Quadratic Functions and Equations

Quadratic Graphs and Their Properties

Objective: To graph quadratic functions of the form y = ax2 and y = ax2 + c.

Objectives

• I can identify a vertex.

• I can grapy 𝑦 = 𝑎𝑥2.

• I can compare widths of parabolas.

• I can graph 𝑦 = 𝑎𝑥2 + 𝑐.

• I can use the falling object model.

Vocabulary• A quadratic function is a type of nonlinear function that models certain

situations where the rate of change is not constant.

• The graph of a quadratic function is a symmetric curve with the highest or lowest point corresponding to a maximum or minimum value.

• Standard Form of a Quadratic Function:• A quadratic function is a function that can be written in the form 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐,

where a ≠ 0. This form is called the standard form of a quadratic function.

• Examples: 𝑦 = 3𝑥2 𝑦 = 𝑥2 + 9 𝑦 = 𝑥2 − 𝑥 − 2

Vocabulary

• The simplest quadratic function 𝑓 𝑥 = 𝑥2 𝑜𝑟 𝑦 = 𝑥2 is the quadratic parent function.

• The graph of a quadratic function is a U-shaped curve called a parabola.

• You can fold a parabola so that the two sides match exactly. This property is called symmetry.

• The fold or line that divides the parabola into two matching halves is called the axis of symmetry.

Vocabulary

• The highest or lowest point of a parabola is its vertex, what is on the axis of symmetry.

• If a > 0 in y = ax2 + bx + c, the parabola opens upward. The vertex is the minimum point, or lowest point, of the parabola.

• If a < 0 in y = ax2 + bx + c , the parabola opens downward. The vertex is the maximum point, or highest point, of the parabola.

Identifying a Vertex

PracticeIdentify the vertex.

Practice

Vocabulary

• You can use the fact that a parabola is symmetric to graph it quickly.

• First, find the coordinates of the vertex and several points on one side of the vertex.

• Then reflect the points across the axis of symmetry.

• For graphs of functions of the form y = 𝑎𝑥2, the vertex is at the origin.

• The axis of symmetry is the y–axis, or x = 0.

Graphing 𝑦 = 𝑎𝑥2

Graph the function. Make a table of values. What are the domain and range?

1. 𝑦 =1

3𝑥2

2. 𝑦 = −3𝑥2

3. 𝑦 = 4𝑥2

Practice

Graph each function. Then identify the domain and range of the function.

1. 𝑦 = −4𝑥2

2. 𝑦 = −1

3𝑥2

3. 𝑓 𝑥 = 1.5𝑥2

4. 𝑓 𝑥 =2

3𝑥2

Vocabulary

• The coefficient of the x2–term in a quadratic function affects the width of the parabola as well as the direction in which it opens.

• When 𝑚 < 𝑛 , the graph of 𝑦 = 𝑚𝑥2 is wider than the graph of 𝑦 = 𝑛𝑥2.

Comparing Widths of Parabolas

Use your calculator to graph. What is the order, from widest to narrowest, of the graphs of the quadratic functions.

𝑓 𝑥 = −4𝑥2, 𝑓 𝑥 =1

4𝑥2, 𝑓(𝑥) = 𝑥2

𝑓 𝑥 = −𝑥2, 𝑓 𝑥 = 3𝑥2, 𝑓 𝑥 = −1

3𝑥2

Practice

Order each group of quadratic functions from widest to narrowest graph.• 𝑦 = 3𝑥2, 𝑦 = 2𝑥2, 𝑦 = 4𝑥2

• 𝑦 = −1

2𝑥2, 𝑦 = 5𝑥2, 𝑦 = −

1

4𝑥2

• 𝑓 𝑥 = 5𝑥2, 𝑓 𝑥 = −3𝑥2, 𝑓 𝑥 = 𝑥2

• 𝑓 𝑥 = −2𝑥2, 𝑓 𝑥 = −2

3𝑥2, 𝑓 𝑥 = −4𝑥2

Vocabulary

• The y – axis is the axis of symmetry for graphs of functions of the form 𝑦 = 𝑎𝑥2 + 𝑐.

• The value of c translates the graph up or down.

Graphing 𝑦 = 𝑎𝑥2 + 𝑐

What is the relationship of the following graphs?1. 𝑦 = 2𝑥2 + 3 𝑎𝑛𝑑 𝑦 = 2𝑥2

2. 𝑦 = 𝑥2 𝑎𝑛𝑑 𝑦 = 𝑥2 − 3

3. 𝑦 = −1

2𝑥2 𝑎𝑛𝑑 𝑦 = −

1

2𝑥2 + 1

Practice

Graph each function.1. 𝑓 𝑥 = 𝑥2 + 4

2. 𝑓 𝑥 = −𝑥2 − 3

3. 𝑓 𝑥 =1

2𝑥2 + 2

Vocabulary

As an object falls, its speed continues to increase, so its height above the ground decreases at a faster and faster rate.

Ignoring air resistance, you can model the object’s height with the function h = –16t2 + c.

The height h is in feet, the time t is in seconds, and the object’s initial height c is in feet.

Using the Falling Object Model

An acorn drops from a tree branch 20 feet above the ground. The function h = –16t2 + 20 gives the height h of the acorn (in feet) after t seconds. What is the graph of this quadratic function? At what time does the the acorn hit the ground? t h = –16t2 + 20

Using the Falling Object Model

An acorn drops from a tree branch 70 feet above the ground. The function h = –16t2 + 70 gives the height h of the acorn (in feet) after t seconds. What is the graph of this quadratic function? At what time does the the acorn hit the ground? t h = –16t2 + 70

Practice

A person walking across a bridge accidentally drops an orange into the river below from a height of 40 feet. The function ℎ = −16𝑡2 + 40 gives the orange’s approximate height h above the water, in feet, after t seconds. In how many seconds will the orange hit the water?

A bird drops a stick to the ground from a height of 80 feet. The function ℎ =−16𝑡2 + 80 gives the stick’s approximate height h above the ground, in feet, after t seconds. Graph the function. At about what time does the stick hit the ground?

Quadratic Functions

Objective: To graph quadratic functions of the form y = ax2 + bx + c.

Objectives

• I can graph y = 𝑎𝑥2 + 𝑏𝑥 + 𝑐.

• I can use the vertical motion model.

Vocabulary• In the quadratic function y = ax2 + bx + c, the value of b affects the position of the axis of

symmetry.

• The axis of symmetry changes with each equation because of the change in the b-value. The equation of the axis of symmetry is related to the ratio

𝑏

𝑎.

• The equation of the axis of symmetry is 𝑥 = −1

2

𝑏

𝑎𝑜𝑟 𝑥 =

−𝑏

2𝑎.

• Graph of a Quadratic Function

o The graph of y = ax2 + bx + c, where a ≠ 0, has the line 𝑥 =−𝑏

2𝑎as its axis of symmetry.

The x–coordinate of the vertex is −𝑏

2𝑎.

Vocabulary

• When you substitute x = 0 into the equation y = ax2 + bx + c, you get y = c. So the y–intercept of a quadratic function is c.

• You can use the axis of symmetry and the y–intercept to help you graph a quadratic function.

Graphing y = ax2 + bx + c

What is the graph of the function? Show the axis of symmetry.1. 𝑦 = 𝑥2 − 6𝑥 + 4

2. 𝑦 = −𝑥2 + 4𝑥 − 2

3. 𝑦 = 2𝑥2 + 3

4. 𝑦 = −3𝑥2 + 12𝑥 + 1

5. 𝑓 𝑥 = 𝑥2 + 4𝑥 − 5

6. 𝑓 𝑥 = −4𝑥2 + 11

Practice

What is the graph of the function? Show the line of symmetry. 1. 𝑦 = 2𝑥2 − 6𝑥 + 1

2. 𝑓 𝑥 = 2𝑥2 + 4𝑥 − 1

3. 𝑦 = 6𝑥2 + 6𝑥 − 5

4. 𝑓 𝑥 = −5𝑥2 + 3𝑥 + 2

5. 𝑦 = −2𝑥2 − 10𝑥

6. 𝑦 = −4𝑥2 − 16𝑥 − 3

Vocabulary

• You have used h = –16t2 + c to find the height h above the ground of an object falling from an initial height c at time t.

• If an object projected into the air given an initial upward velocity vcontinues with no additional force of its own, the formula h = –16t2 + vt + c givens its approximate height above the ground.

Using a Vertical Motion Model

During halftime of a basketball game, a sling shot launches T–shirts at the crowd. A T–shirt launched with an initial upward velocity of 72 feet per second. The T–shirt is caught 35 feet above the court. The T–shirt is launched from a height of 5 feet.

a. How long will it take the T–shirt to reach its maximum height?

b. What is the maximum height?

c. What is the range of the function that models the height of the T–shirt over time?

Using a Vertical Motion Model

During halftime of a basketball game, a sling shot launches T–shirts at the crowd. A T–shirt launched with an initial upward velocity of 64 feet per second. The T–shirt is caught 35 feet above the court. The T–shirt is launched from a height of 5 feet.

a. How long will it take the T–shirt to reach its maximum height?

b. What is the maximum height?

c. What is the range of the function that models the height of the T–shirt over time?

Practice

A baseball is thrown into the air with an upward velocity of 30 feet per second. Its height h, in feet, after t seconds is given by the function h = –16t2 + 30t + 6.

a. How long will it take the ball to reach its maximum height?

b. What is the ball’s maximum height?

c. What is the range of the function?

Solving Quadratic Equations

Objective: To solve quadratic equations by graphing and using square roots.

Objectives

• I can solve by graphing.

• I can solve using square roots.

• I can choose a reasonable solution.

Vocabulary• Standard Form of a Quadratic Equation:

A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where a ≠ 0. This form is called the standard form of a quadratic equation.

• Quadratic equations can be solved by a variety of methods, including graphing and finding square roots.

• One way to solve a quadratic equation ax2 + bx + c = 0 is to graph the related quadratic function y = ax2 +bx + c. The solutions of the equation are the x – intercepts of the related function.

Vocabulary

• A quadratic equation can have two, one, or no real-number solutions.

• The solutions of a quadratic equation and the x–intercepts of the graph of the related function are often called roots of the equation or zeros of the function.

Solving by Graphing

What are the solutions of each equation? Use a graph of the related function.

1. x2 – 1 = 0

2. x2 = 0

3. x2 + 1 = 0

4. x2 – 16 = 0

5. 3x2 + 6 = 0

6. x2 – 25 = –25

PracticeSolve each equation by graphing the related function. If the equation has no real–number solution, write no solution.

1. 𝑥2 − 9 = 0

2. 3𝑥2 = 0

3.1

3𝑥2 − 3 = 0

4. 𝑥2 + 5 = 5

5. 𝑥2 − 10 = −10

6. 2𝑥2 − 18 = 0

Vocabulary

• You can solve equations of the form x2 = k by finding the square roots of each side.

Solving Using Square Roots

What are the solutions?

1. 3x2 – 75 = 0

2. m2 – 36 = 0

3. 3x2 + 15 = 0

4. 4d2 + 16 = 16

5. t2 = 144

6. y2 − 225 = 0

7. x2 − 25 = 0

8. 2x2 − 8 = 0

Practice

Solve each equation by finding square roots. If the question has no real–number solution, write no solution.

1. 𝑛2 = 81

2. 𝑤2 − 36 = −64

3. 64𝑏2 = 16

4. 5𝑞2 − 20 = 0

5. 144 − 𝑝2 = 0

6. 3𝑎2 + 12 = 0

7. 𝑟2 + 49 = 49

8. 𝑘2 − 196 = 0

Vocabulary

• You can solve some quadratic equations that model real – world problems by finding square roots.

• In many cases, the negative square root may not be a reasonable solution.

Choosing a Reasonable Solution

1. An aquarium is designing a new exhibit to showcase tropical fish. The exhibit will include a tank that is a rectangular prism with a length l that is twice the width w. The volume of the tank is 420 ft3. What is the width of the tank to the nearest tenth of a foot? (𝑉 = 𝑙𝑤ℎ), h = 3 ft

2. Suppose that the tank has a height of 4 feet and a volume of 500 ft3. What is the width of the tank to the nearest tenth of a foot.

Practice

• You have enough paint to cover an area of 50 ft². What is the side length of the largest square that you could paint? Round your answer to the nearest tenth of a foot.

• Find the length of a square with an area of 75 ft². Round to your answer to the nearest tenth of a foot.

Factoring to Solve Quadratic Equations

Objective: To solve quadratic equations by factoring.

Objectives

• I can use the zero-product property.

• I can solve by factoring.

• I can write in standard form first.

• I can use factoring to solve real-world problems.

Vocabulary

• In the previous lesson, you solved quadratic equations ax2 + bx + c = 0 by finding square roots. This method works if b = 0.

• You can solve some quadratic equations, including equations where b ≠ 0, by using the Zero–Product Property.

• The Multiplication Property of Zero states that for any real number a, a × 0 = 0. This is equivalent to the following statement: For any real numbers a and b, if a = 0 or b = 0, then ab = 0.

• The Zero–Product Property reverses this statement.

Vocabulary

Zero–Product Property

• For any real numbers a and b, if ab = 0, then a = 0 or b = 0.

• Example: If (x + 3)(x + 2) = 0, then x + 3 = 0 or x + 2 = 0.

Using the Zero–Product Property

What are the solutions of the equation?

1. (4t + 1)(t – 2) = 0

2. (x + 1)(x – 5) = 0

3. (2x + 3)(x – 4) = 0

4. (2y + 1)(y + 14) = 0

5. (7n – 2)(5n – 4) = 0

6. (v – 4)(v – 7) = 0

Practice

Use the Zero – Product Property to solve each equation.1. (x – 9)(x – 8) = 0

2. (4k + 5)(k + 7) = 0

3. n(n + 2) = 0

4. –3n(2n – 5) = 0

5. (7x + 2)(5x – 4) = 0

6. (4a – 7)(3a + 8) = 0

Vocabulary

• You can also use the Zero–Product Property to solve equations of the form ax2 + bx + c = 0 if the quadratic expression ax2 + bx + c can be factored.

Solving by Factoring

What is the solutions of the equations?

1. x2 + 8x + 15 = 0

2. m2 – 5m – 14 = 0

3. p2 + p – 20 = 0

4. 2a2 – 15a + 18 = 0

5. t2 + 3t – 54 = 0

6. 3y2 – 17y + 24 = 0

Practice

Solve by factoring.1. 𝑥2 + 11𝑥 + 10 = 0

2. 𝑔2 + 4𝑔 − 32 = 0

3. 3𝑞2 + 𝑞 − 14 = 0

4. 𝑝2 − 4𝑝 = 21

5. 2𝑤2 − 11𝑤 = −12

6. 16𝑏2 = 81

Vocabulary

• Before solving a quadratic equation, you may need to add or subtract terms from each side in order to write the equation in standard form.

• Then factor the quadratic expression.

Writing in Standard Form First

What are the solutions?

1. 4x2 – 21x = 18

2. x2 + 14x = –49

3. p2 – 4p = 21

4. 2w2 – 11w = –12

5. 3h2 + 17h = –10

6. 9b2 = 16

Practice

• Solve by factoring.1. 𝑥2 + 13𝑥 = −42

2. 𝑐2 = 5𝑐

3. 𝑡2 = −3𝑡 + 54

4. 3𝑦2 − 17𝑦 = −24

5. 7𝑛2 + 16𝑛 + 15 = 2𝑛2 + 3

6. 4𝑞2 + 3𝑞 = 3𝑞2 − 4𝑞 + 18

Using Factoring to Solve a Real–World Problem

You are constructing a frame for a rectangular photo (17 in by 11 in).You want the frame to be the same width all the way around and the total area of the frame and photo to be 315 in2. What should the outer dimensions of the frame be?

Using Factoring to Solve a Real–World Problem

You are constructing a frame for a rectangular photo (17 in by 11 in).You want the frame to be the same width all the way around and the total area of the frame and photo to be 391 in2. What should the outer dimensions of the frame be?

Practice

• A box shaped like a rectangular prism has a volume of 280 in³. Its dimensions are 4 in by (n + 2) in by (n + 5) in. Find n.

• You are building a rectangular deck. The area of the deck should be 250 ft2. You want the length of the deck to be 5 feet longer than twice its width. What should the dimensions of the deck be?

Completing the Square

Objective: To solve quadratic equations by completing the square.

Objective

• I can find c to complete the square.

• I can solve 𝑥2 + 𝑏𝑥 + 𝑐 = 0.

• I can find the vertex by completing the square.

• I can complete the square when a ≠ 1.

Vocabulary

• You can solve any quadratic equation by first writing it in the form m2 = n.

• In general, you can change the expression x2 + bx into a perfect–square trinomial

by adding (𝑏

2)2 to x2 + bx. This process is called completing the square.

• The process is the same whether b is positive or negative.

Finding c to Complete the Square

What is the value of c so that it is a perfect–square trinomial?

1. x2 – 16x + c

2. x2 + 20x + c

3. g2 + 17g + c

4. q2 – 4q + c

5. k2 – 5k + c

6. x2 + 18x + c

Practice

Find the value of c such that each expression is a perfect–square trinomial.1. 𝑥2 + 4𝑥 + 𝑐

2. 𝑎2 − 7a + c

3. 𝑏2 + 12𝑏 + 𝑐

4. 𝑤2 + 18𝑤 + 𝑐

5. 𝑔2 − 20𝑔 + 𝑐

6. 𝑛2 − 9𝑛 + 𝑐

Vocabulary

• To solve an equation in the form x2 + bx + c = 0, first subtract the constant term c from each side of the equation.

Solving 𝑥2 + 𝑏𝑥 + 𝑐 = 0

What is the solution of the equation?1. x2 – 14x + 16 = 0

2. x2 + 9x + 15 = 0

3. m2 + 7m – 294 = 0

4. m2 + 16m = –59

5. g2 + 7g = 144

6. z2 – 2z = 323

Practice

What are the solutions for the following?

1. x2 + 6x = 216

2. t2 – 6t = 247

3. r2 – 4r = 30

4. p2 + 5p – 7 = 0

5. m2 + 12m + 19 = 0

6. w2 – 14w + 13 = 0

Vocabulary

• The equation 𝑦 = (𝑥 − ℎ)2+𝑘 represents a parabola with vertex (h, k).

• You can use the method of completing the square to find the vertex of quadratic functions of the form 𝑦 = 𝑥2 + 𝑏𝑥 + 𝑐.

Finding the Vertex by Completing the Square

Find the vertex by completing the square.1. 𝑦 = 𝑥2 + 6𝑥 + 8

2. 𝑦 = 𝑥2 + 4𝑥 + 10

3. 𝑦 = 𝑥2 + 12𝑥 + 34

4. 𝑦 = 𝑥2 + 18𝑥 − 307

5. 𝑦 = 𝑥2 + 12𝑥 − 468

Practice

Find the vertex of each parabola by completing the square.1. 𝑦 = 𝑥2 + 4𝑥 − 16

2. 𝑦 = 𝑥2 + 6𝑥 − 7

3. 𝑦 = 𝑥2 + 2𝑥 − 28

4. 𝑦 = 𝑥2 − 2𝑥 − 323

Vocabulary

• The method of completing the square works when a = 1 in ax2 + bx + c = 0.

• To solve an equation when a ≠ 1, divide each side by a before completing the square.

Completing the Square When a ≠ 1

You are planning a flower garden consisting of three square plots surrounded by a 1–foot border. The total area of the garden and the border is 100 ft2. What is the side length x of each square?

x

x

1 1

1

1

xx

Completing the Square When a ≠ 1

You are planning a flower garden consisting of three square plots surrounded by a 1–foot border. The total area of the garden and the border is 150 ft2. What is the side length x of each square? Round to the nearest hundredth.

1

1

x x x

x

Practice

Solve each equation by completing the square. If necessary, round to the nearest hundredth.

1. 4a2 – 8a = 24

2. 2y2 – 8y – 10 = 0

3. 5n2 – 3n – 15 = 0

4. 4w2 + 12w – 44 = 0

5. 3r2 + 18r = 21

6. 2v2 – 10v – 20 = 8

The Quadratic Formula and the Discriminant

Objective: To solve quadratic equations using the quadratic formula. To find the number of solutions of a quadratic equation.

Objectives

• I can use the quadratic formula.

• I can find approximate solutions.

• I can choose an appropriate method.

• I can use the discriminant.

Vocabulary

• You can find the solution(s) of any quadratic equation using the quadratic formula.

• Quadratic Formula:

• Algebra: If ax2 + bx + c = 0, and a ≠ 0, then 𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎.

• Example: Suppose 2𝑥2 + 3𝑥 − 5 = 0. Then a = 2, b = 3, and c = –5. Therefore

𝑥 =−(3)± (3)2−4(2)(−5)

2(2)

• Be sure to write a quadratic in standard form before using the quadratic formula.

• Here’s Why It Works: If you complete the square for the general equation ax2 + bx + c = 0, you can derive the quadratic formula.

• Step 1: Write ax2 + bx + c = 0 so that the coefficient of x2 is 1.

• 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

• 𝑥2 +𝑏

𝑎𝑥 +

𝑐

𝑎= 0 Divide each side by a.

• Step 2: Complete the square.

• 𝑥2 +𝑏

𝑎𝑥 = −

𝑐

𝑎Subtract

𝑐

𝑎from each side.

• 𝑥2 +𝑏

𝑎𝑥 + (

𝑏

2𝑎)2= −

𝑐

𝑎+ (

𝑏

2𝑎)2 Add (

𝑏

2𝑎)2 to each side.

• (𝑥 +𝑏

2𝑎)2= −

𝑐

𝑎+

𝑏2

4𝑎2 Write the left side as a square.

• (𝑥 +𝑏

2𝑎)2= −

4𝑎𝑐

4𝑎2 +𝑏2

4𝑎2 Multiply −𝑐

𝑎by

4𝑎

4𝑎to get like denominators.

• (𝑥 +𝑏

2𝑎)2=

𝑏2−4𝑎𝑐

4𝑎2 Simplify the right side.

• Step 3: Solve the equation for x.

• (𝑥 +𝑏

2𝑎)2=±

𝑏2−4𝑎𝑐

4𝑎2 Take square roots of each side.

• 𝑥 +𝑏

2𝑎= ±

𝑏2−4𝑎𝑐

2𝑎Simplify the right side.

• 𝑥 = −𝑏

2𝑎±

𝑏2−4𝑎𝑐

2𝑎Subtract

𝑏

2𝑎from each side.

• 𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎Simplify

Using the Quadratic Formula

What are the solutions using the quadratic formula?1. x2 – 8 = 2x

2. x2 – 4x = 21

3. 2x2 + 5x + 3 = 0

4. 3x2 + 19x = 154

5. 18x2 – 45x – 50 = 0

Practice

Use the quadratic formula to solve each equation.1. 3𝑥2 − 41𝑥 = −110

2. 5𝑥2 + 16𝑥 − 84 = 0

3. 2𝑥2 − 𝑥 − 120 = 0

4. 3𝑥2 + 44𝑥 = −96

5. 5𝑥2 − 47𝑥 = 156

Vocabulary

• When the radicand in the quadratic formula is not a perfect square, you can use a calculator to approximate the solutions of an equation.

Finding Approximate Solutions

1. In the shot put, an athlete throws a heavy metal ball through the air. The arc of the ball can be modeled by the equation y = –0.04x2 + 0.84x + 2, where x is the horizontal distance, in meters, from the athlete and y is the height, in meters, of the ball. How far from the athlete will the ball land?

2. A batter strikes a baseball. The equation y = –0.005x2 + 0.7x + 3.5 models its path, where x is the horizontal distance, in feet, the ball travels and y is the height, in feet, of the ball. How far from the batter will the ball land? Round to the nearest tenth of a foot.

Practice

Use the quadratic formula to solve each equation. Round your answer to the nearest hundredth.

1. 𝑥2 + 8𝑥 + 11 = 0

2. 5𝑥2 + 12𝑥 − 2 = 0

3. 2𝑥2 − 16𝑥 = −25

4. 6𝑥2 + 9𝑥 = 32

Vocabulary

There are many methods for solving a quadratic equation.

Method When to Use

Graphing Use if you have a graphing calculator handy

Square Roots Use if the equation has no x-terms

Factoring Use if you can factor the equation easily

Completing the Square Use if the coefficient of x2 is 1, but you cannot easily factor the equation

Quadratic Formula Use if the equation cannot be factored easily or at all

Choosing an Appropriate Method

Which method(s) would you choose to solve each equation? Explain your reasoning.

1. 3x2 – 9 = 0

2. x2 – x – 30 = 0

3. 6x2 + 13x – 17 = 0

4. x2 – 5x + 3 = 0

5. –16x2 – 50x + 21 = 0

6. x2 – 8x + 12 = 0

7. 169x2 = 36

8. 5x2 + 13x – 1 = 0

Practice

Which method(s) would you choose to solve each equation? Justify your reasoning.

1. 𝑥2 + 4𝑥 − 15 = 0

2. 9𝑥2 − 49 = 0

3. 4𝑥2 − 41𝑥 = 73

4. 3𝑥2 − 7𝑥 + 3 = 0

5. 𝑥2 + 4𝑥 − 60 = 0

6. −4𝑥2 + 8𝑥 + 1 = 0

Vocabulary

• Quadratic equations can have two, one, or no real-number solutions.

• Before you solve a quadratic equations, you can determine how many real-number solutions it has by using the discriminant.

• The discriminant is the expression under the radical sign in the quadratic

formula. The discriminant is 𝑏2 − 4𝑎𝑐 of 𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎

• The discriminant of a quadratic equation can be positive, negative, or zero.

Using the Discriminant

Discriminant b2 – 4ac > 0 b2 – 4ac = 0 b2 – 4ac < 0

Example x2 – 6x + 7 = 0 The discriminant (-6)2 – 4(1)(7) = 8 which is positive

x2 – 6x + 9 = 0 The discriminant (-6)2 – 4(1)(9) = 0

x2 – 6x + 11 = 0The discriminant (-6)2 – 4(1)(11) = –8

which is negative

Number of Solutions

There are two real-numbersolutions

There is one real-numbersolutions

There are no real-number solutions

The Discriminant

Using the Discriminant

How many real-number solutions does each have?1. 2x2 – 3x = –5

2. 6x2 – 5x = 7

3. x2 + 3x + 11 = 0

4. 9x2 + 12x + 4 = 0

5. x2 – 15 = 0

Practice

Find the number of real–number solutions of each equation.1. 𝑥2 − 2𝑥 + 3 = 0

2. 𝑥2 + 7𝑥 − 5 = 0

3. 𝑥2 + 2𝑥 = 0

4. 3𝑝2 + 4𝑝 = 10