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
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 1 175
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)
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 2 177
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.
178 Section 1, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 2 179
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
180 Section 1, Answer Key, Lesson 2 Principles of Mathematics 11
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)
Principles of Mathematics 11 Section 1, Answer Key, Lesson 2 181
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
182 Section 1, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 3 183
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)
184 Section 1, Answer Key, Lesson 3 Principles of Mathematics 11
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 .
Principles of Mathematics 11 Section 1, Answer Key, Lesson 3 185
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
Equation of axis ofsymmetry
Domain
Range
x-intercepts
y-intercepts
Direction of opening
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
186 Section 1, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 3 187
Module 2
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 4 189
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
190 Section 1, Answer Key, Lesson 4 Principles of Mathematics 11
Module 2
= 2y x( )= − 21
42
y x
4. Complete the table.
Vertex
Equation of axis ofsymmetry
Domain
Range
x-intercepts
y- intercepts
Direction of opening
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 4 191
Module 2
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 5 193
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
Direction of opening
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)
194 Section 1, Answer Key, Lesson 5 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 5 195
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
196 Section 1, Answer Key, Lesson 5 Principles of Mathematics 11
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)
Principles of Mathematics 11 Section 1, Answer Key, Lesson 5 197
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
Lesson 6Answer Key
1. x2 + 5x + 4
2.
Principles of Mathematics 11 Section 1, Answer Key, Lesson 6 199
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
200 Section 1, Answer Key, Lesson 6 Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 1, Answer Key, Lesson 6 201
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
202 Section 1, Answer Key, Lesson 6 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 6 203
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.
204 Section 1, Answer Key, Lesson 6 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 6 205
Module 2
vi) Minimum value is − 154
v) y-intercept is –3
aba
⋅ FHGIKJ2
2
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)
Principles of Mathematics 11 Section 1, Answer Key, Lesson 7 207
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.
208 Section 1, Answer Key, Lesson 7 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 7 209
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, ∞)
210 Section 1, Answer Key, Lesson 7 Principles of Mathematics 11
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
i) x-coordinate of vertex =
y-coordinate
V(2, –64)
ii) Axis of symmetry equation: x = 2
iii) x-intercepts
iv) y-intercept (let x = 0) = –60
v) Opens upward since a > 0
vi) Since the parabola opens upward, the function has aminimum value at –64.
vii) D: (–∞, ∞)R: [–64, ∞)
Principles of Mathematics 11 Section 1, Answer Key, Lesson 7 211
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
i) x-coordinate of vertex =
y-coordinate
V(–2, –18)
ii) Axis of symmetry equation: x = –2
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
212 Section 1, Answer Key, Lesson 7 Principles of Mathematics 11
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
i) x-coordinate of vertex =
y-coordinate
V(–4, –27)
ii) Axis of symmetry equation: x = –4
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, ∞)
Principles of Mathematics 11 Section 1, Answer Key, Lesson 7 213
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
i) x-coordinate of vertex =
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
,
214 Section 1, Answer Key, Lesson 7 Principles of Mathematics 11
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
ii) Axis of symmetry equation:
iii) x-intercepts
iv) y-intercept (let x = 0) = 4
v) Opens downward because a < 0
vi) Because the parabola opens downward, the
maximum value is
vii) D: (–∞, ∞)
R:
Principles of Mathematics 11 Section 1, Answer Key, Lesson 7 215
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
i) x-coordinate of vertex =
y-coordinate
ii) Axis of symmetry equation:
iii) x-intercepts
iv) y-intercept (let x = 0) = –2
v) Opens downward since a < 0
vi) Because the parabola opens downward, the maximum
value is
vii) D: (–∞, ∞)
R:
216 Section 1, Answer Key, Lesson 7 Principles of Mathematics 11
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
i) x-coordinate of vertex =
y-coordinate
ii) Axis of symmetry equation:
iii) x-intercepts
iv) y-intercept (let x = 0) = 1
v) Opens downward since a < 0
vi) Because the parabola opens downward, the
maximum value is
vii) D: (–∞, ∞)
R:
Principles of Mathematics 11 Section 1, Answer Key, Lesson 7 217
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
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 8 219
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
220 Section 1, Answer Key, Lesson 8 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Lesson 8 221
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
222 Section 1, Answer Key, Lesson 8 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Review 225
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
226 Section 1, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Review 227
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
.
.
.
.
.
228 Section 1, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 1, Answer Key, Review 229
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
230 Section 1, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 1 231
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}
232 Section 2, Answer Key, Lesson 1 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 1 233
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
234 Section 2, Answer Key, Lesson 1 Principles of Mathematics 11
Module 2
Lesson 2Answer Key
Principles of Mathematics 11 Section 2, Answer Key, Lesson 2 235
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
+ − =
+ − =
− + − − =− − =
236 Section 2, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 2 237
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.
238 Section 2, Answer Key, Lesson 2 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 2 239
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
b gb gx x= − =11 5or
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
b gb gx x= − =10 6or
The two numbers are 6 and 10.
240 Section 2, Answer Key, Lesson 2 Principles of Mathematics 11
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
b gb gx x= − =41 6or
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 3 241
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
242 Section 2, Answer Key, Lesson 3 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 3 243
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.
244 Section 2, Answer Key, Lesson 3 Principles of Mathematics 11
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)
Principles of Mathematics 11 Section 2, Answer Key, Lesson 3 245
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}.
246 Section 2, Answer Key, Lesson 3 Principles of Mathematics 11
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)
Principles of Mathematics 11 Section 2, Answer Key, Lesson 3 247
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
248 Section 2, Answer Key, Lesson 3 Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 249
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.
250 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 251
Module 2
b) xx
− − − =12
3 02
x x
x x
− − − =
− + − =
1 2 6 0
2 7 0
2
2
Because b2 – 4ac < 0, there are no real roots.
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− + =
Because b2 – 4ac < 0, there are no real roots.
a b c= = − =6 1 2, ,
b ac2 24 1 4 6 2
1 4847
− = − −
= −= −
b g b gb g
252 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 4 253
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
254 Section 2, Answer Key, Lesson 4 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 5 255
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
256 Section 2, Answer Key, Lesson 5 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 5 257
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
258 Section 2, Answer Key, Lesson 5 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 5 259
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
Lesson 6
Answer Key
1. a) True
b) False.
c) True
d) True
Principles of Mathematics 11 Section 2, Answer Key, Lesson 6 261
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
262 Section 2, Answer Key, Lesson 6 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Lesson 6 263
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
264 Section 2, Answer Key, Lesson 6 Principles of Mathematics 11
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.
Principles of Mathematics 11 Section 2, Answer Key, Lesson 6 265
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
266 Section 2, Answer Key, Lesson 6 Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Review 267
Module 2
2. a) x2 – 2x – 8 = 0
b) x2 – 8x + 15 = 0
x
y
(4 , −1)
x
y
(1 , −9)
268 Section 2, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Review 269
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
270 Section 2, Answer Key, Review Principles of Mathematics 11
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
− =
− − == = − = −
− = − − −= +=
Principles of Mathematics 11 Section 2, Answer Key, Review 271
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
272 Section 2, Answer Key, Review Principles of Mathematics 11
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:
Principles of Mathematics 11 Section 2, Answer Key, Review 273
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
274 Section 2, Answer Key, Review Principles of Mathematics 11
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
Principles of Mathematics 11 Section 2, Answer Key, Review 275
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
276 Section 2, Answer Key, Review Principles of Mathematics 11
Module 2
We reject the negative answer (question 14says that the two numbers are positive).