Chapter 7 Using Quadratic Functions to Model Data

Chapter 7
Using Quadratic Functions to Model Data
Homework 7.1
1. 16 = 4⋅ 4 = 4 ⋅ 4 = 2⋅ 2 = 4
3. 8 = 4⋅ 2 = 4 ⋅ 2 = 2 2
5. 12 = 4⋅ 3 = 4 ⋅ 3 = 2 3
7. 4 = 2 ⋅ 2 = 2 ⋅ 2 =
2
9 3 3 3 3 3
9. 6 = 6 =
6
49 49 7
11. 5 ⋅ 2 = 5 2 =
5 2
2 2 4 2
13. 9 ⋅ 6 = 9 6 = 9 6 =
3 6
6 6 36 6 2
27. x 2 = 32
x = ±
32
x = ± 16⋅
2
x = ± 4 2
29. x 2 = 300
31.
x = ±
300
x = ± 100⋅3
x = ± 10 3
2
x − =
48 0
x
33. x 2 = 9
2
= 48
x = ±
48
x = ± 16⋅3
x = ± 4 3
; there are no real number solutions.
15. 3 = 3 = 3 ⋅ 2 = 3 2 = 3 2 =
3 2
32 16⋅
2 4 2 2 4 4 4⋅
2 8
35. x 2 + 25 = 0 means
number solutions.
2
x = −
25 ; there are no real
17. 3 = 3 ⋅ 2 = 6 =
6
2 2 2 4 2
19.
21. x 2 = 36
11 11 11 11 5 55 55
= = = ⋅ = =
20 20 4⋅5 2 5 5 2 25 10
x = ±
x = ± 6
23. x 2 = 5
x = ±
36
5
25. x 2 − 3 = 0
2
x = 3
x = ±
3
37. 4x 2 = 36
x
2
= 9
x = ±
x = ± 3
39. 5x 2 = 8
x
=
5
2 8
9
8 4⋅
2 2 2
x = ± = ± = ±
5 5 5
2 2 5 2 10
x = ± ⋅ = ±
5 5 25
x = ±
2 10
5
184

SSM: Intermediate Algebra Homework 7.1
41. 3x 2 − 14 = 0
51.
2
3(4 7) 9
x +
= −
43.
45.
47.
49.
2
3x
14
x
2
=
=
14
3
14 14
x = ± = ±
3 3
14 3 42
x = ± ⋅ = ±
3 3 9
x = ±
42
3
2
( x − 4) = 25
x − 4 = ± 25
x − 4 = ± 5
x − 4 = 5 or x − 4 = −5
x = 4 + 5 or x = 4 − 5
x = 9 or x = −1
2
(8x + 3) = 36
8x
+ 3 = ± 36
8x
+ 3 = ± 6
8x
+ 3 = 6 or 8x
+ 3 = −6
8x
= 3 or 8x
= −9
3 9
x = or x = −
8 8
2
(9x − 7) = 0
9x
− 7 = 0
9x
= 7
x =
7
9
2
(8x − 3) + 6 = 90
2
(8x
3) 84
8x
− 3 = ± 84
8x
− 3 = ± 4 ⋅21
8x
− 3 = ± 2 21
8x
= 3 ± 2 21
x =
− =
3 ± 2 21
8
53.
55.
2
(4 7) 3
x +
= − Since the square of a number is
nonnegative, there are no real number
solutions.
2
5( x − 6) + 3 = 23
− =
2
5( x 6) 20
− =
2
( x 6) 4
x − 6 = ± 4
x − 6 = ± 2
x − 6 = 2 or x − 6 = −2
x = 8 or x = 4
− − + = −
2
3(4x
9) 5 31
− − = −
2
3(4x
9) 36
2
(4x
9) 12
4x
− 9 = ± 12
4x
− 9 = ± 4 ⋅3
4x
− 9 = ± 2 3
4x
= 9 ± 2 3
x =
− =
9 ± 2 3
4
57. Solve for x when f(x) = 0
2
0 x 17
x
2
= −
= 17
x = ±
17
− .
The x-intercepts are ( 17,0 ) and ( 17,0)
59. Solve for x when h(x) = 0
2
0 = ( x − 2) − 4
− =
2
( x 2) 4
x − 2 = ± 4
x − 2 = ± 2
x − 2 = 2 or x − 2 = −2
x = 4 or x = 0
The x-intercepts are ( 4,0 ) and ( )
0,0 .
185

Homework 7.1
SSM: Intermediate Algebra
61. Solve for x when f(x) = 0
= − + +
2
0 3( x 1) 12
2
3( x 1) 12
+ =
+ =
2
( x 1) 4
x + 1 = ± 4
x + 1= ± 2
x + 1 = 2 or x + 1 = −2
x = 1 or x = −3
The x-intercepts are ( 1,0 ) and ( 3,0)
63. Solve for x when h(x) = 0
= + +
2
0 2( x 5) 1
2
2( x 5) 1
2
+ = −
( x 5)
+ = −
1
2
− .
Since the square of a number is nonnegative, there
are no real number solutions.
65. Solve for x when f(x) = 0
= − −
2
0 ( x 5) 9
− =
2
( x 5) 9
− = ±
2
( x 5) 9
− = ±
2
( x 5) 3
x − 5 = ± 3
x − 5 = 3 or x − 5 = −3
x = 8 or x = 2
The x-intercepts are ( 8,0 ) and ( )
2,0 . Since these
points are symmetric, the average of the x-
coordinates, 8 + 2 10 = = 5 , is the x-coordinate of
2 2
the vertex of f(x). Substitute 5 for x in the function
to find the y-coordinate of the vertex:
2
f ( x ) = (5 − 5) − 9 = − 9
Therefore, the vertex is (5, -9).
67. Solve for x when h(x) = 0
= − + +
2
0 2( x 7) 9
2
2( x 7) 9
2
+ =
( x 7)
+ =
x + 7 = ±
x + 7 = ±
x + 7 = ±
x = − 7 ±
9
2
9
2
9
2
3
2
3
2
Rationalize the denominator:
3 2 3 2
⋅ =
2 2 2
3 2
x = − 7 ±
2
x ≈ −4.88 or x ≈ −9.12
So, the x-intercepts are ( − 4.88,0)
and ( 9.12,0)
− .
Since these points are symmetric, the average of
− 4.88 + ( −9.12)
the x-coordinates,
= − 7 , is the x-
2
coordinate of the vertex of f(x). Substitute -7 for x
in the function to find the y-coordinate of the
vertex:
2
h( x ) = −2( − 7 + 7) + 9 = 9
Therefore, the vertex is (-7, 9).
186

SSM: Intermediate Algebra Homework 7.2
73.
1
2
1
2
a ⎛ a ⎞ a a
= ⎜ ⎟ = =
1
b ⎝ b ⎠
2
b
b
75. Answers may vary. Example:
69. Solve for x when g(x) = 0
2
0 x 31
x
2
= −
= 31
x = ±
31
x ≈ 5.57 or x ≈ −5.57
So, the x-intercepts are ( − 31,0)
and ( )
31,0 .
Since these points are symmetric, the average of
31 + ( − 31)
the x-coordinates,
= 0 , is the x-
2
coordinate of the vertex of f(x). Substitute 0 for x in
the function to find the y-coordinate of the vertex:
2
g( x ) = (0) − 31 = − 31
You can solve quadratic equations by extracting
the square root as long as they can be placed in
2
vertex form. (Samples: x − 81 = 0 or
2
2( x − 1) + 9 = 0 )
You can solve quadratic equations by factoring as
long as the quadratic expression contained in the
equation is not prime. (Samples:
2
2
x − 81 = ( x − 9)( x + 9) = 0 or x + 4x
+ 4 = 0 ).
Homework 7.2
1.
3.
2
⎛ 12 2
⎜
⎞ ⎟ = 6 = 36 =c
⎝ 2 ⎠
This expression is
form is ( x + 6) 2
.
2
⎛ −14
⎞ 2
⎜ ⎟ = ( − 7) = 49 = c
⎝ 2 ⎠
2
x + 12x
+ 36 and its factored
Therefore, the vertex is (0, -31).
This expression is
form is ( x − 7) 2
.
2
x − 14x
+ 49 and its factored
5.
2 2
⎛ −7 ⎞ ( −7) 49
⎜ ⎟ = c
2 = =
⎝ 2 ⎠ 2 4
This expression is
⎛ 7 ⎞
form is ⎜ x − ⎟
⎝ 2 ⎠ .
2
2 49
x − 7x
+ and its factored
4
71. No, the student did not solve it correctly. They
should have factored the left hand side first.
2
x − 10x
+ 25 = 0
( x − 5)( x − 5) = 0
− =
2
( x 5) 0
x − 5 = 0
x = 5
7.
2 2
⎛ 3 ⎞ 3 9
⎜ ⎟ =
2 = = c
⎝ 2 ⎠ 2 4
This expression is
⎛ 3 ⎞
form is ⎜ x + ⎟
⎝ 2 ⎠ .
2
x + 3x
+ and its factored
4
2 9
187

Homework 7.2
SSM: Intermediate Algebra
9.
11.
2 2 2
⎛ 1 1 ⎞ ⎛ 1 ⎞ 1 1
⎜ ⋅ ⎟ = ⎜ ⎟ = = = c
2
⎝ 2 2 ⎠ ⎝ 4 ⎠ 4 16
2 1 1
This expression is x + x + and its factored
2 16
⎛ 1 ⎞
form is ⎜ x + ⎟
⎝ 4 ⎠ .
2
2 2 2 2
⎛ −4 1 ⎞ ⎛ −4 ⎞ ⎛ −2 ⎞ ( −2) 4
⎜ ⋅ ⎟ = ⎜ ⎟ = ⎜ ⎟ = = = c
2
⎝ 5 2 ⎠ ⎝ 10 ⎠ ⎝ 5 ⎠ 5 25
2 4 4
This expression is x − x + and its factored
5 25
⎛ 2 ⎞
form is ⎜ x − ⎟
⎝ 5 ⎠ .
2
2
⎛ 6 ⎞ 2
13. Since ⎜ ⎟ = 3 = 9 , add 9 to both sides of the
⎝ 2 ⎠
equation.
2
x + 6x
= 1
2
x + 6x
+ 9 = 1+
9
( x )
2
+ 3 = 10
x + 3 = ± 10
x = − 3 ± 10
2
⎛12
⎞ 2
15. Since ⎜ ⎟ = 6 = 36 , add 36 to both sides of the
⎝ 2 ⎠
equation.
2
x + 12x
= 2
2
x + 12x
+ 36 = 2 + 36
( x )
2
+ 6 = 38
x + 6 = ± 38
x = − 6 ± 38
2
⎛ −2
⎞ 2
17. Since ⎜ ⎟ = ( − 1) = 1, add 1 to both sides of the
⎝ 2 ⎠
equation.
2
x − 2x
= 10
2
x − 2x
+ 1 = 10 + 1
( x )
2
− 1 = 11
x − 1 = ± 11
x = 1±
11
19. First, change the equation to the form
2 2
x + bx = p; x − 18x
= 3.
2
⎛ −18
⎞ 2
Since ⎜ ⎟ = ( − 9) = 81, add 81 to both sides of
⎝ 2 ⎠
the equation.
2
x − 18x
= 3
2
x − 18x
+ 81 = 3 + 81
( x )
2
− 9 = 84
x − 9 = ± 84
x − 9 = ± 2 21
x = 9 ± 2 21
21. First, change the equation to the form
2 2
x + bx = p; x + 18x
= 9.
2
⎛18
⎞ 2
Since ⎜ ⎟ = (9) = 81 , add 81 to both sides of
⎝ 2 ⎠
the equation.
2
x + 18x
= 9
2
x − 18x
+ 81 = 9 + 81
( x )
2
+ 9 = 90
x + 9 = ± 90
x + 9 = ± 3 10
x = − 9 ± 3 10
23. First, change the equation to the form
2 2
x + bx = p; x + 4x
= − 8.
2
⎛ 4 ⎞ 2
Since ⎜ ⎟ = (2) = 4 , add 4 to both sides of the
⎝ 2 ⎠
equation.
188

SSM: Intermediate Algebra Homework 7.2
2
x + 4x
= −8
2
x + 4x
+ 4 = − 8 + 4
( x )
2
+ 2 = −4
Since the square of a number is nonnegative, there
are no real number solutions.
2 2
⎛ −7 ⎞ ( −7) 49
25. Since ⎜ ⎟ =
2 =
⎝ 2 ⎠ 2 4
of the equation.
2
x − 7x
= 3
2 49 49
x − 7x
+ = 3 +
4 4
2
⎛ 7 ⎞ 12 49
⎜ x − ⎟ = +
⎝ 2 ⎠ 4 4
⎛ 7 ⎞ 61
⎜ x − ⎟ =
⎝ 2 ⎠ 4
2
7 61
x − = ±
2 4
7 61
x − = ±
2 2
7 61
x = ±
2 2
x =
2 2
7 ± 61
2
⎛ 5 ⎞ 5 25
27. Since ⎜ ⎟ =
2 =
⎝ 2 ⎠ 2 4
the equation.
2
x + 5x
= 4
2 25 25
x + 5x
+ = 4 +
4 4
2
⎛ 5 ⎞ 16 25
⎜ x + ⎟ = +
⎝ 2 ⎠ 4 4
⎛ 5 ⎞ 41
⎜ x + =
2
⎟
⎝ ⎠ 4
2
, add 25
4
, add 49
4
to both sides
to both sides of
5 41
x + = ±
2 4
5 41
x + = ±
2 2
5 41
x = − ±
2 2
x =
− 5 ± 41
2
29. First, change the equation to the form
2 2
x + bx = p; x + x = − 7.
2 2
⎛ 1 ⎞ 1 1
Since ⎜ ⎟ =
2 = , add 1 to both sides of the
⎝ 2 ⎠ 2 4 4
equation.
2
x + x = −7
2 1 1
x + x + = − 7 +
4 4
2
⎛ 1 ⎞ −28 1
⎜ x + ⎟ = +
⎝ 2 ⎠ 4 4
⎛ 1 ⎞ −27
⎜ x + =
2
⎟
⎝ ⎠ 4
2
Since the square of a number is nonnegative, there
are no real number solutions.
2 2 2
⎛ −5 1 ⎞ ⎛ −5 ⎞ ( −5) 25
31. Since ⎜ ⋅ ⎟ = ⎜ ⎟ = =
2
⎝ 2 2 ⎠ ⎝ 4 ⎠ 4 16
both sides of the equation.
2 5 1
x − x =
2 2
2 5 25 1 25
x − x + = +
2 16 2 16
2
⎛ 5 ⎞ 8 25
⎜ x − ⎟ = +
⎝ 4 ⎠ 16 16
⎛ 5 ⎞ 33
⎜ x − ⎟ =
⎝ 4 ⎠ 16
2
, add 25
16 to
189

Homework 7.2
SSM: Intermediate Algebra
5 33
x − = ±
4 16
5 33
x − = ±
4 4
5 33
x = ±
4 4
x =
5 ± 33
4
33. Divide both sides by 2 so that a = 1:
⎛ −4
⎞ 2
x − 4x
= . Since ⎜ ⎟ = ( − 2)
= 4 , add 4 to
2 ⎝ 2 ⎠
both sides of the equation.
2 3
2 3
x − 4x
=
2
2 3
x − 4x
+ 4 = + 4
2
2 3 8
( x − 2)
= +
2 2
2 11
( x − 2)
=
2
x − 2 = ±
x − 2 = ±
11
2
11
Rationalize the denominator:
11 2 22
⋅ =
2 2 2
x = 2 ±
2
22
2
4 22
x = ±
2 2
4 ± 22
x =
2
2
35. Change the equation so that it is in the form
2
2
ax + bx = p : 5x
+ 15x
= − 9 . Divide both sides
by 5 so that a = 1:
2 2
⎛ 3 ⎞ 3 9
⎜ ⎟ =
2 = , add 9
⎝ 2 ⎠ 2 4 4
equation.
2 9
x + 3x
= −
5
2 9 9 9
x + 3x
+ = − +
4 5 4
2
⎛ 3 ⎞ 36 45
⎜ x + ⎟ = − +
⎝ 2 ⎠ 20 20
⎛ 3 ⎞ 9
⎜ x + ⎟ =
⎝ 2 ⎠ 20
2
3 9
x + = ±
2 20
3 9
x + = ±
2 20
3 3
x + = ±
2 2 5
x + 3x
= − . Since
5
2 9
Rationalize the denominator:
3 5 3 5
⋅ =
2 5 5 10
3 3 5
x + = ±
2 10
3 3 5
x = − ±
2 10
15 3 5
x = − ±
10 10
− 15 ± 3 5
x =
10
to both sides of the
190

SSM: Intermediate Algebra Homework 7.2
37. Divide both sides by 3 so that a = 1:
2 2 2
⎛ 4 1 ⎞ ⎛ 2 ⎞ 2 4
Since ⎜ ⋅ ⎟ = ⎜ ⎟ = = , add 4 2
⎝ 3 2 ⎠ ⎝ 3 ⎠ 3 9 9
sides of the equation.
2 4 5
x + x =
3 3
2 4 4 5 4
x + x + = +
3 9 3 9
2
⎛ 2 ⎞ 15 4
⎜ x + ⎟ = +
⎝ 3 ⎠ 9 9
⎛ 2 ⎞ 19
⎜ x + ⎟ =
⎝ 3 ⎠ 9
2
2 19
x + = ±
3 9
2 19
x + = ±
3 9
2 19
x + = ±
3 3
2 19
x = − ±
3 3
x =
− 2 ± 19
3
2 4 5
x + x = .
3 3
to both
39. Change the equation so that it is in the form
2
2
ax + bx = p : 2x
− x = 7 . Divide both sides by 2
so that a = 1:
2 1 7
x − x = . Since
2 2
2 2 2
⎛ −1 1 ⎞ ⎛ −1 ⎞ ( −1) 1
⎜ ⋅ ⎟ = ⎜ ⎟ = =
⎝ ⎠ ⎝ ⎠
sides of the equation.
2
2 2 4 4 16
2 1 7
x − x =
2 2
2 1 1 7 1
x − x + = +
2 16 2 16
2
⎛ 1 ⎞ 56 1
⎜ x − ⎟ = +
⎝ 4 ⎠ 16 16
⎛ 1 ⎞ 57
⎜ x − ⎟ =
⎝ 4 ⎠ 16
2
, add 1
16
to both
1 57
x − = ±
4 16
1 57
x − = ±
4 16
1 57
x − = ±
4 4
1 57
x = ±
4 4
x =
1±
57
4
41. Change the equation so that it is in the form
2
2
ax + bx = p : 5x
+ 7x
= − 3 . Divide both sides by
5 so that a = 1:
2 2 2
2 7 3
x + x = − . Since
5 5
⎛ 7 1 ⎞ ⎛ 7 ⎞ 7 49
⎜ ⋅ ⎟ = ⎜ ⎟ = =
⎝ ⎠ ⎝ ⎠
sides of the equation.
2
5 2 10 10 100
2 7 3
x + x = −
5 5
2 7 49 3 49
x + x + = − +
5 100 5 100
2
⎛ 7 ⎞ −60 49
⎜ x + ⎟ = +
⎝ 10 ⎠ 100 100
⎛ 7 ⎞ −11
⎜ x + ⎟ =
⎝ 10 ⎠ 100
2
, add 49
100
to both
Since the square of a number is nonnegative, there
are no real number solutions.
43. Divide both sides by 8 so that a = 1:
2 2 2
2 5 1
x − x = .
8 2
⎛ −5 1 ⎞ ⎛ −5 ⎞ ( −5) 25
Since ⎜ ⋅ ⎟ = ⎜ ⎟ = = , add
2
⎝ 8 2 ⎠ ⎝ 16 ⎠ 16 256
to both sides of the equation.
2 5 1
x − x =
8 2
2 5 25 1 25
x − x + = +
8 256 2 256
2
⎛ 5 ⎞ 128 25
⎜ x − ⎟ = +
⎝ 16 ⎠ 256 256
⎛ 5 ⎞ 153
⎜ x − ⎟ =
⎝ 16 ⎠ 256
2
25
256
191

Homework 7.2
SSM: Intermediate Algebra
5 153
x − = ±
16 256
5 153
x − = ±
16 256
5 9⋅17
x − = ±
16 16
5 3 17
x = ±
16 16
5 ± 3 17
x =
16
45. First, change the equation so that it is in the form
2
ax + bx = p :
2 x( x − 3) = 5 − x
x − x = − x
2
2 6 5
− =
2
2x
5x
5
Divide both sides by 2, so that a = 1:
2 5 5
x − x =
2 2
2 2 2
⎛ −5 1 ⎞ ⎛ −5 ⎞ ( −5) 25
Since ⎜ ⋅ ⎟ = ⎜ ⎟ = =
2
⎝ 2 2 ⎠ ⎝ 4 ⎠ 4 16
to both sides of the equation.
2 5 5
x − x =
2 2
2 5 25 5 25
x − x + = +
8 16 2 16
2
⎛ 5 ⎞ 40 25
⎜ x − ⎟ = +
⎝ 4 ⎠ 16 16
⎛ 5 ⎞ 65
⎜ x − ⎟ =
⎝ 4 ⎠ 16
2
5 65
x − = ±
4 16
5 65
x − = ±
4 16
5 65
x − = ±
4 4
5 65
x = ±
4 4
x =
5 ± 65
4
, add 25
16
47. First, change the equation so that it is in the form
2
ax + bx = p :
x( x − 1) + 3 = 5( x + 1)
− + 3 = 5 + 5
2
x x x
2
x − 6x
= 2
2
⎛ −6
⎞ 2
= − =
Since ( )
⎜ ⎟
⎝ 2 ⎠
the equation.
2
x − 6x
= 2
2
x − 6x
+ 9 = 2 + 9
( x )
2
− 3 = 11
x − 3 = ± 11
x = 3 ± 11
3 9 , add 9 to both sides of
49. No, the student did not solve the equation
correctly. The student should have first divided
both sides by 4 and then completed the square and
extracted the roots. The correct solution is:
+ =
2
4x
6x
1
2 3 1
x + x =
2 4
2 3 9 1 9
x + x + = +
2 16 4 16
2
⎛ 3 ⎞ 4 9
⎜ x + ⎟ = +
⎝ 4 ⎠ 16 16
⎛ 3 ⎞ 13
⎜ x + ⎟ =
⎝ 4 ⎠ 16
2
3 13
x + = ±
4 16
3 13
x + = ±
4 16
3 13
x + = ±
4 4
3 13
x = − ±
4 4
− 3 ± 13
x =
4
192

SSM: Intermediate Algebra Homework 7.3
51. a.
b.
c.
= + +
2
3 x 6x
13
2
x + 6x
= −10
2
x + 6x
+ 9 = − 10 + 9
( x )
2
+ 3 = −1
Since the square of a number is nonnegative,
there are no real number solutions.
= + +
2
4 x 6x
13
2
x + 6x
+ 9 = 0
( x )
2
+ 3 = 0
x + 3 = 0
x = −3
= + +
2
5 x 6x
13
2
x + 6x
+ 8 = 0
( x )( x )
+ 2 + 4 = 0
x + 2 = 0 or x + 4 = 0
x = − 2 or x = −4
= + +
2
0 x 10x
25
+ =
2
( x 5) 0
x + 5 = 0
x = −5
59. Answers may vary.
Homework 7.3
1. a = 2, b = 5, c = − 2
( )( )
( )
− ± − −
x =
2 2
x =
2
5 5 4 2 2
− 5 ± 41
4
3. a = 3, b = 7, c = − 2
( )( )
( )
− ± − −
x =
2 3
x =
2
7 7 4 3 2
− 7 ± 73
6
53. Solve for x when f(x) = 0:
= − +
2
0 x 8x
3
2
x − 8x
= −3
2
x − 8x
+ 16 = − 3 + 16
− =
2
( x 4) 13
x − 4 = ± 13
x = 4 ± 13
55. Solve for x when f(x) = 0:
= + +
2
0 x 4x
5
2
x + 4x
= −5
2
x + 4x
+ 4 = − 5 + 4
+ = −
2
( x 2) 1
Since the square of a number is nonnegative there
are no real number solutions. Therefore, there are
no x-intercepts.
57. Solve for x when f(x) = 0:
5. a = 2, b = − 5, c = 1
2
( 5) ( 5) 4( 2)( 1)
2( 2)
− − ± − −
x =
x =
5 ± 17
4
7. a = 3, b = − 2, c = 8
2
( 2) ( 2) 4( 3)( 8)
2( 3)
− − ± − −
x =
x =
2 ± −92
6
Since the square root of a negative number is not a
real number, there are no real solutions.
2
9. First write 2x
+ 5x
= 3 in the form
2
2
ax + bx + c = 0 : 2x
+ 5x
− 3 = 0
a = 2, b = 5, c = − 3
193

Homework 7.3
SSM: Intermediate Algebra
( )( )
( )
2
5 5 4 2 3
− ± − −
x =
2 2
− 5 ± 49
x =
4
− 5 ± 7
x =
4
− 5 + 7 −5 − 7
x = or x =
4 4
2 −12
x = or x =
4 4
x = 1 or x = − 3
2
11. a = 3, b = 0, c = − 17
2
( ) ( )( )
2( 3)
0 ± 0 − 4 3 −17
x =
x = ±
x = ±
x = ±
204
6
2 51
6
51
3
2
13. First write 2x
= − 5x
in the form
2
2x
+ 5x
= 0
a = 2, b = 5, c = 0
− ±
x =
− ( )( )
( )
2
5 5 4 2 0
2 2
− 5 ± 25
x =
4
− 5 ± 5
x =
4
− 5 + 5 −5 − 5
x = or x =
4 4
0 −10
x = or x =
4 4
5
x = 0 or x = −
2
2
ax bx c
+ + = 0 :
2
15. First write 4x
= 2x
+ 3 in the form
2
2
ax + bx + c = 0 : 4x
− 2x
− 3 = 0
a = 4, b = − 2, c = − 3
( 2) (
2
2) 4( 4)( 3)
2( 4)
− − ± − − −
x =
2 ± 52
x =
8
2 ± 2 13
x =
8
( ± )
2 1 13
x =
8
1±
13
x =
4
17. First write x( x )
2
ax bx c
+ + = 0 :
( )
− 3x x + 1 = 2
2
3x
3x
2
− − =
2
3x
3x
2 0
− − − =
− 3 + 1 = 2 in the form
a = − 3, b = − 3, c = − 2
( 3) (
2
3) 4( 3)( 2)
2( −3)
− − ± − − − −
x =
3 ± −15
x =
−6
Since the square root of a negative number is not a
real number, there are no real solutions.
2
19. First write ( x )
2
ax bx c
+ + = 0 :
2
( x )
3 − 2 = 5x
2
3x
6 5
− =
2
3x
5x
6 0
3 − 2 = 5x
in the form
x
− − =
a = 3, b = − 5, c = − 6
194

SSM: Intermediate Algebra Homework 7.3
( 5) (
2
5) 4( 3)( 6)
2( 3)
− − ± − − −
x =
5 ± 97
x =
6
21. First write 3( x 4) 2x( x 1)
2
ax bx c
+ + = 0 :
( x + ) = − x( x − )
3 4 2 1
2
3 12 2 2
x + = − x + x
2
2x
x 12 0
+ + =
a = 2, b = 1, c = 12
− ±
x =
+ = − − in the form
− ( )( )
( )
2
1 1 4 2 12
2 2
− 1±
−95
x =
4
Since the square root of a negative number is not a
real number, there are no real solutions.
2
23. First write 3 − 6x = x in the form
2
2
ax + bx + c = 0 : x + 6x
− 3 = 0
a = 1, b = 6, c =− 3
( − ± )
( )( )
( )
2
6 6 4 1 3
− ± − −
x =
2 2
x =
− 6 ± 48
2
− ±
x =
6 4 3
2
2 3 3
x =
2
x = − 3 ± 3
25. a = 2, b = − 5, c = − 4
( 5) (
2
5) 4( 2)( 4)
2( 2)
− − ± − − −
x =
5 ± 57
x =
4
− 5 + 57 −5 − 57
x = or x =
4 4
x ≈ 3.14 or x ≈ −0.64
2
27. First write 2.8x
− 7.1x
= 4.4 in the form
2
2
ax + bx + c = 0 : 2.8x
− 7.1x
− 4.4 = 0
a = 2.8, b = − 7.1, c = −4.4
( 7.1) (
2
7.1) 4( 2.8)( 4.4)
2( 2.8)
− − ± − − −
x =
7.1±
99.69
x =
5.6
7.1+ 99.69 7.1−
99.69
x = or x =
5.6 5.6
x ≈ 3.05 or x ≈ −0.52
29. First write − 5.4 x( x + 9.8) + 4.1 = 3.2 − 6.9x
in the
2
form ax + bx + c = 0 :
2
−5.4x − 52.92x + 4.1 = 3.2 − 6.9x
2
5.4x
26.02x
0.9 0
− − + =
a = − 5.4, b = − 46.02, c = 0.9
( 46.02) (
2
46.02) 4( 5.4)( 0.9)
2( −5.4)
− − ± − − −
x =
46.02 ± 2137.2804
x =
−10.8
46.02 + 2137.28 46.02 − 2137.28
x ≈
or x ≈
−10.8 −10.8
x ≈ −8.54 or x ≈ 0.020
31. x 2 = 25
2
x − 25 = 0
( x + 5)( x − 5) = 0
x + 5 = 0 or x − 5 = 0
x = − 5 or x = 5
195

Homework 7.3
SSM: Intermediate Algebra
33. x( x − 8) = − 16
2
x − 8x
= −16
2
x − 8x
+ 16 = 0
( x − 4)( x − 4) = 0
x − 4 = 0
x = 4
35. 4x 2 = 80
37.
x
2
= 20
x = ±
20
x = ± 2 5
2
(4x + 3) + 2 = 22
( x ) 2
4 + 3 = 20
4x
+ 3 = ± 20
4x
+ 3 = ± 2 5
4x
= − 3 ± 2 5
x =
− 3 ± 2 5
4
39. 2 x(3x − 4) = x − 5
2
6x 8x x 5
2
6x
9x
5 0
a = 6, b = − 9, c = −5
− − ± − − −
x =
2(6)
x =
− = +
− − =
2
( 9) ( 9) 4(6)( 5)
9 ± −39
12
Since the square root of a negative number is not a
real number, there are no real solutions.
41. 9x
2 − 5x
= 0
x(9x
− 5) = 0
x = 0 or 9x
− 5 = 0
x = 0 or 9x
= 5
5
x = 0 or x =
9
43. 5x
2 = 3x
+ 8
2
5x
3x
8 0
− − =
(5x
− 8)( x + 1) = 0
5x
− 8 = 0 or x + 1 = 0
5x
= 8 or x = −1
8
x = or x = − 1
5
45. 3x
2 = 27x
47.
− =
2
3x
27x
0
3 x( x − 9) = 0
x( x − 9) = 0
x = 0 or x − 9 = 0
x = 0 or x = 9
2
2(2x
1) 3
+ = x :
+ =
2
4x
2 3
x
− + =
2
4x
3x
2 0
a = 4, b = − 3, c = 2
− − ± − −
x =
2(4)
x =
2
( 3) ( 3) 4(4)(2)
3 ± −23
8
Since the square root of a negative number is not a
real number, there are no real solutions.
49. x
2 + 12x
+ 36 = 0
( x + 6)( x + 6) = 0
51.
x + 6 = 0
x = −6
− − =
2
24x
18x
60 0
− − =
2
6(4x
3x
10) 0
− − =
2
4x
3x
10 0
(4x
+ 5)( x − 2) = 0
4x
+ 5 = 0 or x − 2 = 0
4x
= − 5 or x = 2
5
x = − or x = 2
4
196

SSM: Intermediate Algebra Homework 7.3
53.
2
( x − 4) − 3 = 7
− =
2
( x 4) 10
x − 4 = ± 10
x = 4 ± 10
55. x( x + 1) = 7 − x
+ = −
2
x x 7 x
2
x + 2x
= 7
2
x + 2x
+ 1 = 7 + 1
+ =
2
( x 1) 8
x + 1 = ± 8
x + 1 = ± 2 2
x = − 1±
2 2
57. 2 x(3x − 1) = 3( x − 2)
61.
63.
3 169
x − = ±
2 4
3 13
x − = ±
2 2
3 13
x = ±
2 2
3 13 3 13
x = + or x = −
2 2 2 2
16 −10
x = =8 or x = = −5
2 2
=
2
5x
35
x
− =
2
5x
35x
0
5 x( x − 7) = 0
5x
= 0 or x − 7 = 0
x = 0 or x = 7
= +
2
50x
5x
105
− = −
2
6x 2x 3x
6
− + =
2
6x
5x
6 0
a = 6, b = − 5, c = 6
2
( 5) ( 5) 4(6)(6)
− − ± − −
x =
2(6)
x =
5 ± −119
12
Since the square root of a negative number is not a
real number, there are no real solutions.
59. 2 x(3x − 1) = 3( x − 2)
− =
2
2x
6x
80
− =
2
2( x 3 x) 80
2
x − 3x
= 40
2 9 9
x − 3x
+ = 40 +
4 4
2
⎛ 3 ⎞ 160 9
⎜ x − ⎟ = +
⎝ 2 ⎠ 4 4
⎛ 3 ⎞ 169
⎜ x − ⎟ =
⎝ 2 ⎠ 4
2
65.
− + =
2
5x
50x
105 0
− + =
2
5( x 25x
21) 0
2
x − 25x
+ 21 = 0
( x − 3)( x − 7) = 0
x − 3 = 0 or x − 7 = 0
x = 3 or x = 7
2
(3x − 1) = 7
3x
− 1 = ± 7
3x
= 1±
7
x =
1±
7
3
67. ( x − 2)( x − 5) = 1
− 5 − 2 + 10 = 1
2
x x x
2
x − 7x
= −9
2 49 49
x − 7x
+ = − 9 +
4 4
2
⎛ 7 ⎞ 36 49
⎜ x − ⎟ = − +
⎝ 2 ⎠ 4 4
⎛ 7 ⎞ 13
⎜ x − ⎟ =
⎝ 2 ⎠ 4
2
197

Homework 7.3
SSM: Intermediate Algebra
69.
71.
7 13
x − = ±
2 4
7 13
x − = ±
2 2
7 13
x = ±
2 2
x =
7 ± 13
2
2 2
( x 1) ( x 2) 0
− + + =
− 2 + 1+ + 4 + 4 = 0
2 2
x x x x
+ + =
2
2x
2x
5 0
2 5
x + x + = 0
2
2 5
x + x = −
2
2 1 5 1
x + x + = − +
4 2 4
2
⎛ 1 ⎞ 10 1
⎜ x + ⎟ = − +
⎝ 2 ⎠ 4 4
⎛ 1 ⎞ 9
⎜ x + ⎟ = −
⎝ 2 ⎠ 4
1 9
x + = ± −
2 4
2
Since the square root of a negative number is not a
real number, there are no real solutions.
2 2
( x 2)( x 5) x 5x 2x 10 x 3x
10
+ − = − + − = − −
73. ( x + 2)( x − 5) = 3
2
x − 3x
− 10 = 3
2
x − 3x
− 13 = 0
a = 1, b = − 3, c = −13
− − ± − − −
x =
2(1)
2
( 3) ( 3) 4(1)( 13)
3 ± 61
x =
2
75.
77.
2
4( x 2) 3 1
− − + = −
−4( x − 2)( x − 2) + 3 = −1
− − − + + = −
2
4( x 2x 2x
4) 3 1
− − + + = −
2
4( x 4x
4) 3 1
− + − + = −
2
4x
16x
16 3 1
− + − =
2
4x
16x
12 0
− − + =
2
4( x 4x
3) 0
2
x − 4x
+ 3 = 0
( x − 3)( x − 1) = 0
x − 3 = 0 or x − 1 = 0
x = 3 or x = 1
− − + = − − − +
2
4( x 2) 3 4( x 2)( x 2) 3
= − − − + +
2
4( x 2x 2x
4) 3
= − − + +
2
4( x 4x
4) 3
= − + − +
2
4x
16x
16 3
= − + −
2
4x
16x
13
79. a. Substitute 3 for f(x):
2
3 x 4x
8
= − +
2
x − 4x
+ 5 = 0
a = 1, b = − 4, c = 5
2 2
b − 4 ac = ( −4) − 4(1)(5) = 16 − 20 = − 4 < 0
So, there are no real number solutions, which
means there are no points on f at y = 3.
b. Substitute 4 for f(x):
= − +
2
4 x 4x
8
2
x − 4x
+ 4 = 0
a = 1, b = − 4, c = 4
2 2
b − 4 ac = ( −4) − 4(1)(4) = 16 − 16 = 0
So, there is one solution to the equation, which
means there is one point on f at y = 4.
198

SSM: Intermediate Algebra Homework 7.3
c. Substitute 5 for f(x):
83. a.
= − +
2
5 x 4x
8
2
x − 4x
+ 3 = 0
a = 1, b = − 4, c = 3
2 2
b − 4 ac = ( −4) − 4(1)(3) = 16 − 12 = 4 > 0
d.
So, there are two real number solutions, which
means there are two points on f at y = 5.
Yes, f does model the data well.
b. In 2007, t = 17. Solve for f(17)
f = − +
2
(17) 30.32(17) 122.15(17) 109.75
= 6795.68
≈ 6796 schools
c. 200 schools per state means 10,000 charter
schools. Solve for t when f(t) = 10,000.
81. Solve for x when f(x) = 2:
= − +
2
2 x 6x
7
2
x − 6x
+ 5 = 0
( x − 5)( x − 1) = 0
x − 5 = 0 or x − 1 = 0
x = 5 or x = 1
Therefore, two points at height 2 are ( 1,2 ) and
( 5,2 ) . Since these points are symmetric and the
average of the x-coordinates at these points is
1 + 5 = 3 the x-coordinate of the vertex is 3.
2
2
Substitute 3 for x in f ( x) = x − 6x
+ 7 to find the
y-coordinate of the vertex:
2
f (3) = 3 − 6(3) + 7 = − 2 So the vertex is ( 3, − 2)
.
85. a.
2
10000 30.32t
122.15t
109.75
2
0 30.32t
122.15t
9890.25
t =
= − +
= − −
2
122.15 (122.15) 4(30.32)( 9890.25)
t ≈ −16.16 or t ≈ 20.19
t ≈ −16 or t ≈ 20
± − −
2(30.32)
The solution t = -16 is model breakdown.
There will be 200 charter schools per state in
2010.
b.
Yes, f does model the data well.
2
f (13) = 23.43(13) − 259.14(13) + 815.77
≈ 1406.65
≈ 1407
This means that in 2003, there will be1407
convictions of police officers.
199

Homework 7.3
SSM: Intermediate Algebra
c.
= − +
2
2000 23.43t
259.14t
815.77
2
0 23.43t
259.14t
815.77
2
0 23.43t
259.14t
1184.23
t =
= − +
= − −
2
259.14 (259.14) 4(23.43)( 1184.23)
2(23.43)
t ≈ −3.48 or t ≈ 14.54
t ≈ −3 or t ≈ 15
± − −
There were 2000 convictions of police officers
in 1987 and 2005.
Since the square root of a negative number is not a
real number, there are no real number solutions,
and therefore, no x-intercepts.
93. a. mx + b = 0
mx + b − b = 0 − b
mx = −b
mx b
= −
m m
b
x = −
m
87. No, the student did not solve the equation correctly
because they did not change the form into
2
ax + bx + c = 0 first. Here is the correct way:
2
2x
5x
1
2
2x
5x
1 0
So, a = 2, b = 5, c = −1
− ± − −
x =
2(2)
x =
+ =
+ − =
2
5 (5) 4(2)( 1)
− 5 ± 33
4
89. Solve for x when f(x) = 0:
2
0 x 4x
6
2
( 4) ( 4) 4(1)( 6)
− − ± − − −
x =
2(1)
x =
x =
= − −
4 ± 40
2
4 ± 2 10
2
x = 2 ± 10
x ≈ 5.16 or x ≈ −1.16
The x-intercepts are (5.16, 0) and (-1.16, 0)
91. Solve for x when h(x) = 0:
2
0 3x
2x
5
− − ± − −
x =
2(3)
x =
= − +
2
( 2) ( 2) 4(3)(5)
2 ± −56
6
b. 7x + 21 = 0
So, m = 7 and b = -21. Using the formula from
part a:
21
x = − = − 3
7
Solving for x in the usual way:
7x
+ 21 = 0
7x
+ 21− 21 = 0 − 21
7x
= −21
7x
21
= −
7 7
x = −3
95. a = k, b = -12, c = 4
2
b − 4ac
= 0
− − =
2
( 12) 4( k)(4) 0
144 − 16k
= 0
97. a = 2, b = k, c = 8
k
2
2
b − 4ac
= 0
− 4(2)(8) = 0
k
2
− 16k
= −144
− 64 = 0
k
2
k = 9
= 64
k = ± 8
200

SSM: Intermediate Algebra Homework 7.4
99. a = 9, b = -6, c = k
2
b − 4ac
= 0
− − =
2
( 6) 4(9)( k) 0
36 − 36k
= 0
− 36k
= −36
k = 1
101. Factoring:
2
x − x − 20 = 0
( x − 5)( x + 4) = 0
x − 5 = 0 or x + 4 = 0
x = 5 or x = −4
Completing the square:
2
x − x = 20
2 1 1
x − x + = 20 +
4 4
⎛ 1 ⎞ 81
⎜ x − ⎟ =
⎝ 2 ⎠ 4
2
1 81
x − = ±
2 4
1 9
x − = ±
2 2
1 9 1 9
x = + or x = −
2 2 2 2
10 8
x = or x = −
2 2
x = 5 or x = −4
Quadratic formula:
2
x − x − 20 = 0
− − ± − − −
x =
2(1)
2
( 1) ( 1) 4(1)( 20)
1±
81
x =
2
1±
9
x =
2
1+ 9 1−
9
x = or x =
2 2
x = 5 or x = −4
103. Answers may vary. Example:
2
First, put the equation in the form ax + bx + c = 0 .
Once you know the value of a, b, and c, substitute
these values into the quadratic formula.
− ±
x =
Homework 7.4
−
2a
2
b b 4ac
1. Substitute the given points into
2
( 1,6 ): 6 = a( 1) + b( 1)
+ c
( 2,11 ):11 2
( 2) ( 2)
( 3,8 ): 8 =
2
( 3) + ( 3)
+
= a + b + c
a b c
Simplify these equations:
a+ b+ c = 6 1
( )
( )
( )
4a + 2b + c = 11 2
9a + 3b + c = 18 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = − 6 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 5 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
8a
+ 2b
= 12 6
Simplify:
4a
+ b = 6 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 1
Next, substitute 1 for a in equation (5):
3 1 + b = 5
( )
b = 2
Then, substitute 1 for a and 2 for b in equation (1):
a+ b+ c = 6
1+ 2 + c = 6
c = 3
Therefore, a = 1, b = 2, and c = 3. So, the equation
2
is y = x + 2x
+ 3 .
201

Homework 7.4
SSM: Intermediate Algebra
3. Substitute the given points into
2
( 1,5 ): 5 = a( 1) + b( 1)
+ c
( 2,11 ):11 2
( 2) ( 2)
( 3,19 ):19 2
( 3) ( 3)
= a + b + c
= a + b + c
Simplify these equations:
a+ b+ c = 5 1
( )
( )
( )
4a + 2b + c = 11 2
9a + 3b + c = 19 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = − 5 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 6 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
8a
+ 2b
= 14 6
Simplify:
4a
+ b = 7 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 1
Next, substitute 1 for a in equation (5):
3 1 + b = 6
( )
b = 3
Then, substitute 1 for a and 3 for b in equation (1):
a+ b+ c = 5
1+ 3 + c = 5
c = 1
Therefore, a = 1, b = 3, and c = 1. So, the equation
2
is y = x + 3x
+ 1.
5. Substitute the given points into
2
( 1,9 ): 9 = a( 1) + b( 1)
+ c
2
( 2,7 ): 7 = a( 2) + b( 2)
+ c
2
( 4, 15 ): 15 ( 4) ( 4)
− − = a + b + c
Simplify these equations:
a+ b+ c = 9 1
( )
( )
( )
4a + 2b + c = 7 2
16a + 4b + c = −15 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
( )
−a − b − c = − 9 4
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = − 2 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
15a
+ 3b
= − 24 6
Simplify:
5a
+ b = − 8 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
2a
= −6
a = −3
Next, substitute -3 for a in equation (5):
3 − 3 + b = −2
( )
b = 7
Then, substitute -3 for a and 7 for b in equation (1):
a+ b+ c = 9
− 3 + 7 + c = 9
c = 5
Therefore, a = -3, b = 7, and c = 5. So, the equation
2
is y = − 3x + 7x
+ 5 .
7. Substitute the given points into
2
( 2, 2 ): 2 = a( 2) + b( 2)
+ c
( 3,11 ):11 =
2
( 3) + ( 3)
+
( 4, 24 ): 24
2
( 4) ( 4)
a b c
= a + b + c
Simplify these equations:
a+ b+ c = 2 1
( )
( )
( )
9a + 3b + c = 11 2
16a + 4b + c = 24 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−4a − 2b − c = − 2 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
5a
+ b = 9 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
12a
+ 2b
= 22 6
Simplify:
6a
+ b = 11 7
( )
( )
202

SSM: Intermediate Algebra Homework 7.4
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 2
Next, substitute 2 for a in equation (5):
5 2 + b = 9
( )
b = −1
Then, substitute 2 for a and -1 for b in equation (1):
4a + 2b + c = 2
4(2) + 2( − 1) + c = 2
8 − 2 + c = 2
c = −4
Therefore, a = 2, b = -1, and c = -4. So, the
2
equation is y = 2x − x − 4 .
9. Substitute the given points into
2
( 1, −3 ): − 3 = ( 1) + ( 1)
+
2
( 3,9 ): 9 = a( 3) + b( 3)
+ c
2
( 5,29 ): 29 ( 5) ( 5)
a b c
= a + b + c
Simplify these equations:
a+ b+ c = −3 1
( )
( )
( )
9a + 3b + c = 9 2
25a + 5b + c = 29 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = 3 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
8a
+ 2b
= 12 5
Simplify:
4a
+ b = 6 6
( )
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
24a
+ 4b
= 32 7
Simplify:
6a
+ b = 8 8
( )
( )
Eliminate b by multiplying equation (6) by -1 and
add each side to the corresponding side of equation
(8):
2a
= 2
a = 1
Next, substitute 1 for a in equation (6):
4(1) + b = 6
b = 2
Then, substitute 1 for a and 2 for b in equation (1):
a+ b+ c = −3
1+ 2 + c = −3
c = −6
Therefore, a = 1, b = 2, and c = -6. So, the equation
2
is y = x + 2x
− 6 .
11. Substitute the given points into
2
( 3,7 ): 7 = a( 3) + b( 3)
+ c
2
( 4,0 ): 0 = a( 4) + b( 4)
+ c
2
( 5, 11 ): 11 ( 5) ( 5)
− − = a + b + c
Simplify these equations:
9a + 3b + c = 7 1
( )
( )
( )
16a + 4b + c = 0 2
25a + 5b + c = −11 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−9a − 3b − c = − 7 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
7a
+ b = − 7 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
16a
+ 2b
= − 18 6
Simplify:
8a
+ b = − 9 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = − 2
Next, substitute -2 for a in equation (5):
7 − 2 + b = −7
( )
b = 7
Then, substitute -2 for a and 7 for b in equation (1):
9a + 3b + c = 7
9( − 2) + 3(7) + c = 7
− 18 + 21+ c = 7
c = 4
Therefore, a = -2, b = 7, and c = 4. So, the equation
2
is y = − 2x + 7x
+ 4 .
203

Homework 7.4
SSM: Intermediate Algebra
13. Substitute the given points into
( 3,2 ): 2 =
2
( 3) + ( 3)
+
( 4,16 ):16 2
( 4) ( 4)
( 5,36 ): 36
2
( 5) ( 5)
a b c
= a + b + c
= a + b + c
Simplify these equations:
9a + 3b + c = 2 1
( )
( )
( )
16a + 4b + c = 16 2
25a + 5b + c = 36 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−9a − 3b − c = − 2 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
7a
+ b = 14 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
16a
+ 2b
= 34 6
Simplify:
8a
+ b = 17 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 3
Next, substitute 3 for a in equation (5):
7 3 + b = 14
( )
b = −7
Then, substitute 3 for a and -7 for b in equation (1):
a+ b+ c = 2
3 − 7 + c = 2
c = −4
Therefore, a = 3, b = -7, and c = -4. So, the
2
equation is y = 3x − 7x
− 4 .
15. Substitute the given points into
2
( 2, 5 ): 5 ( 2) ( 2)
2
( 4,3 ): 3 = a( 4) + b( 4)
+ c
2
( 5,13 ):13 = ( 5) + ( 5)
+
− − = a + b + c
a b c
Simplify these equations:
4a + 2b + c = −5 1
( )
( )
( )
16a + 4b + c = 3 2
25a + 5b + c = 13 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
( )
−4a − 2b − c = 5 4
Adding the left sides and right sides of equations
(2) and (4) gives:
12a
+ 2b
= 8 5
Simplify:
6a
+ b = 4 6
( )
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
21a
+ 3b
= 18 7
Simplify:
7a
+ b = 6 8
( )
( )
Eliminate b by multiplying equation (6) by -1 and
add each side to the corresponding side of equation
(8):
a = 2
Next, substitute 2 for a in equation (6):
6 2 + b = 4
( )
b = −8
Then, substitute 2 for a and -8 for b in equation (1):
4a + 2b + c = −5
4(2) + 2( − 8) + c = −5
8 − 16 = −5
c = 3
Therefore, a = 2, b = -8, and c = 3. So, the equation
2
is y = 2x − 8x
+ 3 .
17. Substitute the given points into
( 1, −1 ): − 1 =
2
( 1) + ( 1)
+
( 2, 1 ): 1
2
( 2) ( 2)
2
( 3,3 ): 3 = a( 3) + b( 3)
+ c
a b c
− − = a + b + c
Simplify these equations:
a+ b+ c = −1 1
( )
( )
( )
4a + 2b + c = −1 2
9a + 3b + c = 3 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = 1 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 0 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
8a
+ 2b
= 4 6
Simplify:
( )
204

SSM: Intermediate Algebra Homework 7.4
( )
4a
+ b = 2 7
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 2
Next, substitute 2 for a in equation (5):
3 2 + b = 0
( )
b = −6
Then, substitute 2 for a and -6 for b in equation (1):
a+ b+ c = −1
2 − 6 + c = −1
c = 3
Therefore, a = 2, b = -6, and c = 3. So, the equation
2
is y = 2x − 6x
+ 3 .
19. Substitute the given points into
( 0, 4 ): 4
2
( 0) ( 0)
( 2,8 ): 8
2
( 2) ( 2)
( 3,1 ):1 =
2
( 3) + ( 3)
+
= a + b + c
= a + b + c
a b c
Simplify these equations:
c = 4 1
( )
( )
( )
4a + 2b + c = 8 2
9a + 3b + c = 1 3
2
y = ax + bx + c .
Since c = 4, substitute 4 for c in equations (2) and
(3):
4a
+ 2b
+ 4 = 8
9a
+ 3b
+ 4 = 1
Simplifying these equations gives:
4a
+ 2b
= 4 4
( )
( )
9a
+ 3b
= −3 5
To eliminate b, multiply both sides of equation (4)
by -3 and both sides of equation (5) by 2:
−12a
− 6b
= −12 6
( )
( )
18a
+ 6b
= −6 7
Adding the left and the right sides of equations (6)
and (7) gives:
6a
= −18
a = −3
Next, substitute -3 for a in equation (4):
4( − 3) + 2b
= 4
2b
= 16
b = 8
Therefore, a = -3, b = 8, and c = 4. So, the equation
2
is y = − 3x + 8x
+ 4 .
21. Substitute the given points into
2
( 0, 1 ): 1 ( 0) ( 0)
2
( 1,3 ): 3 = a( 1) + b( 1)
+ c
2
( 2,13 ):13 ( 2) ( 2)
− − = a + b + c
= a + b + c
Simplify these equations:
c = −1 1
( )
( )
( )
a+ b+ c = 3 2
4a + 2b + c = 13 3
2
y = ax + bx + c .
Since c = -1, substitute -1 for c in equations (2) and
(3):
a + b + ( − 1) = 3
4a
+ 2 b + ( − 1) = 13
Simplifying these equations gives:
a + b = 4 4
( )
( )
4a
+ 2b
= 14 5
To eliminate b, multiply both sides of equation (4)
by -2 and add each side to the corresponding side
in equation (5):
2a
= 6
a = 3
Next, substitute 3 for a in equation (4):
3 + b = 4
b = 1
Therefore, a = 3, b = 1, and c = -1. So, the
2
equation is y = 3x + x − 1.
23. Substitute the given points into
( 0,17 ):17 2
( 0) ( 0)
( 2,11 ):11 2
( 2) ( 2)
( 3,2 ): 2 =
2
( 3) + ( 3)
+
= a + b + c
= a + b + c
a b c
Simplify these equations:
c = 17 1
( )
( )
( )
4a + 2b + c = 11 2
9a + 3b + c = 2 3
2
y = ax + bx + c .
Since c = 17, substitute 17 for c in equations (2)
and (3):
4a
+ 2b
+ 17 = 11
9a
+ 3b
+ 17 = 2
Simplifying these equations gives:
4a
+ 2b
= −6 4
( )
( )
9a
+ 3b
= −15 5
To eliminate b, multiply both sides of equation (4)
by -3 and both sides of equation (5) by 2:
205

Homework 7.4
SSM: Intermediate Algebra
( )
( )
−12a
− 6b
= 18 6
18a
+ 6b
= −30 7
Adding the left and the right sides of equations (6)
and (7) gives:
6a
= −12
a = −2
Next, substitute -2 for a in equation (4):
4( − 2) + 2b
= −6
2b
= 2
b = 1
Therefore, a = -2, b = 1, and c = 17. So, the
2
equation is y = − 2x + x + 17 .
25. Substitute the given points into
( 1,1 ):1 =
2
( 1) + ( 1)
+
( 2, 4 ): 4
2
( 2) ( 2)
( 3,9 ): 9
2
( 3) ( 3)
a b c
= a + b + c
= a + b + c
Simplify these equations:
a+ b+ c = 1 1
( )
( )
( )
4a + 2b + c = 4 2
9a + 3b + c = 9 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = − 1 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 3 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
8a
+ 2b
= 8 6
( )
Simplify:
4a
+ b = 4 7
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 1
Next, substitute 1 for a in equation (5):
3 1 + b = 3
( )
b = 0
Then, substitute 1 for a and 0 for b in equation (1):
a+ b+ c = 1
1+ 0 + c = 1
c = 0
Therefore, a = 1, b = 0, and c = 0. So, the equation
2
is y = x .
27. Substitute the given points into
2
f ( x)
= ax + bx + c .
( 1,1 ):1 =
2
( 1) + ( 1)
+
( 2, 2 ): 2
2
( 2) ( 2)
( 3,3 ): 3
2
( 3) ( 3)
a b c
= a + b + c
= a + b + c
Simplify these equations:
a+ b+ c = 1 1
( )
( )
( )
4a + 2b + c = 2 2
9a + 3b + c = 3 3
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = − 1 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 1 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
8a
+ 2b
= 2 6
Simplify:
4a
+ b = 1 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 0
Next, substitute 0 for a in equation (5):
3 0 + b = 1
( )
b = 1
Then, substitute 0 for a and 1 for b in equation (1):
a+ b+ c = 1
0 + 1+ c = 1
c = 0
Therefore, a = 0, b = 1, and c = 0. So, the equation
f x = x , which is a linear function.
is ( )
2
30. Solve for a in y = a( x − h)
+ k by substituting 5
for h and -7 for k since (5, -7) is the vertex, and
substitute 8 for x and 11 for y since (8, 11) lies on
the parabola:
206

SSM: Intermediate Algebra Homework 7.4
= − +
2
y a( x h)
k
= − + −
2
11 a(8 5) ( 7)
11 = 9a
− 7
18 = 9a
a = 2
2
Since a = 2, y = 2( x − 5) − 7 or, expanding the
solution:
y = 2( x − 5)( x − 5) − 7
= − + −
2
2( x 10x
25) 7
= − + −
2
2x
20x
50 7
= − +
2
2x
20x
43
31. Substitute the given points into
( 0, 4 ): 4
2
( 0) ( 0)
( 1,0 ): 0 =
2
( 1) + ( 1)
+
( 2,0 ): 0
2
( 2) ( 2)
= a + b + c
a b c
= a + b + c
Simplify these equations:
c = 4 1
.
( )
( )
( )
a+ b+ c = 0 2
4a + 2b + c = 0 3
2
y = ax + bx + c .
Since c = 4, substitute 4 for c in equations (2) and
(3):
a + b + 4 = 0
4a
+ 2b
+ 4 = 0
Simplify these equations:
a + b = −4 4
( )
( )
4a
+ 2b
= −4 5
Eliminate b by multiplying equation (4) by -2 and
add each side to the corresponding side of equation
(5):
2a
= 4
a = 2
Next, substitute 2 for a in equation (4):
2 + b = −4
b = −6
Therefore, a = 2, b = -6, and c = 4. So, the equation
2
is y = 2x − 6x
+ 4 .
33. Linear: y = 2x+2 since the slope is 2 and the y-
intercept is (0l, 2);
Quadratic: answers may vary. Example:
2
y = 2x
+ 2 ;
Exponential:
y = 2(2) x .
35. Three possible points are (2, 8), (3, 4), and (6, 4).
2
Substitute the given points into y = ax + bx + c .
( 2,8 ): 8
2
( 2) ( 2)
( 3,4 ): 4
2
( 3) ( 3)
( 6, 4 ): 4
2
( 6) ( 6)
= a + b + c
= a + b + c
= a + b + c
Simplify these equations:
4a + 2b + c = 8 1
( )
( )
( )
9a + 3b + c = 4 2
36a + 6b + c = 4 3
Eliminate c by multiplying both sides of equation
(1) by -1:
−4a − 2b − c = − 8 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
5a
+ b = − 4 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
32a
+ 4b
= − 4 6
Simplify:
8a
+ b = − 1 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
3a
= 3
a = 1
Next, substitute 1 for a in equation (5):
5 1 + b = −4
( )
b = −9
Then, substitute 1 for a and -9 for b in equation (1):
4a + 2b + c = 8
4(1) + 2( − 9) + c = 8
4 − 18 + c = 8
c = 22
Therefore, a = 1, b = -9, and c = 22. So, the
2
equation is y = x − 9x
+ 22 .
2
37. Solve for a in y = a( x − h)
+ k by substituting 5
for h and 8 for k since (5, 8) is the vertex, and
substitute 4 for x and 6 for y since (4, 6) lies on the
parabola:
2
y = a( x − h)
+ k
= − +
2
6 a(4 5) 8
6 = a + 8
a = −2
207

Homework 7.5
SSM: Intermediate Algebra
2
Since a = -2, y = −2( x − 5) + 8 or expanding the
solution:
y = −2( x − 5)( x − 5) − 42
= − − + −
2
2( x 10x
25) 42
= − + − −
2
2x
20x
50 42
2
2x
20x
42
= − + −
.
The quadratic regression equation, which is
2
f ( t) = − 0.66t + 21.49t
− 95.85 , fits the data well.
This equation can be verified by graphing it with
the scattergram:
39. Answers may vary. Example:
( 0, 2 ) , ( 1,1 ), and 6, 4 .
The equation of the parabola through these points
2
is y = x − 2x
+ 2 .
7. a. First draw a scattergram of the data.
Homework 7.5
1. a. A quadratic function would be reasonable.
b. A linear function would be reasonable.
c. An exponential function would be reasonable.
3.
d. None of the mentioned types of functions
would be reasonable for this scattergram.
The linear regression equation is :
f ( t) = 5.09t
+ 11.75
The exponential regression equation is :
f ( t ) = 16.63(1.14) t
The quadratic regression equation is :
2
f ( t) = 0.277t + 1.976t
+ 16.493
b. The exponential model gives the best estimate.
The data does not suggest a quadratic relationship
based on the scattergram above. A quadratic
function is not a reasonable function.
5. First, draw a scattergram of the data.
9. First draw a scattergram of the data:
208

SSM: Intermediate Algebra Homework 7.5
The quadratic regression equation, which is
2
f ( t) = 0.013t − 1.188t
+ 28.239 , fits the data well.
This equation can be verified by graphing it with
the scattergram:
The quadratic function is the best model since
the value of f(d) will eventually return to 0 as
the horizontal distance increases.
15. a. First draw a scattergram for C. The quadratic
regression equation,
2
C( t) = 0.90t − 19.94t
+ 121.38 , is the best
model for this data (as can be seen from the
scattergram below.
11. First draw a scattergram of the data:
The quadratic regression equation is, which is
2
f ( t) = 0.0067t − 0.12t
+ 6.39 , fits the data well.
This equation can be verified by graphing it with
the scattergram:.
Next, draw a scattergram of the data for H.
The quadratic regression equation,
H ( t) = 2.48t
− 1.30 , is the best model for this
data (as can be seen from the scattergram
below.
13. a., b. The quadratic function
2
f ( d) = − 0.000333d + 0.152d
+ 4 , is the best
model for the three points (0, 4), (60.5, 12) and
(127.3, 18).
b. Graph the two regression equations on the
same coordinate axes on your calculator. Use
your calculator to find the points of
intersection.
209

Homework 7.6
SSM: Intermediate Algebra
The intersection points for C and H are
16.8,40.5 . These points
( 8.1,18.8 ) and ( )
mean that the sales of Corona and Heineken
beers were equal in 1988 and in 1997, when
the sales were at 18.8 million cases, and 40.5
million cases, respectively for each company.
b. The parabola has a minimum point at the
vertex, since a is positive. Using a graphing
calculator, find that this minimum point is
approximately (45.77, 1.01).
Homework 7.6
1. a. Solve for t when f ( t ) = 0 :
2
0 0.66t
21.49t
95.85
= − + −
2
21.49 (21.49) 4( 0.66)( 95.85)
− ± − − −
t =
2( −0.66)
t ≈ 5.33 or t ≈ 27.23
The t-intercepts are ( 5.33,0 ) and ( )
27.23,0 .
So, according to this model, no firms
performed drug tests on employees in 1985 or
2027.
So, according to this model, a 46-year-old
driver has 1 fatal crash per 100 million miles.
5. a. Using a graphing calculator, find that the
maximum point of the parabola is
approximately (228, 21.3).
b. Any value of t that cause the percentage of
firms performing drug tests to be negative.
This would occur in years before 1985
5.33
t < 27.23 .
( t < ) and years after 2027 ( )
c. Find the vertex of f:
The t-coordinate of the vertex is
5.33 + 27.23
= 16.28 . Now, find f(16.28):
2
f (16.28)
2
0.66(16.28) 21.49(16.28) 95.85
= − + −
= 79.08
So, the percentage of firms performing drug
tests will reach a maximum of 79% in the year
1996.
3. a. Answers may vary. Example:
So the maximum height is approximately 21.3
feet.
b. Find f(400):
f (400)
2
0.000333(400) 0.152(400) 20.18
= − + +
= 11.52
According to the model, the height of the ball
is approximately 11.5 feet when the ball goes a
horizontal distance of 400 feet. Therefore, the
player did hit a home run.
For a person who is 21 years old, find f(21):
7. a.
2
f (30) = − 0.108(30) + 4.29(30) + 21.83
f (21)
2
0.013(21) 1.19(21) 28.24
= − +
= 8.98
According to the model, a 21-year-old driver
has approximately 9 fatal crashes per 100
million miles.
= 53.33
According to this model, in 2010, 53.33% of
households will have cable. Model breakdown
has likely occurred.
b. Solve for t when f ( t ) = 30
210

SSM: Intermediate Algebra Homework 7.6
2
30 0.108t
4.29t
21.83
= − + +
= − + −
2
0 0.108t
4.29t
8.17
− ± − − − −
t =
2( −0.108)
2
4.29 ( 4.29) 4( 0.108)( 8.17)
t ≈ 2.01 or t ≈ 37.72
According to this model, 30% of households
had cable in 1982 and will have cable in 2018.
Model breakdown has likely occurred for the
2018 prediction.
c. Using a graphing calculator, find that the
maximum point of the parabola is
approximately (19.9, 64.4).
c.
2
0.0067t
0.12t
6.39 300
2
0.0067t
0.12 293.61 0
t =
− + =
− − =
± − −
2
0.12 (0.12) 4(0.0067)( 293.61)
2(0.0067)
t ≈ −200.6 or t ≈ 218.5
So, according to the model, the population of
the U.S. was 300 million in 1589 (model
breakdown) and will be 300 million in 2009.
So, according to the model, the maximum
percentage of households with cable television
will be 64.4% and this will occur in 2000.
d. The vertex of the parabola is approximately
8.96,5.85 . It can be found by graphing the
( )
model with a graphing calculator.
d. Since the percent will never go below 0, model
breakdown occurs for years before 1976. Since
the percent decreases for all values of t greater
than 19.9, model breakdown also occurs for
years after 2000.
9. Solve for t when C(t) = H(t)
− + = −
2
0.90t 19.94t 121.38 2.48t
1.30
− + =
2
0.90t
22.42t
122.68 0
t =
± −
2
22.42 (22.42) 4(0.90)(122.68)
t ≈ 8.1 or t ≈ 16.8
2(0.90)
So, sales of Corona and Heineken were about the
same in 1988 and 1997.
e.
The portion of the parabola to the left of the
vertex suggests that the population decreased
before 1796, which is false. Model breakdown
occurs for years prior to 1796.
P
11. a.
2
f (217) = 0.0067(217) − 0.12(217) + 6.39
≈ 295.85
t
b. Let f ( t ) = 300
211

Homework 7.6
SSM: Intermediate Algebra
14. a.
b.
212
E (209) = 3.9(1.030) ≈ 2053.8
According to the model E, the U.S. population
was 2.1 billion people in 2002. This is a much
higher [prediction than the actual value of
286.7 million.
= − + −
2
0 1.30t
28.99t
61.29
− ± − − −
t =
2( −1.30)
2
28.99 (28.99) 4( 1.30)( 61.29)
t ≈ 2.36 or t ≈ 19.94
So, according to the model, there were no
schools with Internet access in 1992 and there
will be no schools with Internet access in
2010 (model breakdown for the 2010
prediction).
c. f(t) is quadratic, so find the vertex. The vertex
is approximately (11.5, 100.33). It can be
found by graphing the model with a graphing
calculator.
Yes, there is a great difference in these
predictions. The model E gives more accurate
predictions for years between 1790 and 1860
only.
15. a. First, draw a scattergram of the data.
This means that the model predicts that the
percentage of schools with Internet access will
reach a maximum of 100% in 2001.
2
= ( ) = − 1.30 + 28.99 − 61.29 , is the
p f t t t
best model for this data (as can be seen from
the scattergram below.
d. From part b, we know that the model predicts
there will be no Internet access at 1992 ( t =
2.36) or 2010 (t = 19.94) (the latter is likely
model breakdown). Model breakdown occurs
for certain before 1992 (t < 2.36) and after
2010 (t > 19.94) as the percent of schools with
Internet access can not be negative.
e.
Better model
b. Solve for t when f ( t ) = 0 :
212

SSM: Intermediate Algebra
Chapter 7 Review Exercises
17. a. Linear: f ( t) = 1.86t
− 1.61
Exponential: f ( t ) = 1.63(1.26) t
Quadratic:
f t = t − t +
2
( ) 0.25 0.92 2.62
So, the quadratic model predicts that there will
be enough wind energy to meet the needs of
the world in 2049.
e. Answers may vary. Example:
An exponential model gets larger faster than a
quadratic model, so the exponential function
will take less time to obtain enough power to
supply the world.
The quadratic and exponential models fit the
data well.
b. The exponential model, as it gives values
lower than 1.9 (the 1990 capacity) for years
before 1990, whereas the quadratic model
gives values higher than 1.9 for years before
1990.
19. Answers may vary.
Chapter 7 Review Exercises
1. 72 = 36⋅ 2 = 36 2 = 6 2
2. 3 = 3 = 3 ⋅ 5 =
15
5 5 5 5 5
c. Since 1 MW meets the needs of 1000 people.
1 thousand MW meets the needs of
(1000)(1000) = 1 million people. There are 6.2
billion people in the world (or, 6200 million
people). So, 6200 thousand MW are needed.
Let f(t) = 6200 and solve for t:
3.
4.
50 50 25⋅
2 5 2
= = =
49 49 7 7
49 49 7
= =
100 100 10
t
6200 = 1.63(1.26)
6200 t
= 1.26
1.63
t ⎛ 6200 ⎞
log(1.26 ) = log⎜ ⎟
⎝ 1.63 ⎠
⎛ 6200 ⎞
t log(1.26) = log⎜ ⎟
⎝ 1.63 ⎠
⎛ 6200 ⎞
log ⎜ ⎟ 1.63
t =
⎝ ⎠
log(1.26)
t ≈ 35.7
So, according to the exponential model, there
will be enough wind energy to meet the needs
of the world in 2026.
d. Let f(t) = 6200 and solve for t:
2
6200 0.25t
0.92t
2.62
2
0 0.25t
0.92t
6197.38
t =
= − +
= − −
2
0.92 ( 0.92) 4(0.25)( 6197.38)
± − − −
2(0.25)
t ≈ −155.6 (model breakdown) or t ≈ 159.3
5. 6x
2 − 3x
− 2 = 0
− − ± − − −
x =
2(6)
x =
2
( 3) ( 3) 4(6)( 2)
3 ± 57
12
6. 5x 2 = 7
x =
5
2 7
= ±
= ±
7
5
7
5
7 5
= ± ⋅
5 5
= ±
35
5
213

Chapter 7 Review Exercises
SSM: Intermediate Algebra
7.
2
5( x − 3) + 4 = 7
2
5( x 3) 3
2
− =
( x 3)
− =
x − 3 = ±
3
5
3
5
3 5
x − 3 = ± ⋅
5 5
x − 3 = ±
x = 3 ±
x =
8. x 2 = 98
x = ±
15
5
15
5
15 ± 15
5
98
= ± 49⋅
2
= ± 7 2
9. ( x + 1)( x − 7) = 4
10.
− 7 + − 7 = 4
2
x x x
2
x − 6x
− 11 = 0
− − ± − − −
x =
2(1)
=
=
2
( 6) ( 6) 4(1)( 11)
6 ± 80
2
6 ± 4 5
2
= 3 ± 2 5
− + =
2
2( x 4) 9
( x + 4) = −
2
2 9
9
x + 4 = ± −
2
Since the square of a negative number is not a real
number, there are no real solutions.
11. 3x
2 = 6x
− =
2
3x
6x
0
3 x( x − 2) = 0
3x
= 0 or x − 2 = 0
x = 0 or x = 2
12. x
2 − 2x
− 24 = 0
( x − 6)( x + 4) = 0
x − 6 = 0 or x + 4 = 0
x = 6 or x = −4
13. 4x 2 − 25 = 0
(2x
− 5)(2x
+ 5) = 0
2x
− 5 = 0 or 2x
+ 5 = 0
2x
= 5 or 2x
= −5
5 5
x = or x = −
2 2
14. 3x
2 − 7x
+ 2 = 0
(3x
−1)( x − 2) = 0
15.
3x
− 1 = 0 or x − 2 = 0
3x
= 1 or x = 2
1
x = or x = 2
3
2
25x − 9 = 0
(5x
− 3)(5x
+ 3) = 0
5x
− 3 = 0 or 5x
+ 3 = 0
5x
= 3 or 5x
= −3
3 3
x = or x = −
5 5
16. 3x 2 − 1 = x 2 − 5x
+ 1
2
2x
5x
2 0
− ± − −
x =
2(2)
x =
+ − =
2
5 5 4(2)( 2)
− 5 ± 41
4
17. 2x
2 = 4 − 5x
2
2x
5x
4 0
+ − =
− ± − −
x =
2(5)
2
5 5 4(2)( 4)
− 5 ± 57
x =
10
214

SSM: Intermediate Algebra
Chapter 7 Review Exercises
18.
19.
20.
21.
− =
2
4x
x 1
2
x − 4x
= −1
2
x − 4x
+ 4 = − 1+
4
− =
2
( x 2) 3
x − 2 = ± 3
x = 2 ± 3
− + = − + +
2
2x
− 5 = 0
2 2
3x x 5x x 5
2
2x
5
x
2
=
=
x = ±
x = ±
5
2
5
2
5
2
5 2
x = ± ⋅
2 2
x = ±
10
2
− + = +
2
2 x(2x 5) 13 3x
5
2 2
4x 10x 13 3x
5
2
x − 10x
+ 8 = 0
− − ± − −
x =
2(1)
x =
x =
− + = +
2
( 10) ( 10) 4(1)(8)
10 ± 68
2
10 ± 2 17
2
x = 5 ± 17
2
3 x( x 7) 13 2x
7
− + = −
− + = −
2 2
3x 21x 13 2x
7
2
x − 21x
+ 20 = 0
( x − 20)( x − 1) = 0
x − 20 = 0 or x − 1 = 0
x = 20 or x = 1
22.
23.
24.
25.
2 2
2( x x) 3( x 2 x) 5
− = − +
2 2
2x 2x 3x
6 5
2
x − 4x
+ 5 = 0
− − ± − −
x =
2(1)
x =
− = − +
2
( 4) ( 4) 4(1)(5)
4 ± −4
2
Since the square of a negative number is not a real
number, there are no real solutions.
2 2
( x 2) ( x 3) 2
+ + − =
+ 4 + 4 + − 6 + 9 = 2
2 2
x x x x
2
2x
2x
13 2
2
2x
2x
11 0
2
( 2) ( 2) 4(2)(11)
− − ± − −
x =
2(2)
x =
− + =
− + =
2 ± −84
4
Since the square of a negative number is not a real
number, there are no real solutions.
2
5(5x − 8) = 0
2
25x
40 9
2
25x
49
x
2
=
x = ±
x = ±
− =
=
49
25
49
25
7
5
− =
2
2.7x
5.1x
9.4
− − =
2
2.7x
5.1x
9.4 0
− − ± − − −
x =
2(2.7)
2
( 5.1) ( 5.1) 4(2.7)( 9.4)
x ≈ −1.15 or x ≈ 3.04
215

Chapter 7 Review Exercises
SSM: Intermediate Algebra
26.
2
1.7( x 2.3) 3.4 2.8
− = −
2
1.7x
3.91 3.4 2.8
− = −
+ − =
2
1.7x
2.8x
7.31 0
− ± − −
x =
2(1.7)
2
2.8 2.8 4(1.7)( 7.31)
x ≈ −3.05 or x ≈ 1.41
27. x
2 + 6x
− 4 = 0
2
x + 6x
= 4
2
x + 6x
+ 9 = 4 + 9
+ =
2
( x 3) 13
x + 3 = ± 13
x = − 3 ± 13
28. 8x
2 + 12x
+ 8 = 0
2 3
x + x + 1 = 0
2
2 3 9 9
x + x + = − 1+
2 16 16
2
⎛ 3 ⎞ 7
⎜ x + ⎟ = −
⎝ 4 ⎠ 16
3 7
x + = ± −
4 16
x
x
Since the square of a negative number is not a real
number, there are no real solutions.
29. 2x
2 = − 3x
+ 6
2
2x
+ 3x
= 6
2 3
x + x = 3
2
2 3 9 9
x + x + = 3 +
2 16 16
⎛ 3 ⎞ 57
⎜ x + ⎟ =
⎝ 4 ⎠ 16
3 57
x + = ±
4 16
3 57
x + = ±
4 4
3 57
x = − ±
4 4
x =
2
− 3 ± 57
4
30. x( x − 3) + 2 = 2 x( x + 1) + 1
− 3 + 2 = 2 + 2 + 1
2 2
x x x x
2
1 x 5
2 25 25
x + 5x
+ = 1+
4 4
⎛ 5 ⎞ 29
⎜ x + ⎟ =
⎝ 2 ⎠ 4
2
5 29
x + = ±
2 4
5 29
x + = ±
2 2
x
5 29
x = − ±
2 2
x =
= +
− 5 ± 29
2
31. Solve for x when f ( x ) = 0 :
2
3x
10 0
2
3x
10
x
2
=
x = ±
10
3
10
3
10 3
x = ± ⋅
3 3
x = ±
− =
=
30
3
⎛
The x-intercepts are
⎜
⎝
30 ,0
3
32. Solve for x when g( x ) = 0 :
− + + =
2
2( x 3) 5 0
− + = −
2
2( x 3) 5
2
+ =
( x 3)
x + 3 = ±
5
2
5
2
⎞
⎟
⎠
and
⎛
⎜
−
⎝
30 ,0
3
⎞
⎟
.
⎠
216

SSM: Intermediate Algebra
Chapter 7 Review Exercises
5 2
x + 3 = ± ⋅
2 2
x + 3 = ±
x = − 3 ±
x =
10
2
10
2
− 6 ± 10
2
⎛ − 6 + 10 ⎞
The x-intercepts are
⎜ ,0
2 ⎟
and
⎝ ⎠
⎛ −6 − 10 ⎞
⎜ ,0
2 ⎟
.
⎝ ⎠
33. Solve for x when h( x ) = 0 :
2
3x
2x
2 0
− ± − −
x =
2(3)
x =
x =
x =
+ − =
2
2 2 4(3)( 2)
− 2 ± 28
6
− 2 ± 2 7
6
− 1±
7
3
⎛ −1−
7 ⎞
The x-intercepts are
⎜ ,0
3 ⎟
and
⎝ ⎠
⎛ − 1+
7 ⎞
⎜ ,0
3 ⎟
.
⎝ ⎠
34. Solve for x when k( x ) = 0 :
− + − =
2
5x
3x
1 0
− ± − − −
x =
2( −5)
2
3 3 4( 5)( 1)
− 3 ± −11
x =
−10
Since the square root of a negative number is not a
real number, there are no real number solutions.
Therefore, there are no x-intercepts.
35. Solve for x when f ( x ) = 0 :
2
2x
3x
4 0
− − ± − − −
x =
2(2)
x =
− − =
2
( 3) ( 3) 4(2)( 4)
3 ± 41
4
⎛ 3 + 41 ⎞
The x-intercepts are
,0
and
⎜ 4 ⎟
⎝ ⎠
⎛ 3 − 41 ⎞
,0
.
⎜ 4 ⎟
⎝ ⎠
36. Solve for x when g( x ) = 0 :
− + + =
2
3x
5x
7 0
− ± − −
x =
2( −3)
2
5 5 4( 3)(7)
− 5 ± 109
x =
−6
− 1(5 ± 109)
x =
−6
x =
5 ± 109
6
⎛ 5 + 109 ⎞
The x-intercepts are
⎜ ,0
6 ⎟
and
⎝ ⎠
⎛ 5 − 109 ⎞
⎜ ,0
6 ⎟
.
⎝ ⎠
37. Factoring:
2
x − 2x
− 8 = 0
( x − 4)( x + 2) = 0
x − 4 = 0 or x + 2 = 0
x = 4 or x = −2
Completing the square:
2
x − 2x
− 8 = 0
2
x − 2x
= 8
2
x − 2x
+ 1 = 8 + 1
− =
2
( x 1) 9
x − 1 = ± 9
x − 1= ± 3
x − 1 = 3 or x − 1= −3
x = 4 or x = −2
217

Chapter 7 Review Exercises
SSM: Intermediate Algebra
Quadratic Formula:
2
x − 2x
− 8 = 0
− − ± − − −
x =
2(1)
2
( 2) ( 2) 4(1)( 8)
2 ± 36
x =
2
2 ± 6
x =
2
2 + 6 8 2 − 6 −4
x = = or x = =
2 2 2 2
x = 4 or x = −2
38. Find k when the discriminant equals 0:
2
b − 4ac
= 0
k
k
k
39. a.
2
2
2
− 4(3)(12) = 0
− 144 = 0
= 144
x = ±
x = ± 12
b.
144
− + =
2
3x
6x
7 3
2
3x
6x
4 0
− − ± − −
x =
2(3)
x =
− + =
2
( 6) ( 6) 4(3)(4)
6 ± −12
6
Since the square root of a negative is not a real
number, there are no real number solutions.
There is no such value for x.
− + =
2
3x
− 6x
+ 3 = 0
2
3x
6x
7 4
− − ± − −
x =
2(3)
x
x
2
( 6) ( 6) 4(3)(3)
6 ± 0
=
6
6
= = 1
6
c.
2
3x
− 6x
+ 7 = 5
2
3x
− 6x
+ 2 = 0
− − ± − −
x =
2(3)
x =
x =
x =
2
( 6) ( 6) 4(3)(2)
6 ± 12
6
6 ± 2 3
6
3 ± 3
3
d. Answers may vary.
2
40. Using the equation y = a( x − h)
+ k , substitute -4
for h and 3 for k since the vertex is (-4, 3). Also,
substitute -3 for x and 6 for y since (-3, 6) lies on
the parabola.
2
y = a( x − h)
+ k
2
6 a( 3 ( 4)) 3
= − − − +
= − + +
2
6 a( 3 4) 3
6 = a(1) + 3
3 = a
So, the equation is
y = + +
2
3( x 4) 3
41. Substitute the given points into
2
( 1,3 ): 3 = a( 1) + b( 1)
+ c
( 2,6 ): 6 =
2
( 2) + ( 2)
+
( 3,13 ):13 2
( 3) ( 3)
a b c
= a + b + c
Simplify these equations:
a+ b+ c = 3 1
( )
( )
( )
4a + 2b + c = 6 2
9a + 3b + c = 13 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = − 3 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 3 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
8a
+ 2b
= 10 6
Simplify:
4a
+ b = 5 7
( )
( )
218

SSM: Intermediate Algebra
Chapter 7 Review Exercises
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
a = 2
Next, substitute 2 for a in equation (5):
3 2 + b = 3
( )
b = −3
Then, substitute 2 for a and -3 for b in equation (1):
a+ b+ c = 3
2 + ( − 3) + c = 3
c = 4
Therefore, a = 2, b = -3, and c = 4. So, the equation
2
is y = 2x − 3x
+ 4 .
42. Substitute the given points into
2
( 1,4 ): 4 = a( 1) + b( 1)
+ c
( 3, −2 ): − 2 =
2
( 3) + ( 3)
+
( 4, 11 ): 11
2
( 4) ( 4)
a b c
− − = a + b + c
Simplify these equations:
a+ b+ c = 4 1
( )
( )
( )
9a + 3b + c = −2 2
16a + 4b + c = −11 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = − 4 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
8a
+ 2b
= − 6 5
Simplify:
4a
+ b = − 3 6
( )
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
15a
+ 3b
= − 15 7
Simplify:
5a
+ b = − 5 8
( )
( )
Eliminate b by multiplying equation (6) by -1 and
add each side to the corresponding side of equation
(8):
a = − 2
Next, substitute -2 for a in equation (6):
4 − 2 + b = −3
( )
b = 5
Then, substitute -2 for a and 5 for b in equation (1):
a+ b+ c = 4
− 2 + 5 + c = 4
c = 1
Therefore, a = -2, b = 5, and c = 1. So, the equation
2
is y = − 2x + 5x
+ 1.
43. Substitute the given points into
( 2,9 ): 9 =
2
( 2) + ( 2)
+
( 3,18 ):18 2
( 3) ( 3)
( 5,48 ): 48
2
( 5) ( 5)
a b c
= a + b + c
= a + b + c
Simplify these equations:
4a + 2b + c = 9 1
( )
( )
( )
9a + 3b + c = 18 2
25a + 5b + c = 48 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−4a − 2b − c = − 9 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
5a
+ b = 9 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
21a
+ 3b
= 39 6
Simplify:
7a
+ b = 13 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
2a
= 4
a = 2
Next, substitute 2 for a in equation (5):
5 2 + b = 9
( )
b = −1
Then, substitute 2 for a and -1 for b in equation (1):
4a + 2b + c = 9
4(2) + 2( − 1) + c = 9
8 − 2 + c = 9
c = 3
Therefore, a = 2, b = -1, and c = 3. So, the equation
2
is y = 2x − x + 3 .
219

Chapter 7 Review Exercises
SSM: Intermediate Algebra
44. Substitute the given points into
2
( 0,5 ): 5 = a( 0) + b( 0)
+ c
2
( 2,3 ): 3 = a( 2) + b( 2)
+ c
2
( 4, 15 ): 15 ( 4) ( 4)
− − = a + b + c
Simplify these equations:
c = 5 1
( )
( )
( )
4a + 2b + c = 3 2
16a + 4b + c = −15 3
2
y = ax + bx + c .
Since c = 5, substitute 5 for c in equations (2) and
(3):
4a
+ 2 b + (5) = 3
16a
+ 4 b + (5) = −15
Simplifying these equations gives:
4a
+ 2b
= −2 4
( )
( )
16a
+ 4b
= −20 5
To eliminate b, multiply both sides of equation (4)
by -2 and add each side to the corresponding side
in equation (5):
8a
= −16
a = −2
Next, substitute -2 for a in equation (4):
4a
+ 2b
= −2
4( − 2) + 2b
= −2
2b
= 6
b = 3
Therefore, a = -2, b = 3, and c = 5. So, the
2
equation is y = − 2x + 3x
+ 5 .
45. Substitute the given points into
2
( 0,7 ): 7 = ( 0) + ( 0)
+
2
( 1,8 ): 8 = a( 1) + b( 1)
+ c
2
( 3, 8 ): 8 ( 3) ( 3)
a b c
− − = a + b + c
Simplify these equations:
c = 7 1
( )
( )
( )
a+ b+ c = 8 2
9a + 3b + c = −8 3
2
y = ax + bx + c .
Since c = 7, substitute 7 for c in equations (2) and
(3):
a + b + (7) = 8
9a
+ 3 b + (7) = −8
Simplifying these equations gives:
( )
( )
a + b = 1 4
9a
+ 3b
= −15 5
To eliminate b, multiply both sides of equation (4)
by -3 and add each side to the corresponding side
in equation (5):
6a
= −18
a = −3
Next, substitute -3 for a in equation (4):
a + b = 1
( − 3) + b = 1
b = 4
Therefore, a = -3, b = 4, and c = 7. So, the
2
equation is y = − 3x + 4x
+ 7 .
46. Substitute the given points into
( 0,3 ): 3 =
2
( 0) + ( 0)
+
( 2,13 ):13 2
( 2) ( 2)
( 3,24 ): 24
2
( 3) ( 3)
a b c
= a + b + c
= a + b + c
Simplify these equations:
c = 3 1
( )
( )
( )
4a + 2b + c = 13 2
9a + 3b + c = 24 3
2
y = ax + bx + c .
Since c = 3, substitute 3 for c in equations (2) and
(3):
4a
+ 2 b + (3) = 13
9a
+ 3 b + (3) = 24
Simplifying these equations gives:
4a
+ 2b
= 10 4
( )
( )
9a
+ 3b
= 21 5
To eliminate b, multiply both sides of equation (4)
by -3 and multiply both sides of equation (5) by 2
−12a
− 6b
= −30 6
( )
( )
18a
+ 6b
= 42 7
To solve for a, add the corresponding sides of
equations (6) and (7) together:
6a
= 12
a = 2
Next, substitute 2 for a in equation (4):
4a
+ 2b
= 10
4(2) + 2b
= 10
2b
= 2
b = 1
Therefore, a = 2, b = 1, and c = 3. So, the equation
2
is y = 2x + x + 3 .
220

SSM: Intermediate Algebra
Chapter 7 Review Exercises
47. Linear:
2 − 4
slope = m = = −2
1−
0
y − intercept = (0,4)
So, y = − 2x
+ 4
Exponential:
As the value of x increases by 1, the value of y is
1
multiplied by ½,, so the base b = . The y-
2
intercept is (0, 4). Therefore, the exponential
x
⎛ 1 ⎞
function is y = 4 ⎜ ⎟
⎝ 2 ⎠
2a
+ b = −3
2(1) + b = −3
b = −5
Therefore, a = 1, b = -5, and c = 7. So, the
2
equation is y = x − 5x
+ 7 .
49. a. First, draw a scattergram of the data. The
quadratic regression equation,
2
f ( t) = − 0.0123t + 1.85t
+ 0.64 , is the best
model for this data.
Quadratic:
Answers may vary. Example:
48. Substitute the given points into
( 0,7 ): 7
2
( 0) ( 0)
( 2,1 ):1 2
( 2) ( 2)
( 5,7 ): 7
2
( 5) ( 5)
= a + b + c
= a + b + c
= a + b + c
Simplify these equations:
c = 7 1
( )
( )
( )
4a + 2b + c = 1 2
25a + 5b + c = 7 3
y = − +
2
2x
4
2
y = ax + bx + c .
Since c = 7, substitute 7 for c in equations (2) and
(3):
4a
+ 2 b + (7) = 1
25a
+ 5 b + (7) = 7
Simplifying these equations gives:
4a
+ 2b
= −6 4
( )
( )
25a
+ 5b
= 0 5
Simplify these equations:
2a
+ b = −3 6
( )
( )
5a
+ b = 0 7
To eliminate b, multiply both sides of equation (6)
by -1 and add each side to the corresponding side
in equation (7):
3a
= 3
a = 1
Next, substitute 1 for a in equation (5):
b. Find f(18)
2
f (18) = − 0.0123(18) + 1.85(19) + 0.64 ≈ 29.9
So, about 30% of 18-year-old Americans
voted.
c. Solve for t when f(t) = 50
= − + +
2
50 0.0123t
1.85t
0.64
= − + −
2
0 0.0123t
1.85t
49.36
− ± − − −
t =
2( −0.0123)
2
1.85 1.85 4( 0.0123)( 49.36)
t ≈ 34.7 or t ≈ 115.7
t ≈ 115.7 is model breakdown. So, according
to the model, 50% of people at age 35 vote.
d. Using a graphing calculator, find that the
maximum point of the parabola is
approximately (75.3, 70.2).
So the age of Americans who are most likely
to vote is age 75.
221

Chapter 7 Test
SSM: Intermediate Algebra
50. a.
e. Answers may vary. Example:
Candidates tend to focus on issues for older
people because they are the most likely to
vote.
b. Using quadratic regression:
2
f ( t) = 0.037t − 0.47t
+ 31.84
c.
d.
2
f (37) = 0.037(37) − 0.47(37) + 31.84 ≈ 65.1
This means that in 2001, 65.1% of the grades
at Princeton will be A’s.
= − +
2
100 0.037t
0.47t
31.84
2
0 0.037t
0.47t
68.16
t =
Chapter 7 Test
= − −
2
0.47 ( 0.47) 4(0.037)( 68.16)
± − − −
2(0.037)
t ≈ −37.04 or t ≈ 49.74
In 1933 and in 20020, all grades at Princeton
will be A’s. Model breakdown has occurred.
1. 32 = 16 ⋅ 2 = 16 2 = 4 2
2. 7 = 7 ⋅ 2 =
7 2
2 2 2 2
3.
20 20 2 5 2 5 3 2 15
= = = ⋅ =
75 75 5 3 5 3 3 15
4. x
2 − 3x
− 10 = 0
( x − 5)( x + 2) = 0
x − 5 = 0 or x + 2 = 0
x = 5 or x = −2
5. 6x 2 = 100
2 100
x =
6
2 50
x =
3
6.
7.
x = ±
x = ±
50
3
50
3
5 2 3
x = ± ⋅
3 3
x = ±
5 6
3
− + − =
2
3x
x 4 0
− ± − − −
x =
2( −3)
2
1 1 4( 3)( 4)
1± −47
=
−6
Since the square of a negative number is not a real
number, there are no real solutions.
2
4( x − 3) + 1 = 7
2
4( x 3) 6
2
− =
( x 3)
2
− =
( x 3)
− =
x − 3 = ±
6
4
3
2
3
2
3 2
x − 3 = ± ⋅
2 2
x − 3 = ±
x = 3 ±
6
2
6
2
6 ± 6
x =
2
8. 3x
2 − 21x
= 0
3 x( x − 7) = 0
3x
= 0 or x − 7 = 0
x = 0 or x = 7
222

SSM: Intermediate Algebra
Chapter 7 Test
9. x 2 − 81 = 0
( x − 9)( x + 9) = 0
x − 9 = 0 or x + 9 = 0
x = 9 or x = −9
10. ( x − 3)( x + 5) = 6
2
x x x
+ 5 − 3 − 15 = 6
2
x + 2x
− 21 = 0
2
x + 2x
= 21
2
x + 2x
+ 1 = 21+
1
2
( x 1) 22
+ =
x + 1 = ± 22
x = − 1±
22
14.
2
3.7x
2.4 5.9
= −
2
3.7x
5.9x
2.4 0
x
+ − =
2
5.9 5.9 4(3.7)( 2.4)
− ± − −
x =
2(3.7)
x ≈ −1.93 or x ≈ 0.34
15. x
2 − 8x
− 2 = 0
2
x − 8x
= 2
2
x − 8x
+ 16 = 2 + 16
2
( x 4) 18
− =
x − 4 = ± 18
x − 4 = ± 3 2
x = 4 ± 3 2
11.
2
5( x 4) 10
− + =
2
( x 4) 2
+ = −
x + 4 = ± −2
Since the square of a negative number is not a real
number, there are no real solutions.
12. 2 x( x + 5) = 4x
− 3
2
2x 10x 4x
3
+ = −
2
2x
6x
3 0
+ + =
− ±
x =
x =
x =
x =
2
6 6 4(2)(3)
−
2(2)
− 6 ± 12
4
− 6 ± 2 3
4
− 3 ± 3
2
13. 9x
2 = 16 + 24x
2
9x
− 24x
− 16 = 0
2
( 24) ( 24) 4(9)( 16)
− − ± − − −
x =
2(9)
24 ± 1152
x =
18
x =
24 ± 24 2
18
4 ± 4 3
x =
3
16.
2
2( x 4) 3
− = −
2
2x
8 3
−
= −
2
2x
3x
8
+ =
2 3
x + x = 4
2
2 3 9 9
x + x + = 4 +
2 16 16
2
x
x
⎛ 3 ⎞ 64 9
⎜ x +
4
⎟ = +
⎝ ⎠ 16 16
⎛ 3 ⎞ 73
⎜ x + ⎟ =
⎝ 4 ⎠ 16
3 73
x + = ±
4 16
3 73
x + = ±
4 4
3 73
x = − ±
4 4
− 3 ± 73
x =
4
2
223

Chapter 7 Test
SSM: Intermediate Algebra
17. Solve for x when f ( x ) = 0 :
2
3x
8x
1 0
− − ± − −
x =
2(3)
x =
x =
x =
− + =
2
( 8) ( 8) 4(3)(1)
8 ± 52
6
8 ± 2 13
6
4 ± 13
3
⎛
The x-intercepts are
⎜
⎝
⎛ 4 − 13 ⎞
⎜ ,0
3 ⎟
.
⎝ ⎠
18. Solve for x when f(x) = 0
= − − +
2
0 2( x 3) 5
2
2( x 3) 5
2
− =
( x 3)
− =
x − 3 = ±
5
2
5
2
4 + 13 ,0
3
5
x − 3 = ±
2
Rationalize the denominator:
5 2
x − 3 = ± ⋅
2 2
Solve for x :
x − 3 = ±
10
2
10
x = 3 ±
2
x ≈ 1.42 or x ≈ 4.58
⎞
⎟
and
⎠
So, the x-intercepts are approximately ( 1.42,0 )
and ( 4.58,0 ) . Since these points are symmetric,
the average of the x-coordinates, 1.42 + 4.58 = 3 ,
2
is the x-coordinate of the vertex of f(x). Substitute 3
for x in the function to find the y-coordinate of the
vertex:
224
2
h( x ) = −2(3 − 3) + 5 = 5
Therefore, the vertex is (3, 5).
19. Solve for a when the discriminant equals 0:
2
b − 4ac
= 0
2
( 4) 4( a)(4 a) 0
2
16 16a
0
−
a
− − =
2
16a
16
2
− =
= 1
a = ± 1
= −
20. Substitute the given points into
2
( 1,4 ): 4 = a( 1) + b( 1)
+ c
( 2,9 ): 9 =
2
( 2) + ( 2)
+
( 3,16 ):16 2
( 3) ( 3)
a b c
= a + b + c
Simplify these equations:
a+ b+ c = 4 1
( )
( )
( )
4a + 2b + c = 9 2
9a + 3b + c = 16 3
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = − 4 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 5 5
( )
Simplify:
8a
+ 2b
= 12 6
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
4a
+ b = 6 7
( )
Eliminate b by multiplying equation (6) by -1 and
add each side to the corresponding side of equation
(7):

SSM: Intermediate Algebra
Chapter 7 Test
a = 1
Substitute 1 for a in equation (5):
3 1 + b = 5
( )
b = 2
Then, substitute 1 for a and 2 for b in equation (1):
a+ b+ c = 4
1+ 2 + c = 4
c = 1
Therefore, a = 1, b = 2, and c = 1. So, the equation
2
is y = x + 2x
+ 1.
2
21. Using the equation y = a( x − h)
+ k , substitute 5
for h and 3 for k since the vertex is (5, 3). Also,
substitute 3 for x and 11 for y since (3,11) lies on
the parabola.
2
y = a( x − h)
+ k
2
11 a(3 5) 3
2
11 a(3 5) 3
2
11 a( 2) 3
11 = 4a
+ 3
8 = 4a
a = 2
= − +
= − +
= − +
c.
2
x − 6x
+ 11 = 3
2
x − 6x
+ 8 = 0
2
( 6) ( 6) 4(1)(8)
− − ± − −
x =
2(1)
6 ± 4
x =
2
6 ± 2
x =
2
8 4
x = or x =
2 2
x = 4 or x = 2
23. a. First, draw a scattergram of the data. The
quadratic regression equation,
2
f ( t) = − 1.14t + 25.26t
− 81.2 , is the best
model for this data.
So, the equation is
y = − +
2
2( x 5) 3
22. a.
b.
2
x − 6x
+ 11 = 1
2
x − 6x
+ 10 = 0
− − ± − −
x =
2(1)
x =
2
( 6) ( 6) 4(1)(10)
6 ± −4
2
Since the square root of a negative is not a real
number, there are no real number solutions.
There is no such value for x.
2
x − 6x
+ 11 = 2
2
x − 6x
+ 9 = 0
− − ± − −
x =
2(1)
x
x
2
( 6) ( 6) 4(1)(9)
6 ± 0
=
2
6
= = 3
2
b.
2
f (15) = − 1.14(15) + 25.26(15) − 81.2 ≈ 41.2
So, 41.2% of television shows will be at least
an hour long in 2005.
c. Solve for t when f(t) = 50
= − + −
2
50 1.14t
25.26t
81.2
= − + −
2
0 1.14t
1.85t
131.2
− ± − − −
t =
2( −1.14)
2
25.26 25.26 4( 1.14)( 131.2)
t ≈ 8.31 or t ≈ 13.8
So, according to the model, 50% of television
shows will be at least an hour long in 1996 and
in 2004.
225

Cumulative Review Chapters 1-7
SSM: Intermediate Algebra
d. Solve for t when f(t) = 0
= − + −
2
0 1.14t
25.26t
81.2
2
25.26 25.26 4( 1.14)( 81.2)
− ± − − −
t =
2( −1.14)
t ≈ 3.90 or t ≈ 18.26
So, the t-intercepts are ( 3.90,0 ) and
( 18.26,0 ). This means that according to the
model no television shows were at least an
hour in 1994 and none will be at least an hour
in 2008. (Model breakdown has probably
occurred.
e. Model breakdown will definitely occur for
years before 1994 (t < 3.90) and for years after
2008 (t > 18.26) because there can’t be less
than zero shows that last at least an hour.
24. Using a graphing calculator, find that the
maximum point of the parabola is approximately
(2.50, 103).
So, the maximum height reached by the ball is 103
feet at 2.5 seconds.
Cumulative Review of Chapters 1-7
1. 5x 2 = 24
2 24
x =
5
= ±
= ±
24
5
24
5
2 6 5
= ± ⋅
5 5
= ±
2 30
5
2. 5 x + 1 =
7
6 3 2
5 7 1
x = −
6 2 3
5 21 2
x = −
6 6 6
5 19
x =
6 6
6 5 19 6
⋅ x = ⋅
5 6 6 5
19
x =
5
3. log
4
( x + 3) = 2
x + 3 = 4
2
x + 3 = 16
x = 13
x
4. 3e − 5 = 49
x
3e
= 54
x 54
e =
3
x ⎛ 54 ⎞
ln( e ) = ln ⎜ ⎟
⎝ 3 ⎠
⎛ 54 ⎞
x ln( e) = ln ⎜ ⎟
⎝ 3 ⎠
⎛ 54 ⎞
x = ln ⎜ ⎟ ≈ 2.8904
⎝ 3 ⎠
5. 5b 6 + 4 = 82
6
5b
78
b
6
=
=
78
5
1
6
⎛ 78 ⎞
b = ± ⎜ ⎟ ≈ ± 1.5807
⎝ 5 ⎠
6. 3x
2 − 5x
− 4 = 0
− − ± − − −
x =
2(3)
2
( 5) ( 5) 4(3)( 4)
5 ± 73
x =
6
226

SSM: Intermediate Algebra Cumulative Review Chapters 1-7
7.
8.
4x−5
3(2) 95
(2)
4 x−5
=
95
=
3
⎛ 95 ⎞
= ⎜ ⎟
4x−5
log(2) log
⎝ 3 ⎠
95
⎛ ⎞
(4x
− 5)log(2) = log⎜ ⎟
⎝ 3 ⎠
⎛ 95 ⎞
log ⎜ ⎟ 3
4x
− 5 =
⎝ ⎠
log(2)
⎛ 95 ⎞
log ⎜ ⎟ 3
4x
=
⎝ ⎠
+ 5
log(2)
⎛ 95 ⎞
log ⎜ ⎟
⎝ 3 ⎠ + 5
log(2)
x = ≈ 2.4962
4
2
2(3x
− 10) = − 7x
2
6 20 7
x
2
6 7 20 0
x
−
= −
(3x
− 4)(2x
+ 5) = 0
3x
− 4 = 0 or 2x
+ 5 = 0
x
+ x − =
3x
= 4 or 2x
= −5
4 5
x = or x = −
3 2
9. log
b
(65) = 4
b
4
= 65
1
4
b = 65
b ≈ 2.8394
10. 3x 2 + 5x 2 = 12x
+ 20
2
8 12 20 0
x
− x − =
2
4(2 3 5) 0
x
− x − =
2
2 3 5 0
x
− x − =
(2x
− 5)( x + 1) = 0
2x
− 5 = 0 or x + 1 = 0
2x
= 5 or x = −1
5
x = or x = − 1
2
11. 2(5x
+ 2) − 1 = 9( x − 3)
10x
+ 4 − 1 = 9x
− 27
12.
13.
14.
10x
+ 3 = 9x
− 27
x = −30
3log (4 x ) − 2log (4 x ) = 5
4 3
2 2
log (4 x ) − log (4 x ) = 5
4 3 3 2
2 2
log (64 x ) − log (16 x ) = 5
12 6
2 2
12
⎛ 64x
⎞
2 ⎜ 6 ⎟ =
log 5
⎝ 16x
⎠
2
6
( x )
log 4 = 5
4x
= 2
6 5
6
4 32
x
x
6
=
= 8
x = ± (8)
1
6
x ≈ 1.4142
2 x(3x − 4) + 5 = 3−
x
6x − 8x + 5 = 3 − x
2 2
2
7 8 2 0
x
− x + =
− − ± − −
x =
2(7)
x =
x =
x =
2
( 8) ( 8) 4(7)(2)
8 ± 8
14
8 ± 2 2
14
4 ± 2
7
4 7
3ln(2 ) 2ln(3 ) 8
x
2
+ x =
4 3 7 2
ln(2 ) ln(3 ) 8
12 14
ln(8 ) ln(9 ) 8
26
ln(72 ) 8
72x
x
26
x
x
x
26 8
8
e
=
72
+ x =
+ x =
= e
=
1
8 26
⎛ e ⎞
x = ⎜ ⎟ ≈ 1.1540
⎝ 72 ⎠
227

Cumulative Review Chapters 1-7
SSM: Intermediate Algebra
15.
2
5( x − 3) + 4 = 13
2
5( x 3) 9
2
− =
( x 3)
9
5
x − 3 = ±
9
5
x − 3 = ±
3
5
Rationalize the denominator:
3 5
x − 3 = ± ⋅
5 5
3 5
x − 3 = ±
5
3 5
x = 3 ±
5
x =
− =
15 ± 3 5
5
37x
= 37
x = 1
Substitute 1 for x in one of the original equations to
solve for y:
4(1) − 7 y = −10
− 7 y = −14
y = 2
The solution is (1, 2).
18. Substitute y = 3x
− 1 in 2x
− 3y
= − 11
2x
− 3(3x
− 1) = −11
2x
− 9x
+ 3 = −11
− 7x
= −14
x = 2
Substitute 2 for x in one of the original equations to
solve for y:
y = 3x
−1
= 3(2) −1
= 5
16. 2x
2 + 3x
− 6 = 0
2
2x
+ 3x
= 6
2 3
x + x = 3
2
2 3 9 9
x + x + = 3 +
2 16 16
2
⎛ 3 ⎞ 48 9
⎜ x + ⎟ = +
⎝ 4 ⎠ 16 16
⎛ 3 ⎞ 57
⎜ x + ⎟ =
⎝ 4 ⎠ 16
3 57
x + = ±
4 16
3 57
x + = ±
4 4
3 57
x = − ±
4 4
x =
2
− 3 ± 57
4
17. In order to eliminate y, multiply 4x
− 7 y = − 10 by
4 and 3x
+ 4y
= 11 by 7. This gives:
16x
− 28y
= −40
21x
+ 28y
= 77
Add corresponding sides to obtain x:
The solution is (2, 5).
19. In order to eliminate y, multiply 1 x − y = 5 by
2 2
3
3 3 15
− . This gives − x + y = − . Add the left
5
10 5 10
sides and the right sides of the equations:
3 3 15
− x + y = −
10 5 10
2 3 6
x − y =
5 5 5
This yields:
3 2 15 6
− x + x = − +
10 5 10 5
3 4 15 12
− x + x = − +
10 10 10 10
1 3
x = −
10 10
10 1 3 10
⋅ x = − ⋅
1 10 10 1
x = −3
Substitute -3 for x in one of the original equations
to solve for y:
228

SSM: Intermediate Algebra Cumulative Review Chapters 1-7
1 5
( −3)
− y =
2 2
3 5
− − y =
2 2
3 5
y = − −
2 2
8
y = − = −4
2
So the solution is (-3, -4).
20. 2(3x
− 4) < 5 − 3(6x
+ 5)
2(3x
− 4) < 5 − 3(6x
+ 5)
21.
22.
6x
− 8 < 5 −18x
−15
6x
− 8 < −18x
−10
24x
< −2
1
x < −
12
⎛ 1 ⎞
⎜ −∞,
− ⎟
⎝ 12 ⎠
(2 b c ) (3 b c )
=
4 −5 3 −1 −2 4
3 12 −15 4 −4 −8
(2 b c )(3 b c )
= ⋅
12 −4 −15 −8
(8 81)( b b )( c c )
= 648b c
648b
=
23
c
10b
15b
2b
=
2b
=
2b
=
=
8 −23
8
3 2
−
4 5
c
1 3
−
2 5
c
3 ⎛ 1 ⎞
− −
2 3
⎜ − ⎟ −
4 ⎝ 2 ⎠ 5 5
3 2 1
− + −
4 4 5
1 1
− −
4 5
3
2
1 1
4 5
3b c
3
c
3
c
c
1
−
12
x
23.
24.
25.
26.
⎛ 6b c
⎜
⎝ 8b c
8 −3
−4 −1
⎛ 6b c
⎜
⎝ 8b c
8 −3
−4 −1
36
6
3
⎞
⎟
⎠
8 −( −4) −3 −( −1)
⎛ 3b
c ⎞
= ⎜ ⎟
⎝ 4 ⎠
3
3
12 −2
⎛ 3b c ⎞
= ⎜ ⎟
⎝ 4 ⎠
12
⎛ 3b
⎞
= ⎜ 2 ⎟
⎝ 4c
⎠
27b
=
64c
⎞
⎟
⎠
+
7 5
3ln(2 x ) 2ln(4 x )
= ln(2 x ) + ln(4 x )
= +
3
3
7 3 5 2
21 10
ln(8 x ) ln(16 x )
= ⋅
=
21 10
ln(8x
16 x )
31
ln(128 x )
−
8 3
3log
b
(4 x ) 4log
b
(2 x )
= log (4 x ) − log (2 x )
b
= −
8 3 3 4
b
24 12
log
b
(64 x ) log
b
(16 x )
24
⎛ 64x
⎞
= logb
⎜ 12 ⎟
⎝ 16x
⎠
= log 4
b
12
( x )
2
(3x 4) (3x 4)(3x
4)
− = − −
= − − +
2
9x 12x 12x
16
= − +
2
9x
24x
16
27. (5x
− 7)(5x
+ 7)
= + − −
2
25x 35x 35x
49
2
= 25x
− 49
2 3
28. −3 x( x − 5)( x + 8)
= − + − −
5 2 3
3 x( x 8x 5x
40)
= − − + −
5 3 2
3 x( x 5x 8x
40)
6 4 3
= − 3x + 15x − 24x + 120x
229

Cumulative Review Chapters 1-7
SSM: Intermediate Algebra
29.
30.
31.
2
(2x 3)( x 4x
5)
− + −
= + − − − +
3 2 2
2x 8x 10x 3x 12x
15
= + − +
3 2
2x 5x 22x
15
f x
= − − +
2
( ) 2( x 5) 3
= −2( x − 5)( x − 5) + 3
= − − − + +
2
2( x 5x 5x
25) 3
= − − + +
2
2( x 10x
25) 3
= − + − +
2
2x
20x
50 3
= − + −
2
2x
20x
47
− = − = − +
2 2 2
81x 25 (9 x) 5 (9x 5)(9x
5)
40. h − 1 (7) = 4 since h (4) = 7
41. The x-intercept of k is (2, 0)
42.
32. x 3 − 13x 2 + 40x
2
= x( x − 13x
+ 40)
= x( x − 8)( x − 5)
43.
2
33. 8x + 22x − 21 = (2x + 7)(4x
− 3)
34. x 3 + 4x 2 − 9x− 36 :
2
= x ( x + 4) − 9( x + 4)
= + −
2
( x 4)( x 9)
= ( x + 4)( x − 3)( x + 3)
35. Since the y-intercept is (0, 20) and as x increases
by 1, f(x) decreases by 3: f ( x) = − 3x
+ 20
36. As x increases by 1, g(x) increases by a factor of 3,
so the base b = 3. Substitute a point from the table
x
into g( x)
= ab and solve for a.
12 = a(3)
12 = 9a
2
4
a =
3
So, the equation is:
4
g( x) = (3)
3
x
44. First, change the form into y = mx + b
2x
− 5y
= 20
− 5y
= − 2x
+ 20
2
y = x − 4
5
37. For the function k, as x increases by 1, k(x)
increases by 4, so the slope is 4.
38. g (4) = 108
39. x = 1 since g (1) = 4
230

SSM: Intermediate Algebra Cumulative Review Chapters 1-7
45.
49.
y − y
m x − x
2 1
= = = =
2 1
4 − ( −2) 6
−3 − ( −5) 2
3
46.
50. First, change the form of the equation given to
y = mx + b .
3x
− 4y
= 5
− 4y
= − 3x
+ 5
3 5
y = x −
4 4
4
The slope of the perpendicular line will be − . So
3
the equation of the perpendicular line will be
4
y = − x + b . Substitute the given point into this
3
equation to solve for b.
4
6 = − ( − 2) + b
3
8
6 = + b
3
8
b = 6 − 3
47.
48.
18 8
b = −
3 3
10
b =
3
4 10
So, the equation is y = − x + or
3 3
y = − 1.33x
+ 3.33 .
x
51. The equation is of the form y = ab . Since (0, 78)
is on the curve, a = 78. So, the equation is of the
form: y = 78b
x . Now substitute (9, 13) into the
equation and solve for b.
9
13 = 78b
b
9
=
13
78
1
9
⎛ 13 ⎞
b = ⎜ ⎟ ≈ 0.82
⎝ 78 ⎠
So, the equation is
y = 78(0.82) x .
52. Both points satisfy the equation
the system of equations:
5
83 = ab
27 = ab
2
y
x
= ab ; we have
231

Cumulative Review Chapters 1-7
SSM: Intermediate Algebra
Combining these two equations yields
5
83 ab
=
2
27 ab
83 3
= b
27
1
3
⎛ 83 ⎞
b = ⎜ ⎟ ≈ 1.45
⎝ 27 ⎠
So the equation is of the form:
To find a, substitute (2, 27) into
27 = a(1.45)
2
27 = 2.1025a
a = 12.84
So, the equation is
y = 12.84(1.45) x
53. Substitute the given points into
( 1, −1 ): − 1 =
2
( 1) + ( 1)
+
( 2, 4 ): 4 =
2
( 2) + ( 2)
+
( 4, 20 ): 20
2
( 4) ( 4)
a b c
a b c
= a + b + c
Simplify these equations:
a+ b+ c = −1 1
( )
( )
( )
4a + 2b + c = 4 2
16a + 4b + c = 20 3
y = a(1.45) x
y = a(1.45) x
2
y = ax + bx + c .
Eliminate c by multiplying both sides of equation
(1) by -1:
−a − b − c = 1 4
( )
Adding the left sides and right sides of equations
(2) and (4) gives:
3a
+ b = 5 5
( )
Adding the left sides and right sides of equations
(3) and (4) gives:
15a
+ 3b
= 21 6
Simplify:
5a
+ b = 7 7
( )
( )
Eliminate b by multiplying equation (5) by -1 and
add each side to the corresponding side of equation
(7):
2a
= 2
a = 1
Next, substitute 1 for a in equation (5):
3 1 + b = 5
( )
b = 2
Then, substitute 1 for a and 2 for b in equation (1):
a+ b+ c = −1
1+ 2 + c = −1
c = −4
Therefore, a = 1, b = 2, and c = -4. So, the equation
2
is y = x + 2x
− 4 .
54. a. Linear:
b.
6 − 3
slope = m = = 3
1−
0
y − intercept = (0,3)
So, y = 3x
+ 3
Exponential:
As the value of x increases by 1, the value of y
is multiplied by 2,, so the base b = 2. The y-
intercept is (0, 3). Therefore, the exponential
function is y = 3( 2)
x
Quadratic:
Answers may vary. Example:
55. − x
2 + 6x
− 5 = 3
2
x − 6x
+ 8 = 0
( x − 4)( x − 2) = 0
x − 4 = 0 or x − 2 = 0
x = 4 or x = 2
56. − x
2 + 6x
− 5 = 4
2
x − 6x
+ 9 = 0
( x − 3)( x − 3) = 0
− =
2
( x 3) 0
x − 3 = 0
x = 3
y
2
= 3x
+ 3 .
232

SSM: Intermediate Algebra Cumulative Review Chapters 1-7
57. − x
2 + 6x
− 5 = 5
65.
2
x − 6x
+ 10 = 0
x =
± − −
2
6 ( 6) 4(1)(10)
2(1)
6 ± −4
x =
2
Since the square root of a negative is not a real
number, there are no real number solutions. There
is no such value for x.
58. To find the x-intercepts, let f(x) = 0 and solve for x:
2
− + − =
x
6x
5 0
2
x − 6x
+ 5 = 0
( x − 5)( x − 1) = 0
x − 5 = 0 or x − 1 = 0
x = 5 or x = 1
The x-intercepts are (5, 0) and (1, 0).
59. To find the y-intercept, find f(0)
60.
2
f (0) = − (0) + 6(0) − 5 = − 5
The y-intercept is (0. -5)
61. log
2
(16) = 4 since
62.
4
2 = 16
3
log ( b b
) = 3 since 3 3
b = b
66. g( x ) = 3 x
67.
−1
g x =
3
x
( ) log ( )
2
f ( x) = x + 1
5
Replace f ( x) with y :
2
y = x + 1
5
Solve for x :
2
y − 1 = x
5
5 ( y − 1) = x
2
5 5
x = y −
2 2
−1
Replace x with g ( y) :
−1
5 5
g ( y)
= y −
2 2
Write in terms of x :
−1
5 5
g ( x)
= x −
2 2
68. a. First, draw a scattergram of the data. The
linear regression equation,
f ( t) = 3.97t
+ 3.02 , is the best model for this
data.
1 1
log ⎜ ⎟ = log ⎜ = log 5 = −2
2 ⎟
⎝ 25 ⎠ ⎝ 5 ⎠
⎛ ⎞ ⎛ ⎞
−2
63.
5 5 5 ( )
log(2)
64. log
9
(2) = ≈ 0.3155
log(9)
233

Cumulative Review Chapters 1-7
SSM: Intermediate Algebra
b. f (18) = 3.97(18) + 3.02 = 74.48 . In 2008,
74.48% of companies will offer stock options
to at least half of their employees.
c. 100 = 3.97t
+ 3.02
96.98 = 3.97t
t ≈ 24.43
All companies will offer stock options to at
least half of their employees in 2014.
d. Set f(t) = 0 and solve for t
0 = 3.97t
+ 3.02
− 3.02 = 3.97t
3.02
t = − ≈ −0.76
3.97
So, the t-intercept is (-0.76, 0). This means
that no companies offered stock options to at
least half of their employees in 1989. Model
breakdown has occurred.
e. t < -0.76 (since there can’t be less than 0
percent offering stock options) or t> 24.43
(because there can’t be more than 100 percent
of companies offering stock options).
69. a. Find the different models using the regression
feature on a graphing calculator:
Linear: f ( t) = 0.18t
− 0.61
Exponential: f ( t ) = 0.15(1.22) t
Quadratic:
f t = t − t +
2
( ) 0.0216 0.19 0.88
b. Answers may vary. Example: both the
quadratic and exponential functions model the
data well.
c. Quadratic model
d. Using a graphing calculator, find that the
minimum point of the parabola is
approximately (4.41, 0.46).
e. Since the base of the exponential model is b =
1.22, the rate of growth is approximately 1.22-
1 = 0.22 or 22% per year.
f. Quadratic model:
2
5 0.0216t
0.19t
0.88
2
0 0.0216t
0.19t
4.12
t =
= − +
= − −
± − − −
2
0.19 ( 0.19) 4(0.0216)( 4.12)
t ≈ −10.1 or t ≈ 18.9
2(0.0216)
t = -10.1 is model breakdown, so the year that
the quadratic model predicts that sakes will
reach $5 billion in 2009.
Exponential model:
t
5 = 0.15(1.22)
t
33.3333 = 1.22
t
log(1.22) = log(33.33333)
t log(1.22) = log(33.33333)
t =
log(33.33333)
log(1.22)
t ≈ 17.6
So the year that the exponential model
predicts that sakes will reach $5 billion in
2008.
Answers may vary. Example:
It makes sense that the year predicted by the
exponential model is before the year predicted
by the quadratic model, as the exponential
model is a steeper curve, and rises quicker
than the quadratic model.
70. a. First, draw a scattergram of the data for h. The
linear regression equation, h( t) = 2.4t
+ 10.7 ,
is the best model for this data.
This means that the minimum sales revenues
was 0.46 billion dollars.
234

SSM: Intermediate Algebra Cumulative Review Chapters 1-7
First, draw a scattergram of the data for c. The
linear regression equation, c( t) = t + 53 , is the
best model for this data.
b. The rates of change are the slopes of the
respective models.
Rate of change at Hartford: 2.4 percent per
year.
Rate of change in Connecticut: 1 percent per
year.
c. h (18) = 2.4(18) + 10.7 = 53.9 . This means that
in 2008, the percent of students who will score
above the goals in Hartford will be 53.9%.
c (18) = 18 + 53 = 71. This means that in 2008,
the percent of students who will score above
the goals in Connecticut will be 71%.
d. Solve for t when h( t) = c( t)
2.4t
+ 10.7 = t + 53
1.4t
= 42.3
t ≈ 30.21
So, the percentage of Hartford and
Connecticut students who will score above the
goals will be equal in 2020.
235