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Module 2: Answer Key

Section 1: Quadratic Functions

Lesson 1 Relations and Functions 173

Lesson 2 Linear and Quadratic Functions 177

Lesson 3 Quadratic Functions: y = ax2 and y = ax2 + k 183

Lesson 4 Quadratic Functions: y = a(x – h)2 189

Lesson 5 Quadratic Functions: y = a(x – h)2 + k 193

Lesson 6 Completing the Square 199

Lesson 7 Special Features of the Quadratic Function 207

Lesson 8 Problems Involving Quadratic Functions 219

Review 223

Section 2: Solving Equations

Lesson 1 Solving Quadratic Equations Graphically 231

Lesson 2 Solving Quadratic Equations by Factoring 235

Lesson 3 Solving Quadratic Equations by Completing theSquare and by the Quadratic Formula 241

Lesson 4 The Nature of Roots 249

Lesson 5 Radical Equations 255

Lesson 6 Rational and Absolute Value Equations 261

Review 267

Principles of Mathematics 11 Answer Key, Contents 171

172 Answer Key, Contents Principles of Mathematics 11

Lesson 1Answer Key

1. a) A is not a function because (2, 1) and (2, 3) have the samex-coordinate matched with two different y-values.

b) B is a function since for each x-value there is only oney-value.

c) The equation y = 2x + 3 is a function since no two orderedpairs have the same x-value with different y-values.

d) The equation y = x2 + 2 is a function since no two orderedpairs have the same x-value with different y-values.

2. a) y = 3x, the y-value is 3 times the x-value

b) y = x2, the y-value is the x-value squared

c) the y-value is the x-value

d) y = 2x – 1, the y-value is always 1 less than twice the x-value

3. a) Because a vertical line intersects the graph at only onepoint, the relation is a function.

b) Because a vertical line intersects the graph at more thanone point, the relation is not a function.

c) Because a vertical line intersects the graph in more thanone point, the relation is not a function.

d) Because a vertical line intersects the graph at only onepoint, the relation is a function.

4. The wanted value is the y-coordinate of the pointcorresponding to the x-value, which is in the parentheses.a) f(2) = 6b) f(5) = –3c) f(–1) = 3 d) f(4) is not defined

5. a) h(0) = 0b) h(–1) = 1c) h(2) = 0

12

y x= 12

,

Principles of Mathematics 11 Section 1, Answer Key, Lesson 1 173

Module 2

6. For f(x) = 3 – xa) f(0) = 3 – 0 = 3b) f(1) = 3 – (–1) = 4c) f(3) = 3 – 3 = 0d) f(t) = 3 – t

7.

174 Section 1, Answer Key, Lesson 1 Principles of Mathematics 11

Module 2

f x x x

f

f

f a a a a a

f x x x x x

f x x x x x x

x x

b gb g b g b ge j e j e jb g b g b gb g b g b gb g b g b g d i b g

= − −

− = − − − − = + − =

= − − = − − = −

= − − = − −

= − − = − −

− = − − − − = − + − − −

= − +

2 3 1

1 2 1 3 1 1 2 3 1 4

2 2 2 3 2 1 4 3 2 1 3 3 2

2 3 1 2 3 1

2 2 2 3 2 1 8 6 1

1 2 1 3 1 1 2 2 1 3 1 1

2 4

2

2

2

2 2

2 2

2 2

2

a)

b)

c)

d)

e)

2 3 3 1

2 7 42

− + −

= − +

x

x x

f) To find the value of , first find f f f1 1 2 1 3 1 1

2 3 12

2b gc h b g b g b g= − −

= − −= −

Then replace by and find f f1 2 2b g b g− −

f − = − − − −

= + −=

2 2 2 3 2 1

8 6 113

2b g b g b g

8.

9. Domain Rangea) {2} {1, 3, 5, 6}b) {2, 3, 4, 5} {1}c) {x|x ∈ R} {y|y ∈ R}d) {x|x ∈ R} {y|y ≥ 2, y ∈ R}

10. Domain Rangea) –3 ≤ x ≤ 3 0 ≤ y ≤ 3b) 0 ≤ x ≤ 3 –3 ≤ y ≤ 3c) x ≥ 0 y ∈ Rd) x ∈ R y ≥ 0


Module 2

f x x g x xb g b g= − = −2 2 2

a) f − = − −

= −=

3 3 2

9 27

2b g b g b) g − = − −

=

2 2 2

4

b g b g

f g

g

f

3

3 2 3

6

6 6 2

36 2 34

2

b gc hb g b g

b g b g

= −

= −

− = − −

= − =

g f

f

g

3

3 3 2

7

7 2 7

14

2

b gc hb g

b g b g

= −

== −

= −

c) d)


Module 2

Notes

Lesson 2Answer Key

1. a)

b) Change into the slope-intercept form:y = mx + b–y = –2x – 1y = 2x + 1

c)

x

y

55

5

10

x

y

1

1

x

y

2


Module 2

2. x = 1 or x = –1, from the graph

3. a) is quadratic because it is in the form f(x) = ax2 + bx + cb) is not quadratic since it has the variable x raised to the

third powerc) is quadratic (similar to (a)) d) is not quadratice) is linear, not quadraticf) is quadratic since its graph is a parabola

4. a) Domain: x ∈ Rb) Range: y ≥ –3, y ∈ R (the values are never below –3).c) Vertex: (4, –3) (remember this is the turning point).d) Axis of symmetry equation: x = 4. (This is the equation of

the vertical line that passes through the vertex.)e) Zeros: 2, 6. Remember this is not an ordered pair. It

represents the x-coordinates of the points where the curveintersects the x-axis.

f) x-intercepts: 2, 6g) Maximum value does not exist.h) Minimum value is at –3 (the y-coordinate of the vertex)

5. a) D: x ∈ RR: y ≤ 3, y ∈ RVertex coordinates (0, 3)Axis of symmetry equation: x = 0 Zeros: –3, 3y-intercept: 3 is the y-coordinate of the point where thecurve crosses the y-axisMaximum value at 3 Minimum value does not exist.


Module 2

b) Domain: x ∈ RRange: y ≥ 2.6 (approximate), y ∈ RVertex: (0.5, 2.6) (approximate)Axis of symmetry equation: x = 0.5Zeros: none (the curve does not cross the x-axis)y-intercept: 2.7 (approximate)Maximum value does not existMinimum value at approximately 2.6

c) Domain: x ∈ RRange: y ≥ –5, y ∈ RVertex: (4, –5) Axis of symmetry equation: x = 4Zeros: 1.5, 6.5 approximatelyy-intercept: 5Maximum value: noneMinimum value: –5

6. a) Domain: x ∈ RRange: y ≥ –9, y ∈ RVertex: (2, –9)Axis of symmetry equation: x = 2 Zeros: –1 and 5x-intercepts: –1 and 5y-intercept: –5Maximum value does not exist Minimum value is at –9


Module 2

b) Domain: x ∈ RRange: y ≤ 9, y ∈ RVertex: (–1, 9) Axis of symmetry equation: x = –1Zeros: –4 and 2x-intercepts: –4 and 2y-intercept: 8Maximum value at 9Minimum value does not exist

2 4 6 8 10

2

4

6

2468

2

4

6

8

x

y

2 4 6 8 10

2

4

6

24682

4

6

8

x

y

8

10


Module 2

7. a) ii)b) vi)c) v)d) iii)e) i)f) iv)

8. a) This is one example: reflection of (4, 3) would be (0, 3).

b)

�

��

x

y

axis of symmetry

V (2, 3)

reflection

point (5, 6)

x

y

22(0, 3)


Module 2

c) The axis of symmetry is at the point half way betweenthe reflection points. Therefore the point half waybetween is at (2, 2) and the equation of the axis ofsymmetry is x = 2.

x

y


Module 2

Lesson 3Answer Key

1. State whether each statement is true or false. If thisstatement is false, rewrite it so it is true.a) True, because it rearranges to y = –x2 + 1, which is

quadratic.b) False, because it has no squared term and so is linear.

The equation t(x) = 1 – x is a linear function whereas t(x) = 1 – x2 is a quadratic function.

c) True, since it is in quadratic formd) True, since the second coordinate axis is the y-axis.e) False, the vertex of the parabola pictured at the right is

(2, 1). f) False, since the axis of symmetry is always a vertical line

not a horizontal line as given. An equation of the axis ofsymmetry of the parabola pictured at the right is x = 2.

g) False, since the graph of y = x2 is narrower than .

The graph of is a wider parabola

than the graph of y = x2.h) True, because |5| > 1. i) False, because when a > 0 the parabola opens upward.

The graph of is a parabola that opens

upward or the graph of is a parabola that

opens “downward.”

j) True, because when a < 0, the parabola opens downward.k) True, because the function has a maximum value when

the function opens downward.l) False, because the minimum value is not defined when

the parabola opens downward and continues indefinitely.The minimum value of the function pictured at the rightcannot be stated as the parabola continues indefinitelydownward.

y x= − 14

2

y x= 14

2

y x= 14

2

y x= 14

2


Module 2

m) False, because the range of the function is y ≤ 4, y ∈ R. Therange of the function pictured at the right is y ≤ 4.

2. Complete

3. a) y = x2 is stretched vertically by a factor of 4

b) y = x2 is shrunk vertically by a factor of

c) y = x2 is stretched vertically by a factor of 7 and reflectedover the x-axis

4. a) upward if a > 0b) downward if a < 0 c) if a > 0, a minimum point resultsd) if a < 0, a maximum point results

5. a) 3 b) 1 c) 4 d) 2

6. y = ax2

Substitute in x = 1, y = –4 –4 = a(1)2

–4 = a∴ the equation is y = –4x2

12

Vertex

Equation of axis ofsymmetry

Domain

Range

x-intercepts or zeros

Direction of opening

Maximum y-value

Minimum y-value

y-intercept

y x= 1

32 y x= 2 y x= −3 2

(0, 0)

0

0

0

0

0

0 0

upward upward downward

cannot bedetermined

cannot bedetermined

cannot bedetermined

0 0

x = 0 x = 0 x = 0

y ≥ 0 y ≥ 0 y ≤ 0

x ∈ R x ∈ R x ∈ R

(0, 0) (0, 0)


Module 2

7. A point on the curve is (2, 6).y = ax2

Substitute x = 2, y = 66 = a(2)2

6 = a(4)

8. a) False The graph of y = 3x2 + 4 is the graph of y = 3x2

shifted upward 4 units or the graph of y = 3x2 – 4 isthe graph of y = 3x2 shifted downward 4 units.

b) True.c) True.d) True.e) False An equation of the axis of symmetry of y = x2 – 1 is

x = 0. f) False The graph of y = x2 – 1 opens upward or the graph

of y = –x2 – 1 opens downward.g) Trueh) False The maximum value of y = –2x2 + 3 is 3i) True.j) True.k) True.l) False The graph of y = x2 + 2 and y = –x2 + 2 are mirror

reflections of each other in the line y = 2. y = x2 and y = –x2 are mirror reflections of eachother in the line y = 0 (the x-axis)

m) True.n) True.

32

32

32

2

2

=

=

= −

a

y x

y xThe reflection equation is .


Module 2

9.

10. a) y = x2 is stretched vertically by 5 and shifted 4 unitsvertically upward

b) y = x2 is stretched vertically by 2, reflected over thex-axis and shifted vertically upward 1 unit

c) y = x2 is shrunk vertically by and shifted verticallydownward 1 unit

d) rearrange to y = 3x2 – 2, y = x2 is stretched vertically by3 and shifted 2 units vertically downward.

11. a) The graph opens upward, and the minimum value ispositive.

b) The graph opens upward, and the minimum value isnegative.

c) The graph opens downward, and maximum is positive.d) The graph opens downward, and maximum is negative.

12

Vertex


Domain

Range

x-intercepts

y-intercepts


Maximum value

Minimum value

y x= +2 4 y x= 2 y x= −2 1 y x= − +2 1

(0, 4)

None

None None None

None

1

Upward UpwardUpward Downward

0

0

04

4

–1, 1

–1

–1

–1, 1

1

x = 0

y ≥ 4 y ≥ 0 y ≥ –1 y ≤ 1

x = 0 x = 0 x = 0

(0, 0) (0, –1) (0, 1)

x ∈ R x ∈ R x ∈ R x ∈ R


Module 2

12. If the vertex is at (0,–3), the function has been shiftedvertically downward 3 units. So, k = –3. The curve passes through (1,–1), so y = –1 when x = 1.Substitute into

The equation is y = 2x2 – 3.

y ax k

a

aa

= +

− = −

− = −=

2

21 1 3

1 32

b g


Module 2


Module 2

Notes

Lesson 4

Answer Key

1. a) True.b) False. The graph of y = (x – 7)2 is the graph of y = x2

shifted 7 units to the right or the graph of y = (x + 7)2 is the graph of y = x2 shifted 7 units tothe left.

c) True.d) False. The vertex of the graph of y = 2(x – 1)2 is (1, 0) or

the vertex of the graph of y = 2(x – 2)2 is (2, 0).e) False. An equation of the axis of symmetry of the graph of

or an equation of the axis of

symmetry of the graph of

f) False. An equation of the axis of symmetry of the graph of

g) False. The graph of is a wider parabola

than the graph of y = –3(x – 2)2 or the graph ofy = –3(x – 2)2 is a narrower parabola than the

graph of

h) True.i) True.j) True. g(x) = (x + 4)2 g(x – 4) = (x – 4 + 4)2 = x2

g(–x – 4) = (–x – 4 + 4)2 = (–x)2 = x2

2. a) iib) ivc) iiid) i

( )= − − 212 .

3y x


Module 2

( )= − =213 is 3

2y x x

= − =

21 13 is .2 2

y x x

( )= − − 212

3y x

( )= − + = −213 is 3.

2y x x

3. a) y = x2 shifted horizontally 4 units to the right and shrunkvertically by a factor of

b) y = x2 shifted horizontally 1 unit to the right, stretched by 2and reflected over the x-axis

c) shifted horizontally 1 unit to the left and stretchedvertically by a factor of 2

x

y

2

4

4

x

2

4

4

y

x

y

12


Module 2

= 2y x( )= − 21

42

y x

4. Complete the table.

Vertex


Domain

Range

x-intercepts

y- intercepts


Maximum value

Minimum value

( )= + 22y x ( )= − 2

1y x ( )= − 22 3y x ( )= + 24

63

y x

(–2, 0) (1, 0) (3, 0) (–6, 0)

x = –2 x = 1 x = 3 x = –6

x ∈ R x ∈ R x ∈ R x ∈ R

y ≥ 0 y ≥ 0 y ≥ 0 y ≥ 0

–2 1 3 –6

4 1 18 48

upward upward upward upward

none none none none

0 0 0 0


Module 2


Module 2

Notes

Lesson 5

Answer Key

1. a) False. The vertex of the graph of y = (x – 4)2 + 2 is (4, 2) orthe vertex of the graph of y = (x + 4)2 + 2 is (–4, 2).

b) False. The vertex of the graph of y = 3(x – 5)2 – 2 is (5, –2)or the vertex of the graph of y = 3(x – 5)2 + 2 is (5, 2).

c) False. The vertex of the graph of y = 2x2 + 7 is (0, 7), orthe vertex of the graph of y = 2(x – 2)2 + 7 is (2, 7).

d) True.e) False. The vertex of the graph of y = –2(x – 3)2 is (3, 0), or

the vertex of the graph of y = –2(x + 2)2 + 3 is (–2, 3).

f) False. An equation of the axis of symmetry of the graph ofy = –3(x – 6)2 + 1 is x = 6, or an equation of the axisof symmetry of the graph of y = –3(x – 1)2 + 1 is x = 1.

g) True.h) True.i) False. The graph of y = 4(x – 3)2 – 1 opens upward, or the

graph of y = –4(x – 3)2 – 1 opens downward.j) True.k) True. The range of y = –2(x + 3)2 + 4 is {y|y ≤ 4}, or the

range of y = 2(x + 3)2 + 4 is {y|y ≥ 4}.l) False. The minimum value of y = 3(x – 5)2 – 2 is –2.

2. a) iii b) v c) viii d) iie) i f) vii g) vi h) iv


Module 2

3. Complete the following table.

4.

( )= + 22 1y x

( )−= − +211 6

2y x

( )= + −22 6 10y x

( )= − +26 1 8y x

Direction ofopening

Axis of symmetry

Wider orNarrower

Vertex

U p x = –1 Narrower(–1, 0)

Down x = 1 Wider(1, 6)

U p x = –6 Narrower(–6,–10)

U p x = 1 Narrower(1, 8)

Vertex

Axis of symmetryequation

Domain

Range

y-intercept


Maximum value

Minimum value

( )= − +21 2y x ( )= − − −2

1 2y x ( )= + +21 2y x ( )= − + −2

1 2y x

(1, 2)

3

Upward

None

None None

None

UpwardDownward Downward

–3 3 –3

–2

2 2

–2

x = 1

y|y ≥ 2 y|y ≤ –2 y|y ≤ –2y|y ≥ 2

x ∈ R x ∈ R x ∈ R x ∈ R

x = 1 x = –1 x = –1

(1, –2) (–1, 2) (–1, –2)


Module 2

5. a) y = 2(x + 1)2

y = x2 is shifted horizontal to theleft by 1 unit and stretchedvertically by a factor of 2Vertex: (–1, 0)Axis of symmetry: x = –1

b)y = x2 is shifted horizontally 1unit to the right, shrunkvertically by a factor of ,reflected over the x-axis, andshifted vertically 6 unitsupward.Vertex: (1, 6)Axis of symmetry: x = 1

c) y = 2(x + 6)2 – 10y = x2 is shifted horizontally 6units to the left, stretchedvertically by a factor of 2, andshifted vertically 10 unitsdownwardVertex: (–6, –10)Axis of symmetry: x = –6


Module 2

12

y x= − − +12

1 62b g

x

y

6

4

2

224

y

6

4

2

224x

4 6

2

4

8

y4

2

24x

2

4

6810

6

8

10

d) y = 6 (x – 1)2 + 8y = x2 is shifted horizontally 1unit to the right, stretchedvertically by a factor of 6, andshifted vertically 8 unitsVertex: (1, 8)Axis of symmetry: x = 1


Module 2

g x xb g b g= − − −3 2 12

• maximum value at –1• {y|y ≤ –1, y ∈ R}

c)

y x= + +12

1 42b g

• minimum value at 2• {y|y ≥ 2, y ∈ R}

d)

Multiply into the square bracket first

=

12

12

1 22y x + +b g

y

6

4

2

224x

4

8

10

12

14

f x xb g = − +FHGIKJ +1

22

2

• maximum value at 2• {y|y ≤ 2, y ∈ R}

6. a)

y x= + −2 4 22b g• minimum value at –2• {y|y ≥ –2, y ∈ R}

b)


Module 2

7. y x k

x y

k

k

k

k

k

= − +

= =

= − +

= − +

= +

=

∴

2

1 32

32

1 2

32

1

32

1

12

2

2

2

b g

b g

b g

Substitute and into the above equation

represents the minimum value

minimum value is 12


Module 2

Notes

Lesson 6Answer Key

1. x2 + 5x + 4

2.


Module 2

a) x x k

k

k x x x

2

2

2 2

8

82

16

16 8 16 4

+ +

FHGIKJ = =

∴ = → + + = +b g

b) x x k

k

k x x x

2

2

2 2

8

82

16

16 8 16 4

− +

−FHGIKJ = =

∴ = → − + = −b g

c) x x k

k

k x x x

2

2

2 2

20

202

100

100 20 100 10

+ +

FHGIKJ = =

∴ = → + + = +b g

d) x x k

k

k x x x

2

2

2 2

2

22

1

1 2 1 1

− +

−FHGIKJ = =

∴ = → − + = −b g


Module 2

e) x x k

k

k x x x

2

2

22

5

52

254

254

5254

52

− +

−FHGIKJ = =

∴ = → − + = −FHGIKJ

f) x x k

k

k x x x

2

2

22

7

72

494

494

7494

72

+ +

FHGIKJ = =

∴ = → + + = +FHGIKJ

x x a x r t

x a x r t

22

22

2 2

8 82

82

4 16

− + −FHGIKJ = − + + −F

HGIKJ

− − = − +

b gb g b g

3. a)

a r t= = = −1 4 16

Complete the square

Simplify

x x a x r t

x x a x r t

x a x r t

2 2

22

22

22

5

5 52

52

52

254

+ = − +

+ + FHGIKJ = − + + FHG

IKJ

+FHGIKJ − = − +

b gb g

b g

b)

a r t= = − = −1

52

254

Complete the square

Simplify

ii) V(3, 9); Equation of axis of symmetry: x – 3 = 0iii) y = x2 shifted horizontally 3 units to the right,

reflected over the x-axis and shifted vertically upward9 units

iv) range: y ≤ 9v) To find the y-intercept, let x = 0

vi) Because the graph opens downward, it has amaximum value of 9.


Module 2

12

2 4

12

442

412

42

12

2 2

2 2

22

22

2 2

x x a x r t

x x a x r t

x a x r t

+ + = − +

+ + FHGIKJ

FHG

IKJ = − + − + FHG

IKJ

+ + = − +

b g

b g

b g b g

c)

4. a)

a r t= = − =12

2 2

f x x xb g = − +2 6

f x x x

f x x x

f x x x

f x x

f x x

b gb g d i

b g b gb g

b g b gb g b g

= − +

= − −

+ − − = − − + −FHGIKJ

FHG

IKJ

− = − −

= − − +

2

2

2 22

2

2

6

1 6

1 3 1 66

2

9 1 3

1 3 9

Factor out –1 tomake sure that 1 isthe coefficient of x2

Complete the square

Simplify

Rearrange

in or in the originaly x

y

= − − +

= − − +

= − +=

1 3 9

1 0 3 9

9 90

2

2

b gb g y x x

y

= − += − +

=

2 6

1 0 6 0

0b g b g

i)

( )

÷ =

+

+

2

2

Note: Factor out 1

and complete2

1the square 2 4

21

so 2 2

1factors to 4 .

2

x x

x x


Module 2

b) f x x x: → + −2 5 32

i)

ii)

iii)

y x x

y x x

y x x

y x x

y x

y x

or

f x x

= + −

= +FHGIKJ −

+ + FHGIKJ = + + FHG

IKJ

FHG

IKJ

+ + ⋅ = + +FHG

IKJ

+ + = +FHGIKJ

+ = +FHGIKJ

→ +FHGIKJ −

2 5 3

252

3

3 254

252

54

3 22516

252

2516

3258

254

498

254

254

498

2

2

22

2

2

2

2

2

:

Factor out 2 tomake 1 thecoefficient of x2

Complete thesquare

Simplify

For convenience, replace f:x → with f(x) or y

V x− −FHG

IKJ = −5

4498

54

, ; equation of symmetry is

iv) Range: y ≥ − 498

vi) Because the curve opens upward, it has a

minimum value of − 498

.

v) y x x x

y

= + − =

= + −

= −

2 5 3 0

2 0 5 0 3

3

2

2

let

b g b g

y x= 2 54

is shifted horizontally units to the left,

stretched by 2, and shifted vertically downward498

units

Note: The y-intercept isthe constantin the originalfunction.

ii) V(–2, 4); Equation of axis of symmetry: x = –2iii) y = x2 is shifted horizontally 2 units to the left,

reflected over the x-axis, and shifted verticallyupward 4 units

iv) Range: y ≤ 4v) For the y-intercept, let x = 0, preferably in the given

function.

vi) Maximum value is 4

y x x

y

= − −

= − −

=

2

2

4

0 4 0

0b g b g


Module 2

c) x x x f x x x, − − = − −4 42 2d io t b gis equivalent to

d) f x x xb g = − +2 12 192

i) f x x x

f x x x

f x x x

f x x

b gb g d i

b g

b g b g

= − −

= − +

− = − + + FHGIKJ

FHG

IKJ

= − + +

2

2

22

2

4

4

4 442

2 4

Complete squareand add –4 toeach side

i) f x x x

f x x x

f x x

f x x

b g d i

b g

b g b gb g b g

= − +

− + −FHGIKJ = − + −F

HGIKJ

FHG

IKJ

− = −

= − +

2 6 19

19 262

2 66

2

1 2 3

2 3 1

2

22

2

2

2

Factor out 2 andmake 1 thecoefficient of x2

Complete thesquare and add2(3)2 to each side

Simplify

ii) V(3, 1); Equation of axis of symmetry: x = 3iii) y = x2 is shifted vertically 3 units to the right, stretched

vertically by a factor of 2, and shifted upward 1 unitiv) Range: y ≥ 1v) For the y-intercept, let x = 0, ∴ y = 19vi) Because the function opens upward, it has a minimum

of 1.


Module 2

e) y x x− + = −3 3 32

i) y x x

y x x

y x x

y x x

y x x

y x

y x

− + = −

+ = −

+ = −

+ + ⋅ −FHGIKJ = − + −F

HGIKJ

FHG

IKJ

+ + ⋅ −FHGIKJ = − +FHG

IKJ

+ = −FHGIKJ

= −FHGIKJ −

3 3 3

3 3 3

3 3

3 31

23

12

3 31

23

14

154

312

312

154

2

2

2

22

2

22

2

2

d iRearrangefactor out 3 andmake 1 thecoefficient of x2

Complete the square

and add to

each side

Factor and rearrange

312

2

−FHGIKJ

iii)

iv) Range: y ≥ − 154

y x= 2 12

is shifted horizontally unit to the right,

stretched vertically by a factor of 3, and shifted

vertically downward 154

units

ii) V x12

154

12

, −FHGIKJ =; Equation of axis of symmetry is

5. a) Group the terms containing the variablesb) Rearrangec) Factor out ad) Complete the square and add to each sidee) Factorf) Rearrangeg) Simplify

6. a) From the coordinates of the vertex, h = 2 and k = 6.From the point on the curve, x = 1 and y = 7.Substitute into

b) Since a is positive, the curve opens upward and will havea minimum value at 6.

c) y = 1(x – 2)2 + 6 or y = (x – 2)2 + 6

7. Rearrange the equation into the correct form and completethe square.

The vertex is at V(8, 326), meaning that after 8 seconds themaximum height of 326 metres is reached.

h t t

h t t

h t

h t

= − + +

= − − + + + ⋅

= − − + +

= − − +

5 80 6

5 16 8 6 5 8

5 8 6 320

5 8 326

2

2 2 2

2

2

d i d ib gb g

y a x h k

a

a

a

= − +

= − +

= − +

=

b gb gb g

2

2

2

7 1 2 6

7 1 6

1


Module 2

vi) Minimum value is − 154

v) y-intercept is –3

aba

⋅ FHGIKJ2

2


Module 2

Notes

Lesson 7

Answer Key

1.

2.

3. a) y = x2 – 2x – 3

i) Because it factors readily, find the x-intercepts:

The x-intercepts are –1 and 3.

ii) The x-coordinate of the vertex is the midpoint of the x-intercepts

For y = x2 – 2x – 3, let x = 1

V = (1, –4)


Module 2

a)

b)

c)

d)

e)

−∞ − ∞

−

− ∞

∞ −∞ −

− − ∞

, ,

,

,

, ,

, ,

3 2

10 5

8

5 7

5 2 2

b g gbb gb g b gb g

∪

∪

∪

a)

b)

c)

d)

e)

2 5

2 7

3 10

0 6

6 3 2 8

,

, ,

, ,

,

, ,

bb g b gb g

b g

−∞ − ∞

−∞ − ∞

− −

∪

∪

∪

y x x

x x

x xx x

= − +

− + =

− = + == = −

3 1

3 1 0

3 0 1 03 1

b gb gb gb g

or or

− + = =1 32

22

1

y = − −

= − −= −

1 2 1 3

1 2 34

2 b g

iii) Axis of symmetry equation: x = 1

iv) y-intercepty = x2 – 2x – 3, let x = 0 to find y-intercept

v) Opens upward because a is positivevi) Maximum value is not definedvii) Minimum value corresponding to the y-coordinate of

the vertex = –4

viii) Inequality IntervalD: {x|x ∈ R} (–∞, ∞)R: {y|y ≥ –4} [–4, ∞)

Graph

b) y = –1(x2 + 6x + 8)

i)

The x-intercepts are –4 and –2.


Module 2

y = − −

= −

0 2 0 3

3

2 b g

2

4

6

2 4 62462

4

6

x

y

y x x

x x

xx x

= − + +

− + = + =

− − == − = −

1 4 2

1 4 0 2 0

4 04 2

b gb gb g or

or

ii) x-coordinate at vertex:

y-coordinate at vertex:

V = (–3, 1)

iii) Axis of symmetry equation: x = –3

iv) y-intercept, let x = 0

v) Opens downward since a is negative

vi) Maximum value is 1

vii) No minimum value

viii) Inequality IntervalD: x ∈ R (–∞, ∞)R: y ≤ 1 (–∞, 1]

Graph


Module 2

− + −= − = −

4 22

62

3b g

y x x= − − −

= − − − − −

= − + −=

2

2

6 8

3 6 3 8

9 18 81

b g b g

y x xy

= − − −= −

2 6 88

2 4 62462

4

6

x

y

8

1

4. a) y = x2 + 6x – 7

i) x-coordinate of vertex =

y-coordinate

V(–3, –16)

ii) Axis of symmetry equation: x = –3

iii) x-intercepts

iv) y-intercept (let x = 0) = –7

v) Because a > 0, the parabola opens upward.

vi) Since the the parabola opens upward, the function has a minimum value of –16.

vii) D: (–∞, ∞)R: [–16, ∞)


Module 2

a b

x

= =

= − = −

1 66

2 13b g

= − + − −

= − −= −

3 6 3 7

9 18 716

2b g b g

y x x

x x

x x

= + −

+ − =

= − =

7 1

7 1 0

7 1

b gb gb gb g

or

−ba2

b) y = x2 – 4x – 60


y-coordinate

V(2, –64)

ii) Axis of symmetry equation: x = 2

iii) x-intercepts


v) Opens upward since a > 0

vi) Since the parabola opens upward, the function has aminimum value at –64.

vii) D: (–∞, ∞)R: [–64, ∞)


Module 2

−ba2

a b

x

= = −

=− −

= =

1 4

42 1

42

2b gb g

y = − −

= − −= −

2 4 2 60

4 8 6064

2b g b g

y x x

y x x

x x

x

= − −= − +

− + =

= −

2 4 60

10 6

10 6 0

10 6

b gb gb gb g

or

c) y = 2x2 + 8x – 10


y-coordinate

V(–2, –18)


iii) x-intercepts

iv) y-intercept (let x = 0)

v) Opens upward because a > 0

vi) Because the parabola opens upward, the minimum value is –18.

vii) D: (–∞, ∞)R: [–18, ∞)

− ba2


Module 2

a b

x

= =

= − = −

2 88

2 22b g

y = − + − −

− −

= −

2 2 8 2 10

2 4 16 10

18

2b g b gb g

y x x

x x

x x

x

= − +

− + =

− = + =

= −

2 1 5

2 1 5 0

2 1 0 5 0

1 5

b gb gb gb gb g or

or

y = + −

= −

2 0 8 0 10

10

2b g b g

d) y = 3x2 + 24x + 21


y-coordinate

V(–4, –27)


iii) x-intercepts

iv) y-intercept (let x = 0)

v) Opens upward because a > 0

vi) Because the parabola opens upward, the minimum value is –27.

vii) D: (–∞, ∞)R: [–27, ∞)


Module 2

−ba2

a b

x

= =

= − = −

3 2424

2 34b g

y = − + − +

= − += −

3 4 24 4 21

48 96 2127

2b g b g

y x x

x x

x x

x x

= + +

+ + =

+ = + =

= − = −

3 1 7

3 1 7 0

3 1 0 7 0

1 7

b gb gb gb gb g or

or

y

y

= + +

=

3 0 24 0 21

21

2b g b g

e) y = x2 + 5x + 6


y-coordinate

ii) Axis of symmetry equation:

iii) x-intercepts

iv) y-intercept (let x = 0) = 6

v) Opens upward since a > 0

vi) Because the parabola opens upward, the minimum

value is

vii) D: (–∞, ∞)

R:− ∞LNMIKJ

14

,


Module 2

−ba2

a b

x

= =

= − = −1 5

52 1

52b g

y = −FHGIKJ + −FHGIKJ +

= − +

= − +

= −

52

55

26

254

252

6

254

504

244

14

2

V − −FHG

IKJ

52

14

,

x = − 52

y x x

x x

x xx x

= + +

+ + =

+ = + == − = −

3 2

3 2 0

3 0 2 03 2

b gb gb gb g

or or

− 14

.

f) y = –x2 + 3x + 4

i) x-coordinate of the vertex =

y-coordinate


iii) x-intercepts


v) Opens downward because a < 0

vi) Because the parabola opens downward, the

maximum value is

vii) D: (–∞, ∞)

R:


Module 2

−ba2

a b

x

= − = +

=− +

−= −

−=

1 3

32 1

32

32

b gb g

y = −FHGIKJ + FHG

IKJ +

= − + +

= − + +

=

32

332

4

94

92

4

94

184

164

254

2

V32

254

,FHGIKJ

x = 32

y x x

y x x

x x

x x

x x

= − − −

= − − +

− − + =

− − = + =

= = −

2 3 4

4 1

4 1 0

4 0 1 0

4 1

d ib gb g

b gb gb g or

or

−∞FHGOQP,

254

254

.

g) y = –2x2 – 5x – 2


y-coordinate


iii) x-intercepts


v) Opens downward since a < 0

vi) Because the parabola opens downward, the maximum

value is

vii) D: (–∞, ∞)

R:


Module 2

−ba2

a b

xba

= − = −

= − =− −

−= −

2 5

25

2 254

b gb g

y = − −FHGIKJ − −FHG

IKJ −

= − FHGIKJ + −

= − + −

=

254

554

2

22516

254

2

258

508

168

98

2

V −FHGIKJ

54

98

,

x = − 54

y x x

y x x

x x

x x

x x

= − + +

= − + +

− + + =

− + = + =

= − = −

2 5 2

1 2 1 2

1 2 1 2 0

1 2 1 0 2 0

12

2

2d ib gb g

b gb gb g or

or

98

.

−∞FHGOQP,

98

h) y = –3x2 + 2x + 1


y-coordinate


iii) x-intercepts


v) Opens downward since a < 0

vi) Because the parabola opens downward, the

maximum value is

vii) D: (–∞, ∞)

R:


Module 2

−ba2

a b

xba

= − =

= − = −−

=

3 2

22

2 313b g

y

y

= − FHGIKJ + FHG

IKJ +

= − + +

=

313

213

1

13

23

1

43

2

V13

43

,FHGIKJ

x = 13

y x x

y x x

x x

x

= − − −

= − + −

− + = − =

= −

1 3 2 1

1 3 1 1

1 3 1 0 1 0

13

1

2d ib gb g

b g or

or

43

.

−∞FHGOQP,

43


Module 2

Notes

Lesson 8

Answer Key

1.

The maximum height of 27 m is reached in 1.5 seconds.

2. Let x = width of theswimming area inmetres620 – 2x = length of theswimming area inmetres

Remember, you only need three sides roped off.

∴ Area: A(x) = x(620 – 2x)= 620x – 2x2

The maximum area occurs at the x-value of the quadraticfunction.

When the width is 155 m, the length is 620 – 2(155) = 310 m.

The dimensions of the swimming area are 155 m x 310 m.

xba

= − = −−

=2

6202 2

155b g m


Module 2

h t t t

h t t t

h t t

h t t

t

b g

b g b g

b g

b g

= − + +

− + − −FHGIKJ = − − + −F

HGIKJ

FHG

IKJ

− − = − −FHGIKJ

= − −FHGIKJ +

= − −FHGIKJ +

3 9814

814

33

23 3

32

814

274

332

3 32

1084

332

27

2

22

2

2

2

2

620 2x

x

beach

x

3. Let x = additional number of trees to be planted.Then, the total number of trees is 65 + x.1500 – 20x = yield of apples per tree.

The maximum occurs when

The total number of trees that yield the maximum number ofapples is 65 + 5 = 70 trees.

4. Let x = larger of the two numbersx – 14 = smaller of the two numbers

Since a is positive, a minimum occurs at at x = 7. If x = 7, thenx – 14 = –7.

The two numbers are 7 and –7 giving a minimum product of–49.

5. Honest John’s profit per car = $6400 – $4000 = $2400Let x = the decrease in the number of cars20 – x = number of cars sold2400 + 300x = profit per carP(x) = total profitP(x) = (20 – x)(2400 + 300x)P(x) = –300x2 + 3600x + 48 000


Module 2

f x

f x x x

x x

b gb g b gb g

=

= + −

= − + +

total yield

65 1500 20

20 200 975002

xba

= − = −−

=2

2002 20

5b g .

P x x x

x x

P x x x

P x x

b g b g

b gb g b g

= −

= −+ = − +

= − −

14

14

49 14 49

7 49

2

2

2

Profit per car = 2400 + 300(6) = 4200.

Selling price = cost + profit= 4000 + 4200= $8200

6. x = widthy = lengthTotal lengths = 2yTotal widths = 4x

2y + 4x = 800y = 400 – 2x Divide by two and solve for y

A = (400 – 2x)x

A = –2x2 + 400x

So the largest area is 100 x 200 = 20 000 m2.

7. Maximum income = total number of tickets sold x price perticketx = number of increments of 100 tickets1000 + 100x = total number of tickets sold60 – 3x = price per ticket

Completing the square gives:P(x) = –300(x – 5)2 + 67 500

P x x x

x x

b g b gb g= + −

= − + +

1000 100 60 3

300 3000 60 0002

Using m

m

xba

y

= − = −−

=

=

2400

2 2100

200b g


Module 2

Using xba

= − = −−

=2

36002 300

6b g

x x x x

y

y

Maximum income of $67 500 results with five decreases in theticket price.The price is 60 – 3x = 60 – 3(5) = $45.

8. Let x = number of $2 drops in pricePrice = 60 – 2xNumber sold: 800 + 50xf(x) = maximum cash return

To get the maximum cash return, drop the price seven times,that is, $14 in price.The price becomes 60 – 14 = $46 per radio.The number sold = 800 + 50(7) = 1150.

f x x x

x x x

x x

f x x x

f x x

f x x

b g b gb g

b g b gb g

b g b gb g b g

= − +

= − + −

= − + +

− + − − = − − + −FHGIKJ

FHG

IKJ

− − = − −

= − − +

60 2 800 50

48000 1600 3000 100

100 1400 48000

48000 100 7 100 14142

48000 4900 100 7

100 7 52900

2

2

2 22

2

2


Module 2

Review

Answer Key

1. a) Domain: (–∞, ∞)b) Range: [–4, ∞)c) Vertex: (3, –4)d) Equation of axis of symmetry: x = 3e) Zeros: 1, 5f) x-intercepts: 1, 5g) y-intercept: 5h) No maximumi) Minimum value at –4j) (1, 0) (5, 0) is one pair

2. i) y = –2(x + 1)2 – 3

a) a = –2, h = –1, k = –3

b) x = –1

c) (–∞, ∞)

d) opens downward

e) V(–1, –3)

f) maximum value at –3

g) y = x2 stretched by 2, reflected over the x-axis andshifted one unit to the left and 3 downward

h) Let y = 0

Since a squared number can never be negative, thereare no zeros.

i) Since the graph is below the x-axis, it has no x-intercept.

j) range: (–∞, –3]

− + − =

+ =−

2 1 3 0

132

2

2

x

x

b gb g

Principles of Mathematics 11 Section 1, Answer Key, Review 223

Module 2

k) Let x = 0 to find y-intercept

l)

m) y = –2(x + 4)2 – 1

ii)

a) a = , h = 1, k = 3

b) axis of symmetry: x = 1c) domain: (–∞, ∞)d) opens upwarde) vertex: (1, 3)f) minimum value at 3g) y = x2 shrunk by one-half shifted 1 unit to the right

and 3 units upward

h) Let y = 0

This is impossible since a squared number cannot benegative. Therefore, there are no zeros.

i) Since the graph is totally above the x-axis, there areno x-intercepts.

012

4 5

512

4

2

2

= − +

− = −

x

x

b g

b g

12

y x= − +12

1 32b g

x

y

y

y

= − + −

= −

2 0 1 3

5

2b g

224 Section 1, Answer Key, Review Principles of Mathematics 11

Module 2

j) range: [3, ∞)k) Let x = 0

l)

m)

iii) f(x) = 2(x + 1)2 – 3a) a = 2, h = –1, k = –3b) axis of symmetry: x = –1c) domain: (–∞, ∞)d) opens upwarde) vertex: (–1, –3)f) minimum value at –3g) y = x2 stretched by 2 shifted horizontally to the left

1 unit and vertically downward 3 units

y x= + +12

2 52b g

x

y

y

y

y

= − +

= +

=

12

0 1 3

12

3

72

2b g


Module 2

h) Let f(x) = 0

i) The x-intercepts are at

j) range: [–3, ∞)k) Let x = 0 to find y-intercept

l)

m) f(x) = 2(x + 4)2 – 1

x

y

f xb g b g= + −

= −= −

2 0 1 3

2 31

2

− + − −2 62

2 62

,

2 1 3 0

2 1 3

132

132

16

22 6

2

2

2

2

x

x

x

x

x

+ − =

+ =

+ =

+ = ±

= − ± − ±

b gb gb g

or


Module 2

3. a)

b)

c)

d) y x x

y x x

y x x

y x

y x

y x

= − −

+ = − +FHG

IKJ

+ + ⋅ = − +FHG

IKJ

+ + = −FHGIKJ

+ = −FHGIKJ

= −FHGIKJ −

2 3 7

7 232

7 29

162

32

916

7 98

2 34

658

234

234

658

2

2

2

2

2

2

y x x

y x x

y x x

y x

y x

= + −

+ = + +

+ + ⋅ = + +

+ = +

= + −

12

2 1

112

4

112

412

4 4

312

2

12

2 3

2

2

2

2

2

d i

d i

b g

b g

f x x x

f x x x

f x x x

f x x x

f x x

f x x

b gb g d ib g d ib g d ib g b gb g b g

= − − +

= − + +

− = − + +

− − = − + +

− = − +

= − + +

3 6 2

3 2 2

2 3 2

2 3 3 2 1

5 3 1

3 1 5

2

2

2

2

2

2

f x x x

f x x x

f x x x

f x x

f x x

b gb g

b g d ib g b gb g b g

= − −

+ = − − +

+ + ⋅ = − +

+ = −

= − −

2 8 5

5 2 8 5 5

5 2 4 2 4 4

13 2 2

2 2 13

2

2

2

2

2


Module 2

e)

4.

5. To find the x-coordinate find the x-coordinate of themidpoint.

The corresponding equation isy = (x – 5)(x – 1)

so replace x by 3 to find the y-value

∴ coordinates of the vertex is at (3, –4)

y = − −

= −

= −

3 5 3 1

2 2

4

b gb gb gb g

x x1 2

25 1

23

+ = + =

( ) ( )

( )

( )

( )

− = − + −

− + = − + − + ⋅

− = − + − +

− − = − +

= = = −

22

2 222

22

22

4 5 5

5 54 5 5 4

2 2

54 5 25

2

54 20

25

4, , 202

x x a x h k

x x a x h k

x a x h k

x a x h k

a h k

f x x x

f x x x

f x x x

f x x

f x x

b gb g d i

b g d ib g b gb g b g

= − −

+ = −

+ + = − +

+ = −

= − −

01 2 1

1 01 20

1 10 01 20 100

11 01 10

01 10 11

2

2

2

2

2

.

.

.

.

.


Module 2

6. The x-coordinate =

∴ a = 1, b = –8

Substitute into:

∴ vertex: (4, –14)axis of symmetry: x = 4

7. y = a(x – h)2 + k(h, k) = (2, –8)(x, y) = (4, 2)

8.

∴ Since a > 0, the curve opens upward and has a minimumvalue at 3.

y x p

x y

p

pp

= − +

=

= − +

= +=

1

2 4

4 2 1

4 13

2

2

b gb g b g

b gP , ,

2 4 2 8

2 4 8

10 452

52

2 8

2

2

= − −

= −

=

=

= − −

a

a

a

a

y x

b gb g

b g

f x x x

f

b gb g

= − +

= − ⋅ +

= − += −

2

2

8 2

4 4 8 4 2

16 32 214

x = − − =82 1

4b g

− ba2


Module 2

9. Let x be the additional number of passengers over 100.∴ price of each ticket = $30 – $0.20xtotal number of passengers = 100 + x∴ income (y)

∴ an additional 25 passengers will give a maximum incomeof $3125.∴ total number of passengers = 125

10. Perimeter: 2x + 2y = 200Area: A = xy

Solve 2x + 2y = 200 for y in terms of x

Substitute into A = xy

∴ when x = 50 m, the maximum area is 2500∴ y = 50 mThe rectangle is 50 m x 50 m.

A

A

A

A

A

= −

= −

= − +

= − − + +

= − − +

x x

x x

x x

x x

x

100

100

100

1 100 50 50

1 50 2500

2

2

2 2 2

2

b g

d ib g

2 2 200100100

x yx y

y x

+ =+ =

= −

x

y

y x x

x x x

x x

y x x

y x

= − +

= + − −

= − + +

= − − + + − − ×

= − − +

30 0 20 100

3000 30 20 0 2

0 2 10 3000

0 2 50 25 3000 0 2 25

0 2 25 3125

2

2

2 2 2

2

.

.

.

. .

.

b gb g

b ge j d ib g


Module 2

Lesson 1

Answer Key

1. a)

b)

c)

d)

e)

f)

g)

MultiplySimplify

2 3 2 0

2 4 3 6 0

2 6 0

2

2

x x

x x x

x x

− + =

+ − − =

+ − =

b gb g

Multiply each side by 3Subtract 3x from each side

xx

x x

x x

2

2

2

43

4 3

3 4 0

− =

− =− − =

Expand (x + 3)2

Subtract 5 from each side

Simplify

x

x x

x x

x x

+ − =

+ + − =

+ + =+ =

3 4 5

6 9 4 5

6 5 5

6 0

2

2

2

2

b g

Expand (x + 3)2

Add +4 to each side

x

x x

x x

+ = −

+ + = −

+ + =

3 4

6 9 4

6 13 0

2

2

2

b g

Subtract 7 from each side andrearrange

5 7 3 7

5 3 14 0

2

2

x x

x x

− + =

+ − =

Distribute 3x into the bracket

Subtract 8 from each side

3 2 8

3 6 8

3 6 8 0

2

2

x x

x x

x x

+ =

+ =

+ − =

b g

Add 6 to each side− + = −

− + =

− =

2 10 6

2 16 0

2 16 0

2

2

2

x

x

xor


Module 2

2. a) x2 + 2x – 8 = 0You can use a graphing calculator or complete the square.

b)

c)

4 2

2

4

2

4

x

y

8 610

4 2

2

4

2

4

2 4x

y

4 2

2

4

2

4

2 4x

y

6

8

6

8

9

x = {–4, 2}


Module 2

+ + =2 4 3 0x x

= − −{ 3, 1}x

+ = −2 8 15x x

= − −{ 5, 3}x

d) e)

f)

y

x5

–3

–1 31 7 9

–1

–5

1

V(6,–4)

( )

( )

− = − +

+ =

=

= = =

2

2

2

2

10 12 22

– 12 32 0 (Rearranged)

– 6 – 4 0

– 6 – 4 is shown below. Solutions are 4 and 8.

x x

x x

x

y x x x

2

2

4

2

4

2 4

y

6

8

6

8

10

10

6 84 2

2

4

2

4

2 4 x

y

6

8

10


Module 2

− =29 0x

= −{ 3,3}x

− + =22 12 10 0x x

= {1,5}x

g)

Left Bound: X = 0 Right Bound: 1.0638298 Guess: 0.42553191

Solution #1 = 0.59

Left Bound: X = 2.9787234Right Bound: 4.0425532Guess: 03.6170213

Solution #1 = 3.41

( )= − −

= − +

− + =

= +

2

2

2

21

–1 2 3

–1 4 4 – 3

4 2 0

Define Y X – 4X 2

x

x x

x x


Module 2

Lesson 2Answer Key


Module 2

1. a)

Check a):

Yes

Yes

Yes

Yes

Yes

Yes

Check: d)

x x

x x

2 12 0

4 3 0

− − =− + =b gb g

x xx x

− = + == = −

4 0 3 04 3

or or

d) x x

x x

2 9 18 0

6 3 0

+ + =+ + =b gb g

x xx x

+ = + == − = −

6 0 3 06 3or

or

4 4 12 016 4 12 0

3 3 12 0

9 3 12 0

2

2

− − =− − =

− − − − =

+ − =b g b g

− + − + =

− + =

− + − + =

− + =

6 9 6 18 0

36 54 18 0

3 9 3 18 0

9 27 18 0

2

2

b g b g

b g b g

Check: b)

Multiplyeach sideby (–1)

b) x x

x x

2 20 0

5 4 0

− − =− + =b gb g

x xx x

− = + == = −

5 0 4 05 4

or or

5 5 20 025 5 20 0

4 4 20 0

16 4 20 0

2

2

− − =− − =

− − − − =

+ − =b g b g

Yes

Yes

Check: c)c) − − + =

− + − =

− + − =

+ − =

x x

x x

x x

x x

2

2

2 3 0

1 2 3 0

1 3 1 0

3 1 0

d ib gb gb gb g

x xx x

+ = − == − =

3 0 1 03 1or

or

− − − − + =

− + + =

− − + =

− − + =

3 2 3 3 0

9 6 3 0

1 2 1 3 0

1 2 3 0

2

2

b g b g

b g b g

Yes

Yes

Check: e)

e) 2 3 2 0

2 1 2 0

2x x

x x

+ − =− + =b gb g

2 1 0 2 012

2

x x

x x

− = + =

= = −

or

or

( ) ( )

2

2

1 12 3 2 02 2

1 3 4 02 2 2

2 2 3 2 2 08 6 2 0

+ − =

+ − =

− + − − =− − =


Module 2

2. a) Check:

Yes

Yes

10 7 12 0

5 4 2 3 0

2x x

x x

− − =+ − =b gb g

5 4 0 2 3 045

32

x x

x x

+ = − =

= − =

or

or

1045

745

12 0

101625

285

12 0

325

285

605

0

1032

732

12 0

1094

212

242

0

452

212

242

0

2

2

−FHGIKJ − −FHG

IKJ − =

⋅ + − =

+ − =

FHGIKJ − FHG

IKJ − =

⋅ − − =

− − =

b) Check:

Yes

Yes

5 21 54 0

5 9 6 0

2x x

x x

+ − =− + =b gb g

5 9 0 6 095

6

x x

x x

− = + =

= = −

or

or

595

2195

54 0

815

1895

2705

0

5 6 21 6 54 0

180 126 54 0

2

2

FHGIKJ + FHG

IKJ − =

+ − =

− + − − =

− − =b g b g

c) Check:

Yes

Yes

3 2 1 5 0

3 6 5 0

2 7 5 0

2 5 1 0

2 2

2

x x x x

x x x x

x x

x x

− − + + =

− − − + =

− + =− − =

b g b g

b gb g2 5 0 1 0

52

1

x x

x x

− = − =

= =

or

or

2 7 5 0

2 52

7 52

5 0

252

352

102

0

2 1 7 1 5 0

2 7 5 0

2

2

2

x x− + =

FHGIKJ − FHG

IKJ + =

− + =

− + =

− + =b g b g


Module 2

d) Check:

Yes

Yes

x x

x x

x x

x x

2

2

2

92

212

0

92

52

0

2 9 5 0

2 1 5 0

+ − =

+ − =

+ − =− + =b gb g

2 1 0 5 012

5

x x

x x

− = + =

= = −

or

or

x x2

2

2

92

52

0

12

92

12

52

0

14

94

104

0

592

552

0

502

452

52

0

+ − =

FHGIKJ + FHG

IKJ − =

+ − =

− + FHGIKJ − − =

− − =

b g b g

f) Check:

Yes

16 64 0

16 4 0

16 2 2 0

2 2 0

2

2

x

x

x x

x x

− =

− =

− + =

− + =

d ib gb gb gb g

x xx x

− = + == = −

2 0 2 02 2

or or

16 2 64 0

64 64 0

16 2 64 0

64 64 0

2

2

b g

b g

− =

− =

− − =

− =

h) Check:

Yes

x x

x x

2 12 28 0

14 2 0

+ − =+ − =b gb g

x xx x

+ = − == − =

14 0 2 014 2or

or

− + − − =

− − =

+ − =

+ − =

14 12 14 20 0

196 168 28 0

2 12 2 28 0

4 24 28 0

2

2

b g b g

b g

e) x 2 9 0+ =

Does not factor — no solution in the set of real numbers

g) 3 16 10 02x x+ + =

Does not factor — no solution that can be found byfactoring.


Module 2

4.

x

x + 7 Let x = length of side of smallerboardx + 7 = length of side of largerboard

x x

x x x

x x

x x

x x

2 2

2 2

2

2

7 169

14 49 169

2 14 120 0

7 60 0

12 5 0

+ + =

+ + + =+ − =

+ − =+ − =

b g

b gb gx x= − =12 5or

–12 is impossible as a length∴ x = 5 cmx + 7 = 12 cmThe length of the sides of thesmaller board are 5 cm and thelength of the sides of the largerboard are 12 cm.

3. A

BC x

x + 7 17

Let x = shorter side of the righttrianglex + 7 = larger side of the righttriangle

Because the triangle has a rightangle,

a b c

x x

x x x

x x

x x

x x

x x

2 2 2

2 2 2

2 2

2

2

2

7 17

14 49 289

2 14 240 0

2 7 120 0

7 120 0

15 8 0

+ =

+ + =

+ + + =+ − =

+ − =

+ − =+ − =

b g

d i

b gb gx xx x

+ = − == − =

15 0 8 015 8

or or

x = –15 is not possible as a length∴ x = 8 m is the shorter sidex + 7 = 15 m is the longer side


Module 2

5. Let x = first odd integerx + 2 = second odd integerx + 4 = third odd integer

If x = –11 is first integerx + 2 = –9 is second integerx + 4 = –7 is third integer

If x = 5 is first integerx + 2 = 7 is second integerx + 4 = 9 is third integer

x x

x x

x x

x x

+ + =

+ + =

+ − =+ − =

2 4 63

6 8 63

6 55 0

11 5 0

2

2

b gb g


6. Let x = first positive numberx + 4 = second positive number

Only positive numbers are asked for∴ x = 6x + 4 = 10

x x

x x x

x x

x x

x x

2 2

2 2

2

2

4 136

8 16 136

2 8 120 0

4 60 0

10 6 0

+ + =

+ + + =+ − =+ − =

+ − =

b g


The two numbers are 6 and 10.


Module 2

8. x + 5

33

3

3

x – 6

1

33

x

x

(x + 5 – 6)

Let x = width of rectanglex + 5 = length of rectangle3 = height of boxx – 1 = length of boxx – 6 = width of box

450 1 6 3

150 1 6

150 7 6

0 7 144

0 16 9

2

2

= − −

= − −

= − += − −= − +

x x

x x

x x

x x

x x

b gb gb gb g

b gb gx x= = −16 9or

Volume = lwh

Since the width cannot be negative, x = 16 cm wide and x + 5 = 21 cm long.

7.

40

30

x

Let x = width of the pathway in m 2x + 40 = length of the gardenand pathway2x + 30 = width of the garden andpathway

2 40 2 30 984 1200

4 140 1200 984 1200

4 140 984 0

35 246 0

41 6 0

2

2

2

x x

x x

x x

x x

x x

+ + − =

+ + − =+ − =

+ − =+ − =

b gb g


Area of pathway and garden – area of pathway = area of garden

Since the width cannot be negative, x = 6 is the width of thepathway.

x

x

x

Lesson 3

Answer Key

1. a) x2 – 2x – 5 = 0 b) 3x2 – 2x + 5 = 0a = 1, b = –2, c = –5 a = 3, b = –2, c = 5

c) 5x2 – 3x – 8 = 0 d) 2(x2 – 2x) – 1 = 0a = 5, b = –3, c = –8 2x2 – 4x – 1 = 0

a = 2, b = –4, c = –1

e) 5x2 = 9x f) 4 – 2x2 = 9x5x2 – 9x = 0 –2x2 – 9x + 4 = 0 a = 5, b = –9, c = 0 a = –2, b = –9, c = 4 or

2x2 + 9x – 4 = 0a = 2, b = 9, c = –4

2. a) x2 + 2x – 15 = 0 b) 2w2 – 3w + 1 = 0a = 1, b = 2, c = –15 a = 2, b = –3, c = 1


Module 2

xb b ac

a

x

= − ± −

=− ± − −

= − ± +

= − ±

= − ±

= − + − −

= −

2

2

42

2 2 4 1 15

2 1

2 4 602

2 642

2 82

2 82

2 82

3 5

b gb gb g

or

or

wb b ac

a

w

w

w

w

= − ± −

=+ ± − −

= ± −

= ±

=

2

2

42

3 3 4 2 1

2 2

3 9 84

3 14

112

b g b gb gb g

or

c) 7w2 – 3w = 0 d) 5x2 – 1 = 0a = 7, b = –3, c = 0 a = 5, b = 0, c = –1

e) x2 – 0.1x – 0.06 = 0 f) –x2 – 7x – 1 = 0a = 1, b = –0.1, c = –0.06 a = –1, b = –7, c = –1


Module 2

wb b ac

a

w

w

w

w

= − ± −

=+ ± − −

= ± −

= ±

=

2

2

42

3 3 4 7 0

2 7

3 9 014

3 314

037

b g b gb gb g

or

xb b ac

a

x

x

x

x

= − ± −

=± − −

= ±

= ±

= ±

2

2

42

0 0 4 5 1

2 5

20102 510

55

b gb gb g

xb b ac

a

x

x

x

x

x

x

= − ± −

=± − − −

= ± +

= ±

= ±

= −

= −

2

2

42

01 01 4 1 0 06

2 1

01 0 01 0 242

01 0 252

01 052

0 62

0 42

03 0 2

. . .

. . .

. .

. .

. .

. .

b g b gb gb g

or

or

xb b ac

a

x

x

x

x

= − ± −

=+ ± − − − −

−

= ± −−

= ±−

= ±−

2

2

42

7 7 4 1 1

2 1

7 49 42

7 452

7 3 52

b g b gb gb g

3. a) 3x2 – 6x – 5 = 0 b) 2x2 – 4x – 1 = 0a = 3, b = –6, c = –5 a = 2, b = –4, c = –1

c) 9x2 – 8x – 7 = 0 d) 2x2 – x – 3 = 0a = 9, b = –8, c = –7 a = 2, b = –1, c = –3


Module 2

xb b ac

a

x

x

x

x

x

x

= − ± −

=+ ± − − −

= ± +

= ±

= ±

=±

= ±

2

2

42

6 6 4 3 5

2 3

6 36 606

6 966

6 4 66

2 3 2 6

2 3

3 2 63

b g b gb gb g

e jb g

x

x

x

x

x

x

=+ ± − − −

= ± +

= ±

= ±

=±

= ±

4 4 4 2 1

2 2

4 16 84

4 244

4 2 64

2 2 6

2 2

2 62

2b g b gb gb g

e jb g

xb b ac

a

x

x

x

x

x

= − ± −

=+ ± − − −

= ± +

= ±

= ±

= ±

2

2

42

8 8 4 9 7

2 9

8 64 25218

8 31618

8 2 7918

4 799

b g b gb gb g

xb b ac

a

x

x

x

x

x

= − ± −

=+ ± − − −

= ±

= ±

= + −

= −

2

2

42

1 1 4 2 3

2 2

1 254

1 54

1 54

1 54

32

1

b g b gb gb g

or

or

4. a) f(x) = 5x2 – x – 3 b) 0 = 2x2 + 6x – 10 = 5x2 – x – 3 a = 2, b = 6, c = –1a = 5, b = –1, c = –3

5. 3x2 – 5x – 1 = 0a = 3, b = –5, c = –1

Use calculator for decimals: x ≈ 1.8, –0.2.


Module 2

xb b ac

a

x

x

x

x

= − ± −

=+ ± − − −

= ± +

= ±

= + −

2

2

42

1 1 4 5 3

2 5

1 1 6010

1 6110

1 6110

1 6110

b g b gb gb g

or

xb b ac

a

x

x

x

x

x

x

= − ± −

=− ± − − −

= − ± +

= − ±

= − ±

= − ±

= − + − −

2

2

42

6 6 4 2 1

2 2

6 36 84

6 444

6 2 114

3 112

3 112

3 112

b g b gb gb g

or

xb b ac

a

x

x

x

x

= − ± −

=+ ± − − −

= ± +

= ±

= + −

2

2

42

5 5 4 3 1

2 3

5 25 126

5 376

5 376

5 376

b g b gb gb g

or

6. a) 6x2 + 5x – 6 = 0 b)a = 6, b = 5, c = –6

Preference: Factoring is much more efficient.

7. a) 0 = x2 + 8x + 15 b) a = 1, b = 8, c = 150 = (x + 3) (x + 5)x = –3, –5

c)


Module 2

xb b ac

a

x

x

x

x

= − ± −

=− ± + − −

= − ±

= − ±

= −

2

2

42

5 5 4 6 6

2 6

5 16912

5 1312

23

b g b gb gb g

or 32

6 5 6 0

2 3 3 2 0

2x x

x x

+ − =+ − =b gb g

2 3 02 3

32

xx

x

+ == −

= −

3 2 03 2

23

xx

x

− ==

=

x

x

x

x

= − ± −

= − ±

= − ±

= − −

8 64 602

8 42

8 22

3 5or

4 2

2

4

2

4

x

y

8 6x = –3, –5

8. a) 3(x + 3)2 + 5(x + 3) + 8 = 0

b)

c) (x2 – 1)2 – 6(x2 – 1) – 7 = 0

d)

e) no quadratic pattern

9. a)

Replace p by x2 + 2x

The solution set is {–3, –1, 1}.


Module 2

2 3 3 2 3 0

2 3 3 2 3 02

x x

x x

− + − =

− + − =e j

x x

x x

+ =

+ − =

6 4

6 4 0

12

12

12

2

e j

x x x x

p x x

2 2 2

2

2 2 2 3 0

2

+ − + − =

= +

d i d iLet

p p

p p

2 2 3 0

3 1 0

− − =− + =b gb g

p p= = −3 1or

x x

x x

x x

x x

2

2

2 3

2 3 0

3 1 0

3 1

+ =

+ − =+ − =

= − =b gb g

or

x x

x x

x x

x

2

2

2 1

2 1 0

1 1 0

1

+ = −

+ + =+ + =

= −b gb g

or

b)

c)


Module 2

2 4 3 0

2 4 3 0

2 1

1 2 1

x x

x x

− −

− −

+ + =

+ + =d iLet p x

p pa b c

=

+ + == = =

−1

22 4 3 02 4 3, ,

pb b ac

a= − ± −

=− ± −

= − ± −

= − ± −

2

2

42

4 4 4 2 3

2 2

4 16 244

4 84

b gb gb g

∴ there are no real solutions

x x

x x

x x

p x

p pa b c

= −

− + =

− + =

=− + =

= = − =

6 2

6 2 0

6 2 0

6 2 01 6 2

2

2

e jLet

, ,

2 42

6 36 82

6 282

6 2 7 or 3 72

3 7 or 3 7 now square both sides

b b acpa

x x

− ± −=

± −=

±=

±= ±

= + = −

x x

x x

= + + = − +

= + = −

9 6 7 7 9 6 7 7

16 6 7 16 6 7

or

or

Solution set = ±16 6 7o t


Module 2

Let

or

px

xp p

p p

p

= −

− + =− − =

=

1

3 2 0

2 1 0

2 1

2

b gb g

d)

or

Replace by px

x−1

xx

x xx

− =

− =− =

12

1 21

xx

x x

− =

− =− =

11

11 0

∴ no solution

∴ −solution set 1l q

Lesson 4

Answer Key

1. a) F — The discriminant of 2x2 + 5x + 6 is –23.b) Tc) T

d) F

e) F — The equation x2 + 2x + 1 has one real root since b2 – 4ac = 0.


Module 2

− + = − +6 3210

3 2 25

2. Discriminanta) –15b) 25c) –9d) 0e) 50

Characteristics of Rootsa) no real rootsb) 2 real rational rootsc) no real rootsd) 1 real roote) 2 real irrational roots

3. y = ax2 + bx + c = 0If discriminant value isa) negativeb) zeroc) positive

Graph intercepts x-axisno timesonce (is tangent to x-axis)two times

4. a) x x2 8 16 0− + =

a b c= = − =1 8 16, ,

b ac2 24 8 4 1 16

64 640

− = − −

= −=

b g b gb g

Because b2 – 4ac = 0, there is one real root.


b) a a2 2 7 0+ + =

a b c= = =1 2 7, ,

b ac2 24 2 4 1 7

4 2824

− = −

= −= −

b g b gb g

Because b2 – 4ac < 0, there are no real roots.

c) p2 16 0− =

a b c= = = −1 0 16, ,

b ac2 24 0 4 1 16

64

− = − −

=b gb g

Because b2 – 4ac > 0 and equal to a perfect square, there aretwo real rational roots.

d) 2 5 02x x+ − =

a b c= = = −2 1 5, ,

b ac2 24 1 4 2 5

1 4041

− = − −

= +=

b gb g

Because b2 – 4ac > 0 and equal to a non-perfect square, thereare two real irrational roots.

5. a) xx

2

24 4 0+ + =

x x2 8 8 0+ + =

Because b2 – 4ac > 0 and not a perfect square, there aretwo irrational roots.

Multiply through by 2:

a b c= = =1 8 8, ,

b ac2 24 8 4 1 8

64 3232

− = −

= −=

b gb g


Module 2

b) xx

− − − =12

3 02

x x

x x

− − − =

− + − =

1 2 6 0

2 7 0

2

2


Multiply each term by 2:

a b c= − = = −2 1 7, ,

b ac2 24 1 4 2 7

1 5655

− = − − −

= −= −

b gb g

c) 2 3 4

2 6 4

2 4 6 0

2

2

2

x x

x x

x x

− =

− =− − =

d i

Because b2 – 4ac > 0 and a perfect square, it has two realrational roots.

a b c= = − = −2 4 6, ,

b ac2 24 4 4 2 6

16 4864

− = − − −

= +=

b g b gb g

d) 6 2 02x x− + =


a b c= = − =6 1 2, ,

b ac2 24 1 4 6 2

1 4847

− = − −

= −= −

b g b gb g


Module 2

e) 4 12 9 02x x− + =

Because b2 – 4ac = 0, there is one real root.

a b c= = − =4 12 9, ,

b ac2 24 12 4 4 9

144 1440

− = − −

= −=

b g b gb g

6. 3 3 02x mx− + =

If the roots are not real, b2 – 4ac < 0 a b m c= = − =3 3, ,

− − <

− <

m

m

b g b gb g2

2

4 3 3 0

36 0

− < < −6 6 6m mor is in the interval ( , 6)

7. a) kx x2 6 2 0− + =If there is one root, b2 – 4ac = 0 a k b c= = − =, ,6 2

− − =

− =− = −

6 4 2 0

36 8 08 36

2b g b gb gk

kk

k = 368

92

or

b) x k x2 8 9 0+ − + =b gIf there is one root, b2 – 4ac = 0 a b k c= = − =1 8 9, ,

k

k k

k k

k k

− − =

− + − =

− + =− − =

8 4 1 9 0

16 64 36 0

16 28 0

2 14 0

2

2

2

b g b gb g

b gb gk k= =2 14or


Module 2

8.

9. a) Two real rootsb) No real rootsc) One root

10.

2 4 2 02 2x x k k+ + − − =d iThe equation has one root when b2 – 4ac = 0.

a b c k k= = = − −2 4 2 2, ,

4 4 2 2 0

16 16 8 8 0

8 8 0

8 1 0

2 2

2

2

− − − =

− + + =

+ =+ =

b gd i

b g

k k

k k

k k

k k

8 0 1 00 1

k kk k

= + == = −

or or

x

y

11.

x

y


Module 2

12. a) If the equation has two real roots, b2 – 4ac > 0a = 3, b = –2, c = k

b) If the equation has a double root, b2 – 4ac = 0.a = 1, b = k, c = k + 2

c) If the equation has no realroots, then b2 – 4ac < 0.a = k, b = 8, c = –4

b ac

k

kk

k

2

2

4 0

2 4 3 0

4 12 012 4

13

− >

− − >

− >− > −

<

b g b g

Remember that when dividing aninequality by a negative number,the inequality sign reversesdirection.

b ac

k k

k k

2

2

2

4 0

4 1 2 0

4 8 0

− =− + =

− − =

b gb g

a = 1, b = –4, c = –8

kb b ac

a= − ± −

=± − − −

= ± +

= ±

= ±

= ±

2

2

42

4 4 4 1 8

2 1

4 16 322

4 482

4 4 32

2 2 3

b g b gb gb g

b ac

k

kkk

2

2

4 0

8 4 4 0

64 16 016 64

4

− <− − <

+ << −< −

b gb g

Lesson 5

Answer Key

1. a) True

b) False. For each real x, is a real number if andonly if x ≥ 3.

c) False. For each real x,

d) False. The real number solution set of is theempty set.

e) True

f) False. For each real number x, if x2 = 16 then

g) False. Squaring both sides of


Module 2

2.

3. a)

x − 3

x x2 =| |.

x 2 3= −

x x= = −16 16or .

x x+ =2 3 yields

x x x+ + =4 4 9 2.

a)

b)

c)

2 1 2 1

5 25 10

2 5 2 5 2 5

4 4 5 5

4 5 1

2

2

2

x x

x x x

x x x

x x

x x

− = −

+ = + +

+ − = + − + −

= + − + −

= − + −

e je je j e je j

x x

x x x

x x

x x

+ = +

+ + = +

+ − =+ − =

2 2 7

4 4 2 7

2 3 0

3 1 0

2 2

2

2

b g e j

b gb g

Check:

x = −3 1,

x = −

− + = − +

− =− ≠

3

3 2 2 3 7

1 11 1

b g b g

Check:x =

+ = +

==

1

1 2 2 1 7

3 93 3

b g b g

∴ =x 1reject


Module 2

b) x x

x x

x x x

x x

x

x

= − −

− = − −

− + = −− + =

− =

=

2 2 5

2 2 5

4 4 2 5

6 9 0

3 0

3

2 2

2

2

2

b g e j

b g

Check:

No real number solutions or ∅.

x =

= − −

= −≠ −

3

3 2 2 3 5

3 2 13 2 1

b g

c) 2 3 1 1

2 3 1 1 1

2 3 1 2 1 1

2 3 2 2 1

1 2 1

1 2 1

2 1 4 4

2 3 0

3 1 0

3 1

2 2

2 2

2

2

x x

x x

x x x

x x x

x x

x x

x x x

x x

x x

x

+ − + =

+ = + +

+ = + + + +

+ = + + +

+ = +

+ = +

+ + = +

− − =− + =

= −

e j e j

b g e j

b gb gor

Check:x =

+ − + =

− =

3

2 3 3 3 1 1

3 2 1

b g

Check:x = −

− + − − + =

− =

1

2 1 3 1 1 1

1 0 1

b g

∴ = −x 3 1or


Module 2

d) x

x

x

xx

2

22 2

2

2

3 1 0

3 1

3 1

42

− + =

− = −

− =

== ±

e j b g

Check:x =

− + =+ ≠

2

2 3 1 01 1 0

2

Check:x = −

− − + =

+ ≠

2

2 3 1 0

1 1 0

2b g

e) x x

x x

x x x

x x

x x

x

= − +

− = −

− + = −− + =

− − =

=

3 2 2

2 3 2

4 4 3 2

7 6 0

6 1 0

6 1

2 2

2

2

b g e j

b gb gor

Check:x =

= − +

= += +

6

6 3 6 2 2

6 16 26 4 2

b g

Check:x =

= − +

= +≠ +

1

1 3 1 2 2

1 1 21 1 2

b g

∴ ∅ or empty set

∴ =x 6


Module 2

f)x x

x x

x x

x x

22

2

22 2

2

6 2

6 4

6 16 0

8 2 0

+FH

IK =

+ =

+ − =+ − =

b g

e j b g

b gb g

Check:x = −

− + − =

− =

=

8

8 6 8 2

64 48 2

16 2

2b g b g

Check:x =

+ =

+ =

=

=

2

2 6 2 2

4 12 2

16 2

4 2

2 b g

x x= − =8 2or

g) 3 2 3 2

3 2 9 6 2 2

3 9 2 2 6 2

66

6 26

2

2

2 0

2 0

2 2

2 2

2

2

x x

x x x

x x x

x x

x x

x x

x x

x x

+ = −

+ = − +

− + − = −

−−

= −−

=

=

− =− =

e j e j

b g e j

b g

Check:x =

+ = −

≠ −

0

3 0 2 3 0 2

2 2

b g

Check:x =

+ = −

=

=

2

3 2 2 3 2 2

8 2 2

2 2 2 2

b g

x x= =0 2or

∴ = − =x x8 2or

∴ =x 2


Module 2

h) 1 1

1 2 1

2

4 4

4 5 0

4 5 0

2 2

2

22 2

2 2

2

− + = +

− + − + = +

− =

− =− =− =

x x x

x x x x x

x x x

x x x

x x

x x

e j e j

e j b g

b g

Check:x =

− + =

0

1 0 0 1

Check:

145

45

45

1

15

45

95

35

35

− + = +

+ =

=

x x x= − = ⇒ =0 4 5 045

or

∴ = =x x045

or


Module 2

Notes

Lesson 6

Answer Key

1. a) True

b) False.

c) True

d) True


Module 2

2. a) x xx

x x

x x

x x

x x

− = −−

− ≠ +

− = −− + =

− − =

323

3 3

3 2

3 2 0

2 1 0

2

2

b g b g b g

b gb g

, if

x x= =2 1

If , then x xx

xx x≠ −

−+F

HGIKJ = + −3 2 3

63 2

12 32b g .

Check:

If x = 2

xx

= −−23

22

2 3

221

= −−

= −−

If x = 1

12

1 3

122

= −−

= −−

b) 2 97 2

57

2 7 2 97

2 72

5 2 77

xx

xx

x xx

x x xx

−−

+ =−

− −−

+−

=−

−b gb gb g

b gb g b gb gb g

4 18 7 10

3 18 10 0

3 28 0

7 4 0

2

2

2

x x x

x x

x x

x x

− + − =− − − =

− − =− + =b gb g

LCD is 2(x – 7) and x – 7 ≠ 0, so x ≠7

x x= = −7 4or

It is not possible for x = 7.∴ x = –4

∴ = =x x2 1or

Check x = – 4

2 4 94 7

42

54 7

1711

2511

1711

2211

511

− −− −

+ − =− −

−−

− =−

− = −

b g


Module 2

c) x x xx

x

x x

x x

− = −FHGIKJ ≠

− = −

− + =

41

0

4 1

4 1 0

2

2

b g

a b c= = − =1 4 1, ,

Not factorable

xb b ac

a

x

= − ± −

=+ ± − −

= ± −

= ±

= ±

= ±

2

2

42

4 4 4 1 1

2 1

4 16 42

4 4 32

4 2 32

2 3

b g b gb gb g

d) 33 1

3 1 2 3 12 13 1

3 1

3 6 2 2 1

3 8 3 0

3 1 3 0

2

2

2

xx

x xxx

x

x x x

x x

x x

+FHGIKJ + − + =

++

+

− − = +− − =

+ − =

b g b g b gb g b g

b gb g3 1 0 3 0x x+ = − =

3 1 033

131

3

xx

x

+ ≠

≠ −

≠ −

because the denominator ≠ 0

Check:

x = +

+ − = −+

− =− −

+ −

− = − +

− − = −−

− − =− +

− +

− − = − −

2 3

2 3 41

2 3

3 21 2 3

2 3 2 3

3 2 2 3

2 3 41

2 3

3 21 2 3

2 3 2 3

3 2 2 3

e je je j

e je je j

∴ = + −x 2 3 2 3or

Check: x = 3

3 33 3 1

22 3 13 3 1

2710

2 710

710

710

2⋅+

− =++

− =

=

b gb gb g

∴ =x 3

Check: x = −2 3


Module 2

e) 23

3 2 31

2 32 3 3

3 92 3 3

2 3 3 0 2 3 3

xx

x xx

x x

xx x

x x x x

−− + +

++ − +

++ −

+ − = + −

b g b gb g b g b gb gb gb gb g b gb g b gb g

4 6 3 3 9 0

4 10 6 022

2 5 302

2 3 1 0

2

2

2

x x x x

x x

x x

x x

+ + − + + =+ + =

+ + =

+ + =

d ib gb g

2 3 03

2

x

x

+ =

= −or

because the denominator ≠ 0

f)

3. a)

xx

xx

x

x x

x x

x x

2

2

2

123

73

3

12 7

7 12 0

4 3 0

+−

=−

≠+ =

− + =− − =b gb g

x x= =4 3

3 12

3 12 3 124 4

x

x xx x

=

= = −= = −

or or

b) 2 1 17

2 18

2 18 2 189 9

x

x

x xx x

− =

=

= = −= = −

or or

Check: x = –1

2 11 3

12 1 3

3 1 9

2 1 3 1 90

24

164

0

32

32

0

1

2

−− −

+− +

+− +

− − − −=

−−

+ +−

=

− =

∴ = −

b gb gb g

b g b g

x

x

x x

+ =

= − ≠ −1 0

13

2, but

Check: x = 4

4 124 3

7 44 3

28 28

2 +−

=−

=

b g

∴ =x 4

Check:

Check:

3 4 12

12 12

b g =

=

2 9 1 17

18 1 17

⋅ − =

− =

or

or

3 4 12

12 12

− =

=

b g

2 9 1 17

18 1 17

− − =

− =

b g


Module 2

c) 5 2 3x + = −

(Taking the absolute value of any quantity will neverresult in a negative answer.)

d) x x

x x

x x

2

2

4 12 0

4 12 0

6 2 0

+ − =

+ − =+ − =b gb g

x x= − =6 2or

Check:36 24 12 0

4 8 12 0

− − =

+ − =

e) x

x x

234

112

122

34

112 12

34

112

12

− =

−FHGIKJ = FHG

IKJ −FHG

IKJ = −

12 or 12b g b g

6 9 16 10

10653

xx

x

x

− ==

=

=

6 9 166

8643

xx

x

− = −

=

=

Check:

532

34

112

56

34

112

10 912

112

− =

− =

− =

or 432

34

112

46

34

8 912

112

− =

− = − =

The solution set is φ where φ means empty set.


Module 2

f)

4. Let x = smaller positive integerx + 4 = larger positive integer

Lowest common denominator: 15x(x + 4)

x x

x xx

x

− = +

− = +− =

− =

5 3 7

5 3 712 26

Check: Check:− − = − +

− = − +

− = −

6 5 3 6 7

11 18 7

11 11

b g

x x

x xx

x

− = − +

− = − −

= −

= −

5 3 7

5 3 744

241

2

b g

− − = −FHGIKJ +

− =

12

5 31

27

512

512

1 14

115x x

−+

=

15 41

15 41

415 4

115

15 4 15 4

15 60 15 4

0 4 60

0 10 6

10 6

2

2

x xx

x xx

x x

x x x x

x x x x

x x

x x

x

+ ⋅ − + ⋅+

= + ⋅

+ − = +

+ − = +

= + −= + −

= −

b g b g b gb g b g

b gb gor

or

∴ = − = −x x612

or

Because you are considering positive integers, x ≠ –10, sox = 6x + 4 = 10∴ the two positive integers are 6 and 10


Module 2

Let x = number > 2.

Lowest common denominator: x

5.

11 4

xx+ =

xx

x x x

x x

x x

⋅ + ⋅ =

+ =

− + =

14

1 4

4 1 0

2

2

a b c= = − =1 4 1, ,

xb b ac

a= − ± −

=− − ± − −

= ± −

= ±

= ±

= ±

2

2

42

4 4 4 1 1

24 16 4

24 12

24 2 3

22 3

b g b g b gb g

If the number is , then > = +2 2 3x .

Review

Answer Key

1. a) b)

c) d)

e) f)

x2 + 1 = 0 has no solution inthe set of reals

x = 1x = 5

2xx

== −

11

3 6 5 0

2 7 5 0

2 5 1 0

2 2

2

x x x x

x x

x x

− − − + =− + =

− − =b gb g

x

x x

x x x

4

2 2

2

1 0

1 1 0

1 1 1 0

− =

− + =

− + + =

d id ib gb gd i

( )( )− + =

− = + =−= =

−= = =

2 1 2 1 0

2 1 0 2 1 01 12 2

1 10, ,2 2

x x

x x

x x

x x x

x = 12

x = −43

8 00

xx

==

Multiply by −+ − =

+ − =

1

6 5 4 0

3 4 2 1 0

2x x

x xb gb g

32 8 0

8 4 1 0

3

2

x x

x x

− =

− =d i

xx

+ == −

2 02

2 9 092

x

x

− =

=

xx

− ==

2 02

3 00

xx

==

2 5 18 0

2 9 2 0

2x x

x x

− − =− + =b gb g

3 6 0

3 2 0

2x x

x x

− =− =b g


Module 2

2. a) x2 – 2x – 8 = 0

b) x2 – 8x + 15 = 0

x

y

(4 , −1)

x

y

(1 , −9)


Module 2

roots are at –2 and 4

roots are at 3 and 5

c) x2 – 4 = 0

3. a)

b) 2

2

2

2

2

2 3 8 02 3 832 8 Dividing both sides by 223 9 942 16 16

3 734 16

3 734 4

3 734 43 73

4

x xx x

x x

x x

x

x

x

x

− − =

− =

− + =

− + = +

− =

− = ±

= ±

±=

x x

x x

x x

x

x

x

2

2

2

2

6 2 0

6 2

6 9 2 9

3 11

3 11

3 11

+ − =+ =

+ + = +

+ =

+ = ±

= − ±

b g

x

y


Module 2

roots are at –2 and 2

4. a) a = –1, b = –7, c = –1

b) a = 2, b = 4, c = 1

5. a) Find the value of b2 – 4aca = 4, b = –12, c = 9

There is 1 real root.

b ac

b ac

2 2

2

4 12 4 4 9

144 1440

4 0

− = − −

= −=

− =

b g b gb g

x =− ± −

⋅

= − ±

= − ±

= − ±

4 4 4 2 1

2 2

4 84

4 2 24

2 22

2 b gb g

xb b ac

a

x

= − ± −

=− − ± − − − −

−

= ± −−

= ±−

= ±−

2

2

42

7 7 4 1 1

2 1

7 49 42

7 452

7 3 52

b g b g b gb gb g


Module 2

b)

Since b2 – 4ac is a perfect square, there are 2 rational roots.

c)

Since b2 – 4ac < 0, there are no real roots.

d)

Since b2 – 4ac < 0, there are no real roots.

6.

k = 0, k = –1

( )

( )( )

( )

2 2

2

2

2 2

2

2

3 6 3 0

3, 6, 34 0

6 4 3 3 0

36 36 12 12 012 12 012 1 0

x x k k

a b c k kb ac

k k

k kk kk k

+ + − − =

= = = − −− =

− − − =

− + + =

+ =+ =

( ) ( )( )

2

22

6 2 06, 1, 2

4 1 4 6 21 48 47

x xa b c

b ac

− + == = − =

− = − −= − = −

( ) ( )( )

2

2

22

1 2 6 02 7 02, 1, 7

4 1 4 2 71 56

55

x xx x

a b c

b ac

− − − =

− + − == − = = −

− = − − −= −= −

( ) ( )( )

2

2

22

2 6 42 4 6 0

2, 4, 6

4 4 4 2 616 4864

x xx x

a b c

b ac

− =

− − == = − = −

− = − − −= +=


Module 2

7. 3x2 + kx + 12 = 0 If there are two different real roots then b2 – 4ac > 0a = 3, b = k, c = 12

8. 2x2 – 5x = k If there are no real roots then b2 – 4ac < 0a) b) The graph does not cross

the x-axis.

9. 7x2 + 5 = 10x Rearrange to 7x2 – 10x + 5 = 0a = 7, b = –10, c = 5discriminant is b2 – 4ac > (–10)2 – 4(7)(5) = 100 – 140 = –40

− − == = − = −

− <

− − − <+ <

< −−<

2

2

2

2 5 02 5

4 0

( 5) 4(2)( ) 025 8 0

8 25258

x x ka b c k

b ac

kkk

k

− >

− >

− >

>> < −

2

2

2

2

4 0

4(3)(12) 0

144 0

14412 or 12

b ac

k

k

kso k k


Module 2

10. a)

Multiply each term by LCM and simplify.

x = –3, x = 1x ≠ 1 because the denominator would become 0∴ x = –3

b)

Check:

Solution: x = 29

∴ x = 2 is extraneous

x =

⋅ + = ⋅ +

= +≠

2

2 2 5 2 2 2 1

9 4 13 5

x =

⋅ + = ⋅ +

= +

= ⋅ +

=

29

229

5 2 229

1

499

249

1

73

223

1

73

73

x x= =29

2,

square both sidesisolate radical termsquare

divide by 4

2 5 2 2 1

2 5 8 4 2 1

4 6 4 2

16 48 36 32

36 80 16 0

9 20 4 0

9 2 2 0

2

2

2

x x

x x x

x x

x x x

x x

x x

x x

+ = +

+ = + +

− =− + =

− + =

− + =− − =b gb g

1 1 1 2 1 2 2

1 3 2 2 4

2 3 0

3 1 0

2

2

x x x x

x x x x

x x

x x

− + + + = +

− + + + = +

+ − =+ − =

b g b gb g b g

b gb g

12 1

11

21 1

2 1 1

x x x x x

x x x

+ ++

−=

− +

+ + −

b gb g b gb gb gb gb gLCM:


Module 2

c)

x = 6, x = 2Check: x = 6 Check: x = 2

x = 6 or x = 2

d)

Multiply by x(x + 3)(x + 1)

x = 2 or x = 3

e)

Let y = x2 + 2x

y = –4 or y = 3or x x

x x

x

2 2 3

3 1 0

3 1

+ =+ − =

= −b gb g

or

x x

x x

2

2

2 4

2 4 0

+ = −

+ + =φ

y y

y y

2 12 0

4 3 0

+ − =+ − =b gb g

x x x x2 2 22 2 12+ + + =d i d i

5 1 2 3 1 6 3

5 5 2 4 3 6 18

5 5 2 8 6 6 18

5 6 0

2 3 0

2 2 2

2 2 2

2

x x x x x x

x x x x x x

x x x x x x

x x

x x

+ + + + = +

+ + + + = +

+ + + + = +− + =

− − =

b g b gb g b gd i

b gb g

53

2 61x x x+

+ =+

6 2 1 4 32 2

− = + −=

18 2 12 3 14 3 1

− − − =− =

3 2 2 3 1

3 2 1 2 3

3 2 1 2 2 3 2 3

2 2 3

2 2 3

4 2 3

8 12 0

6 2 0

22

2

2

x x

x x

x x x

x x

x x

x x

x x

x x

− − − =

− = + −

− = + − + −

= −

= −

= −

− + =− − =

e jb g

b gb g


Module 2

f)

x = 4 or –2Check: x = 4 Check: x = –2|42 – 2(4) – 8| = 0 |(–2)2 – 2(–2) – 8| = 0x = 4 x = –2

g)

Check:

11. Let x = first positive integerx + 1 = second positive integer

x = –9 or x = 8Since x is positive, then the two integers are 8 and 9.

x x

x x

x x

+ =

+ − =+ − =

1 72

72 0

9 8 0

2

b g

b gb g

∴ = −x 8

23

or

− + = −FHGIKJ −

= −

=

23

5 22

33

133

133

133

133

8 5 16 3

13 13

+ = −

=

x x

x xx

x

+ = − −

+ = − += −

= −

5 2 3

5 2 33 2

23

b gx x

x xxx

+ = −

+ = −− = −

=

5 2 3

5 2 38

8

x x

x x

2 2 8 0

4 2 0

− − =− + =b gb g


Module 2

12.

x = 9 or x = 1x ≠ 9 since this would result in the side measuring (6 – x) to be negative, and a side cannot have a negative length∴ x = 1

13. Let x be the lesser number and x + 1 be the greater.

( )

( ) ( ) ( )

( ) ( ) ( )( )

( )( )( )

+ = = ++

+ + + = + + + + = +

+ + = +

+ =

± ± + ±= = = =

+= = = = = =

+ = + = +

2

2

2

1 1 8L.C.D. 3 1

1 31 1 8

3 1 3 1 3 11 3

3 1 3 1 1 8

3 3 3 8 8

8 2 – 3 0

–2 2 – 4 8 –3 –2 4 96 –2 100...

2 8 16 16

–2 10 8 –2 – 10 1 –2 – 10 –12 –3... or = .

16 16 16 2 16 16 41 1 2

1 12 2

x xx x

x x x x x xx x

x x x x

x x x x

x x

x

x = 3.

2 2

1 3Therefore, the numbers are and .

2 2

14 13 6

196 28 169 26 36 12

0 10 9

9 1 0

2 2 2

2 2 2

2

− = − + −

− + = − + + − +

= − +− − =

x x x

x x x x x x

x x

x x

b g b g b g

b gb g


Module 2

We reject the negative answer (question 14says that the two numbers are positive).

Module 2: Answer Key - Open School BCmedia.openschool.bc.ca/.../pdf/pma11_mod2_SMkey.pdf · Module...

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