseball is described by the quadratic equation h = -16t2 + 64, where h is the height in feet and t is the time in seconds. Find the time t when the ball hits the ground.
SOLUTION
Graph the related function h(t) = -16t2 + 64 on a graphing calculator.
Height h is zero when the ball hits the ground. Use the Zero function of the graphing calculator to determine the zeros of this function.
There are two zeros for the given parabola: t = 2 and t = -2. Only values greater than or equal to zero are considered. So, t = 2 is the only solution.
The baseball hits the ground in 2 seconds.
Lesson Practice
Solve each equation by graphing the related function.
a. 3x2 - 147 = 0 x = 7 and x = -7
b. 5x2 + 6 = 0 no solution
c. x2 - 10x + 25 = 0 x = 5
Solve each equation by graphing the related function on a graphing calculator.
d. x2 + 64 = 16x x = 8
e. x2 + 4 = 2x no solution
f. Round to the nearest tenth: -7x2 + 3x = -7. x = -0.8 and 1.2
g. Marcus shot an arrow while standing on a platform. The path of its movement formed a parabola given by the quadratic equation h = -16t2 + 2t + 17, where h is the height in feet and t is the time in seconds. Find the time t when the arrow hits the ground. Round to the nearest hundredth. t = 1.10 seconds
(Ex 1)(Ex 1)
(Ex 2)(Ex 2)
(Ex 2)(Ex 2)
(Ex 2)(Ex 2)
(Ex 3)(Ex 3)
Solve.
1. x(2x - 11) = 0 0, 11_2
2. 12 _ x - 6
= 4 _ x
x = -3
*3. Generalize Using the path of a ball thrown into the air as an example, describe in mathematical terms each part of the graph the path of the ball creates.
*4. Generalize What does the graph of a quadratic equation look like when there is no solution? one solution? two solutions? See Additional Answers.
(98)(98) (99)(99)
(100)(100)See Additional Answers.See Additional Answers.
(100)(100)
Hint
The time t is plotted on the x-axis. The height h is plotted on the y-axis.
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Lesson 100 673
Lesson Practice
Problem a
Scaff olding To locate points on the parabola, fi rst have students fi nd the vertex and axis of symmetry. Then have them fi nd two x-values to the left or right of the vertex and refl ect them over the axis of symmetry.
Problem d
Error Alert Some students may not set the equation equal to zero. Remind them to write the equations in standard form.
Check for Understanding
The questions below help assess the concepts taught in this lesson.
“Where on the graph of a quadratic equation are the solutions located?” Sample: where the graph intersects the x-axis
“What is another name for the solutions of a quadratic equation?” roots or x-intercepts
Practice3
Math ConversationsDiscussion to strengthen understanding
Problem 3
Guide the students by asking them the following questions.
“What is the name of the graph of a quadratic equation?” parabola
“What is the highest or lowest point on the graph called?” vertex
“What are the points where the graph crosses the horizontal axis?” x-intercepts or solutions
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Saxon Algebra 1674
5. Given that y varies directly with x, identify the constant of variation such that when x = 15, y = 30. k = 2
*6. Basketball Ramero shoots a basketball into the air. The ball’s movement forms a parabola given by the quadratic equation h = -16t2 + 7t + 7, where h is the height in feet and t is the time in seconds. Find the maximum height of the path the basketball makes and the time t when the basketball hits the ground. Round to the nearest hundredth. h = 7.77 feet and t = 0.92 seconds
*7. Multiple Choice What is the equation of the axis of symmetry of the parabola defined by y = 1 _
4 (x - 4)2 + 5? B
A x = 1 B x = 4 C x = 5 D x = -4
*8. Solve -7x2 - 10 = 0 by graphing. no solution
*9. Solve 6 _ x
= 8 _
x + 7 . x = 21
10. A deck of ten cards has 5 red and 5 black cards. Cards are replaced in the deck after each draw. Use an equation to find the probability of drawing a black card twice and rolling a 6 on a number cube. P(black, black, 6) = ( 1_
2 )2
· 1_6
= 1_24
11. Geometry The altitude of the right triangle divides the hypotenuse into segments of lengths x units and 5 units. To find x, solve the equation x + 5
_ 6 = 6 _
x . 4 units
*12. Multi-Step Henry starts working a half-hour before Martha. He can complete the job in 4 hours. Martha can complete the same job in 3 hours. a. Let t represent the total time they work together. In terms of t, how long does
Henry work? t + 0.5
b. Use an equation to find how long they work together to complete the job. 1 hour 30 minutes
c. How long does Henry work? 2 hours
13. Find the quotient of a2 + 10a - 24 __
a - 2 . 14. Simplify √ �� 49y5 . 7y2
√ � y
15. Profit An entrepreneur makes $3 profit on each object sold. She would like to make $270 plus or minus $30 total. What is the minimum and maximum number of objects she needs to sell? 80 objects,100 objects
16. Data Analysis A student knows there will be 4 tests that determine her semester grade. She wants her average to be an 85, plus or minus 5 points. What is the minimum and maximum number of points she needs to earn during the semester? 320 points, 360 points
17. Solve the equation �10x� - 3 = 87. {-9, 9}
18. Exercise Tom ran a total of 7x _ x2 + 3x - 18
miles in August and 2x + 1
_ 7x + 42
miles in
September. How many more miles did he run in August?
19. Graph the function y = 5x2 - 10x + 5.
*20. Verify A boundary line is a vertical line. The inequality contains a < symbol. Which half-plane should be shaded on the graph? Sample: Shade the half-plane to the left of the vertical line.
(Inv 7)(Inv 7)
(100)(100)
(100)(100)
(100)(100)
(99)(99)
(80)(80)
(99)(99) 5
x6
5
x6(99)(99)
(93)(93) (61)(61)(a + 12)(a + 12)
(94)(94)
(94)(94)
(94)(94)
(95)(95)
-2x2 + 54x + 3__7(x - 3)(x + 6)
miles -2x2 + 54x + 3__7(x - 3)(x + 6)
miles
(96)(96)
19.
x
y
O
28
2 4-2-4
19.
x
y
O
28
2 4-2-4
(97)(97)
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CHALLENGE
Graph to fi nd the ordered pair that is a solution to all the equations.
2 x 2 + 3x - 5 = 0
3x - 2y = 3
4 x 2 + 3x - 4 = 3(1, 0)
674 Saxon Algebra 1
Problem 12
Extend the Problem
Suppose Martha and Henry work the same job, but that this time, Martha begins one hour before Henry starts. How long do they work together? How long does Martha work? 1 1 __ 7 hours; 2 1 __ 7 hours
Problem 15
Error Alert Make sure that students answer the question asked: the range of the number of objects she needs to sell, not the range of the amount of money she wants to make.
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Lesson 100 675
21. Multiple Choice Which point does not satisfy the inequality x + 2y < 5? DA (0, 0) B (2, 1) C (3, -4) D (-1, 3)
*22. Ages A boy is b years old. His father is 23 years older than the boy. The product of their ages is 50. How old is each person? The boy is 2 years old and the father is 25 years old.
*23. Error Analysis Two students find the roots of 3x2 - 6x = 24. Which student is correct? Explain the error.
Student A Student B
3x2 - 6x = 24 3x2 - 6x = 243x(x - 2) = 24 3x2 - 6x - 24 = 03x = 0 x - 2 = 0 3(x2 - 2x - 8) = 0
x = 0 x = 2 3(x - 4)(x + 2) = 0x - 4 = 0 x + 2 = 0
x = 4 x = -2
24. Does the graph of y + 2x2 = 12 + x open upward or downward? downward
25. Do the side lengths 18, 80, and 82 form a Pythagorean triple? yes
26. Multi-Step The volume of a prism is 3x3 + 12x2 + 9x. What are the possible dimensions of the prism? a. Factor out common terms. 3x( x2 + 4x + 3)
b. Factor completely. 3x(x + 1)(x + 3)
c. Find the dimensions. 3x, x + 1, x + 3
27. Travel The Jackson family drove 480 miles on Saturday and 300 miles on Sunday. Their average rate on Sunday was 10 miles per hour less than their rate was on Saturday. Write a simplified expression that represents their total driving time.
28. Multi-Step At the carnival, a man says that he will guess your weight within 5 pounds. a. You weigh 120 pounds. Write an absolute-value inequality to show the range of
acceptable guesses. �x - 120� ≤ 5
b. Solve the inequality to find the actual range of acceptable guesses. 115 ≤ x ≤ 125
29. Verify If the numerator of a rational expression is a polynomial and the denominator of the rational expression is a different polynomial, will factoring the polynomials always provide a way to simplify the expression? Verify your answer by giving an example.
30. If a 9% decrease from the original price resulted in a new price of $227,500, what was the original price? $250,000
(97)(97)
(98)(98)
(98)(98)23. Student B; Sample: Student A did not put the equation in standard form before factoring.
23. Student B; Sample: Student A did not put the equation in standard form before factoring.
(84)(84)
(85)(85)
(87)(87)
(90)(90)
60(13x - 80) _
x(x - 10)
60(13x - 80) _
x(x - 10)
(91)(91)
(92)(92)29. no; Sample: If there are no common factors, the expression is in the simplest form; x2 - 4__
2x2 + 12x + 18=
(x - 2)(x + 2)
__2(x + 3)(x + 3)
29. no; Sample: If there are no common factors, the expression is in the simplest form; x2 - 4__
2x2 + 12x + 18=
(x - 2)(x + 2)
__2(x + 3)(x + 3)
(47)(47)
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LOOKING FORWARD
Solving quadratic equations by graphing prepares students for
• Lesson 102 Solving Quadratic Equations Using Square Roots
• Lesson 104 Solving Quadratic Equations by Completing the Square
• Lesson 110 Using the Quadratic Formula
• Lesson 113 Interpreting the Discriminant
Lesson 100 675
Problem 21
Guide the students by asking them the following questions.
“Which choice can you eliminate immediately? Why?” Sample: choice A; The result of any calculations with two zeros will be less than 5.
“Which choice would you eliminate next? Why?” Sample: choice C; 3 + 2(-4) is a lot less than 5.
“What is the correct choice?” Sample: choice D; It is equal to 5, which means it is not less than 5.
Problem 25
Error Alert Some students may believe that only multiples of the most common Pythagorean triple—3, 4, and 5—qualify as other Pythagorean triples. Remind students that any set of whole numbers that makes the Pythagorean Theorem true is considered a Pythagorean triple.
Problem 27
Guide the students by asking them the following questions.
“What formula do you have to use to solve the problem?” distance equals rate times time: d = rt
“How can you use this formula to fi nd the driving time?” Rewrite the formula so that the time is isolated on one side of the equation: d __ r = t.
“What LCD will you use to simplify the equation?” (x)(x - 10)
Problem 28
Extend the Problem
After looking at you, the man believes you weigh between 123 and 135 pounds. What is the probability of him choosing a weight from this range that is within 5 pounds of your actual weight? 3
___ 13
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10INVESTIGATION
Saxon Algebra 1676
A quadratic function is a function that can be written in the form f(x) = ax2 + bx + c, where a is not equal to 0.
In Investigation 6, linear functions were graphed as transformations of the parent function f(x) = x. Similarly, you can graph a quadratic function as a transformation of the quadratic parent function f(x) = x2.
Parameter Changes
Complete the table of values for f(x) = x2 and graph the quadratic parent function.
x f (x)
-3 (-3)2 = 9
-2 (-2)2 = 4
-1 (-1)2 = 1
0 02 = 0
1 12 = 1
2 22 = 4
3 32 = 9
As is the case for the linear parent function, the quadratic parent function f(x) = x2 can be written as f(x) = ax2, where a = 1. The graph changes when other values are substituted for a.
1. Graph y = x2 and y = 2x2 on the same set of axes. Compare the two graphs. See Additional Answers.
2. Graph y = x2 and y = 1 _ 2 x2 on the same set of axes. Compare the
two graphs. See Additional Answers.
3. Graph y = x2 and y = -x2 on the same set of axes. Compare the two graphs. See Additional Answers.
4. Generalize What is the effect of a on the graph of y = ax2?
5. Predict How will the graph of f(x) = 2 _ 3 x2 change in relation to the
quadratic parent function? Sample: It’s wider than the parent function.
6. Predict How will the graph of f(x) = -4x2 change in relation to the quadratic parent function?
The graph of a function of the form f(x) = ax2 always crosses the y-axis at (0, 0). When c ≠ 0, the graph of the function f(x) = ax2 + c does not pass through the point (0, 0).
7. Graph the quadratic parent function and the function f(x) = x2 + 1 on the same set of axes. Compare the two graphs. See Additional Answers.
8. Graph the quadratic parent function and the function f(x) = x2 - 2 on the same set of axes. See Additional Answers.
x
y
4
6
2
2 4-2-4
x
y
4
6
2
2 4-2-4
See Additional Answers.See Additional Answers.
Sample: It’s narrower than the parent function and opens downward.Sample: It’s narrower than the parent function and opens downward.
Transforming Quadratic Functions
Online Connection
www.SaxonMathResources.com
Math Reasoning
Analyze If a = 0, would the function still be quadratic? If not, what type of function would f(x) be? no; linear
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INVESTIGATION RESOURCES
Reteaching Master Investigation 10
Technology Lab Master Investigation 10
MATH BACKGROUND
Understanding the connection between equations and graphs is a critical component to understanding functions. Identifying the parent function as a control allows comparisons. By comparing the graphs of quadratic functions with different parameters, the impact of the parameters in the equation on the graphs can be determined.
Specifi cally, in quadratic equations, a determines how wide or narrow the graph is. The constant term, c, determines the vertical shift of the graph.
676 Saxon Algebra 1
10INVESTIGATION
Materials
• graphing calculator• graph paper
Discuss
In this lesson, students learn the effects of parameter changes in quadratic functions. They graph the parent function and other transformations on the same set of axes to see the changes in the graphs.
Defi ne parent function and quadratic function.
Parameter Changes
Extend the Problem
Have students write equations for quadratic functions given a description of the graphs (i.e. opens down, does not cross the x-axis, and is narrower than the parent function).
Error AlertStudents may want to say that when 0 < a < 1, the graph is narrower than the parent function because the coeffi cient is small. Have students clarify that the greater a is, the narrower the graph.
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Investigation 10 677
9. Predict How will the graph of f(x) = x2 + 7 compare to the graph of the quadratic parent function? Sample: The new function moved up 7 units from the parent function.
Combinations of Parameter Changes
Predict How will each graph compare to the graph of the quadratic parent function? Verify your answer with a graphing calculator.
10. f(x) = -x2 + 2 See Additional Answers.
11. f(x) =1_2
x2 - 3 See Additional Answers.
Investigation Practice
Describe how the graph for the given values of a and c changes in relation to the graph of the quadratic parent function. Verify your answer with a graphing calculator.
a. f(x) = ax2 + c for a = 2 and c = 1 The graph is narrower and moved 1 unit up.
b. f(x) = ax2 + c for a = -3 and c = -2 The graph is narrower, opens downward, and moved 2 units down.
c. f(x) = ax2 + c for a =1_2 and c = 2 The graph is wider and moved 2 units up.
d. f(x) = ax2 + c for a = - 1_2
and c = -1 The graph opens downward, is wider, and moved 1 unit down.
Write an equation for the transformation described. Then graph the original function and the graph of the transformation on the same set of axes.
e. Shift f(x) = 2x2 - 4 up 2 units.
f. Shift f(x) = 3x2 + 5 down 4 units and open it downward. f(x) = -3x2 + 1;
x
4
4 8-4-8
3x2+ 5
3x2+ 1
e. f(x) = 2x2 - 2;
x
y8
4
2 4-2-4
-8
2x2 2
2x2 4
e. f(x) = 2x2 - 2;
x
y8
4
2 4-2-4
-8
2x2 2
2x2 4
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Discuss that children inherit some traits from their parents, but not all of them. Similarly, explain that a parent function has characteristics of any function of that type, but that other functions may look a little different. There are mathematical hints in the equation that indicate how different a transformed function will look.
Transforming quadratic functions will prepare students for
• Lesson 107 Graphing Absolute-Value Functions
• Lesson 114 Graphing Square-Root Functions
• Lesson 119 Graphing and Comparing Linear, Quadratic, and Exponential Functions
Investigation 10 677
Discuss
The effect of each parameter in the equation on the graph of the function has been identifi ed. Students will describe the transformations in the graph when multiple parameters are changed.
Investigation Practice
Math ConversationsDiscussion to strengthen understanding
Problem a
Scaff olding Have students tell whether the function is wider or narrower, opens upward or downward, and moves up or down.
Problem b
Error Alert Students may forget the effect of the negative value of a. Remind students that a parabola opens down when a < 0.
Problems e and f
Guide the students by asking them the following questions.
“What value makes the graph move up or down?” c
“To move it down, will you add or subtract c?” subtract
“How do you make a parabola open downward?” Make a negative.
ENGLISH LEARNERS LOOKING FORWARD
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