Chapter 9 Radical Functions

Chapter 9
Radical Functions
Homework 9.1
x
= x
1. 5 2 2/ 5
x
= x
3. 4 3 3/ 4
5.
x = x
1/ 2
3/ 7 3
7
7. ( 2x
+ 9) = ( 2x
+ 9)
4 4/ 7
9. 7 ( 3 x + 2 ) = ( 3 x + 2 )
7x
+ 4 = 7x
+ 4
11. ( ) 1/ 2
13. 49 = 7
15. 50 = 25⋅ 2 = 25 2 = 5 2
8 8/ 2 4
33. ( 2x + 5) = ( 2x + 5) = ( 2x
+ 5)
5 4
35. ( 6x + 3) = ( 6x + 3) ( 6x
+ 3)
37. 3 27 = 3
( x )
= 6 + 3 ⋅ 6x
+ 3
( x )
39. 6 x 6 = x 6/ 6 = x 1 = x
= 6 + 3 6x
+ 3
41. 3 8x 3 = 3 8 ⋅ 3 x 3 = 2x
43. 5 32x 5 = 5 32 ⋅ 5 x 5 = 2x
12 12 3
45. 4 81x = 4 81⋅ 4 x = 3x
2
4
17.
19.
21.
23.
25.
27.
29.
31.
2 2/ 2 1
x = x = x = x
8 8/ 2 4
x = x = x
6 6 3
36x = 36 ⋅ x = 6x
2 2
5x = 5 x = x 5
9 8 8 4
x = x ⋅ x = x ⋅ x = x x
5 4
24x = 4⋅6⋅ x ⋅ x
2
4
= 4 ⋅ 6 ⋅ x ⋅
= 2x
6x
3 8 2 8
80x y = 16⋅5⋅ x ⋅ x ⋅ y
4
x
2 8
= 16 ⋅ 5 ⋅ x ⋅ x ⋅ y
= 4xy
5x
3 5 2 4
200x y = 100⋅ 2⋅ x ⋅ x ⋅ y ⋅ y
2
2 4
= 100 ⋅ 2 ⋅ x ⋅ x ⋅ y ⋅ y
= 10xy
2xy
47. 6 17 6 12 5 6 12 6 5 2 6 5
x = x ⋅ x = x ⋅ x = x x
17 15 2
49. 3 125x = 3 125⋅ x ⋅ x
51.
3
5 3 2
3 15 3 2
= 125 ⋅ x ⋅
= 5x
5 40 7 5 40 5 2
64x y = 32 ⋅ 2⋅ x ⋅ y ⋅ y
x
x
5 5 5 40 5 5 5 2
= 32 ⋅ 2 ⋅ x ⋅ y ⋅ y
8 5 2
= 2x y 2y
5 5/5 1
5 6 xy = 6 xy = 6 xy = 6 xy
53. ( ) ( ) ( )
4 4/ 4 1
4 3x + 6 = 3x + 6 = 3x + 6 = 3x
+ 6
55. ( ) ( ) ( )
20 20/ 5 4
57. 5 ( 4x + 7) = ( 4x + 7) = ( 4x
+ 7)
24 24/ 4 6
4
59. ( x + 7) = ( x + 7) = ( x + 7)
275

Homework 9.1
SSM: Intermediate Algebra
61. 6 31 30
( 2x + 9) = 6 ( 2x + 9) ( 2x
+ 9)
( x )
6
30 6
= + ⋅ x +
2 9 2 9
( x )
5 6
= 2 + 9 2x
+ 9
75. a.
3h 3h 2 3h ⋅2 6h
d = = ⋅ = =
2 2 2 2 ⋅2
2
6 6/8 3/ 4 3
63. 8 x = x = x =
4 x
4 4/ 6 2/3 2
65. 6 x = x = x =
3 x
10 10/12
12
67. ( 2x
+ 7) = ( 2x
+ 7)
69.
71.
73.
6 14 14/ 6
6
x
= x
= x
=
7 /3
3 7
x
3 6
= x ⋅ x
3 6 3
= x ⋅
= x
2 3
6 3
27 = 3
= 3
= 3
3/ 6
1/ 2
= 3
( 2x
7)
= +
5/ 6
( 2x
7)
6
= +
x
x
10 8 10 10 8
16x
= 16 ⋅
10 2 10 8
= 4 ⋅
5
2 /10 8/10
= 4 ⋅ x
1/ 5 4 / 5
= 4 ⋅ x
= 4 ⋅
=
5 4
4x
5 4
x
x
x
5
b.
c.
d =
=
=
=
6h
2
( )
6 1450
2
8700
2
10 87
2
= 5 87 ≈ 46.64 miles
The distance to the horizon from the top of
the Sears Tower is roughly 46.64 miles.
d =
=
=
=
6h
2
( )
6 30,000
2
180,000
2
300 2
2
= 150 2 ≈ 212.13 miles
The distance to the horizon from the plane is
roughly 212.13 miles.
77. Answers will vary.
79. n abc = n ⋅
n n
=
a bc
a
a
bc
b c
n n
= ⋅
=
a b c
n n n
276

SSM: Intermediate Algebra Homework 9.2
81. a.
Homework 9.2
b.
i.
ii.
iii.
4 x + 5 x = 4 + 5 x = 9 x
1. ( )
2.3 x − 4.8 x = 2.3 − 4.8 x = − 2.5 x
3. ( )
5.
3 5x + 2 3x − 6 3x + 7 5x
( 3 5x 7 5x ) ( 2 3x 6 3x
)
= + + −
( ) x ( )
= 3 + 7 5 + 2 − 6 3x
= 10 5x
− 4 3x
3 3
2 x + 5 x − 5 x = 2 x − 5 x + 5 x
7. ( )
( )
= 2 − 5 x + 5
= − 3 x + 5
3
x
3
x
c.
The graphs in (a) coincide with the graph of
y = x .
i.
9.
11.
3 3
6 x −1 − 3 x −1 − 2 x −1
( )
3
3
= 6 − 3 x −1 − 2 x −1
= 3 x −1 − 2 x −1
3 3
5 x + 3 x + 4 x + 2 x
( 5 3) x ( 4 2)
= + + +
= 8 x + 6
3
x
3
x
d.
ii.
iii.
The graphs in (c) coincide with the graph of
y = x .
13.
4 4
6 6
3.7 x −1.1 x − 4.2 x + 4.2 x
4
( 3.7 1.1) x ( 4.2 4.2)
= − + − +
= 2.6
4
x
15. 3( 7 − x + 2) − ( x + 2)
= 3⋅7 − 3⋅ x + 3⋅2 − x − 2
= 21− 3 x + 6 − x − 2
= −3 x − x + 21+ 6 − 2
( 3 1) x ( 21 6 2)
= − − + + −
= − 4 x + 25
− 5 2 + 4 = −5⋅ 2 x − 5⋅
4
17. ( x )
= −10 x − 20
6
x
83. Answers will vary. See Key Points.
277

Homework 9.2
SSM: Intermediate Algebra
3 3
19. 7( x + 1) − 7( x −1)
3 3
3 3
3 3
( )
= 7 ⋅ x + 7 ⋅1− 7 ⋅ x − 7 −1
= 7 x + 7 − 7 x + 7
= 7 x − 7 x + 7 + 7
= 14
21. 25x + 4x = 25 x + 4 x
= 5 x + 2 x
= 7 x
23. 3 20x + 2 45x = 3 4⋅ 5x + 2 9⋅5x
= 3 4 5x
+ 2 9 5x
= 3⋅ 2 5x
+ 2⋅3 5x
= 6 5x
+ 6 5x
= 12 5x
25.
27.
3 2
5 x − x 49x = 5 x ⋅ x − x 49 ⋅ x
2
= 5 x x − x 49 x
= 5x x − 7x x
( 5 7)
= −
2 2
= −2x
x
3 81x − 2 100x = 3⋅9x − 2⋅10x
x
x
= 27x
− 20x
= 7x
33.
35.
37.
4 11 4 7 4 8 3 4 4 3
16x − 3x x = 16x ⋅ x − 3x x ⋅ x
4 8 4 3 4 4 4 3
= 16x x − 3x x x
2 4 3 2 4 3
= 2x x − 3x x
( 2 3)
= −
= −x
2
x
2 4 3
3 x ⋅ 2 x = 3⋅ 2⋅ x ⋅ x
= 6 x ⋅ x
= 6
x
x
2
2 4 3
= 6x
−2 5x ⋅ 4 3x = −2⋅ 4⋅ 5x ⋅ 3x
= −8 5x
⋅3x
39. ( )
= −8 15x
= −8 15 x
= −8x
15
2 7x 7x + 2x = 2 7x ⋅ 7x + 2 7x ⋅ 2x
2
x
= 2 7x ⋅ 7x + 2 7x ⋅ 2x
2 2
= 2 49x
+ 2 14x
2 2
= 2 49 x + 2 14
= 2⋅ 7x
+ 2x
14
= 14x
+ 2x
14
x
29.
31.
3 2
12x + x 75x = 4x ⋅ 3x + x 25⋅3x
2
= 4x 3x + x 25 3x
= 2x 3x + 5x 3x
( )
= 2 + 5 x 3x
= 7x
3x
3 5 3 2 3 3 2 3 2
27x − x 8x = 27x ⋅ x − x 8⋅
x
3 3 3 2 3 3 2
= 27x x − x 8 x
3 2 3 2
= 3x x − 2x x
( 3 2)
= −
= x
3 2
x
x
3 2
x
41. ( 6 − 2 x )( 5 x − 4)
2
( )
= 6⋅5 x − 2 x ⋅5 x − 6⋅ 4 − 2 x −4
= 30 x −10 x − 24 + 8 x
= − 10x + 30 x + 8 x − 24
= − 10x
+ 38 x − 24
43. ( 2 x + 1)( x − 4)
= 2 x ⋅ x + 1⋅ x − 2 x ⋅ 4 −1⋅
4
2
= 2 x + x − 8 x − 4
( )
= 2x
+ 1− 8 x − 4
= 2x
− 7 x − 4
278

SSM: Intermediate Algebra Homework 9.2
2
45. ( )( ) ( ) 2
1− x 1+ x = 1 − x = 1−
x
2 2
x + 5 x − 5 = x − 5 = x − 25
47. ( )( ) ( ) ( )
2 2
2
49. ( 5 + 6 x ) = 5 + 2( 5)( 6 x ) + ( 6 x )
2
2
= 25 + 60 x + 6 ( x )
= 25 + 60 x + 36x
= 36x
+ 60 x + 25
2 2 2
51. ( 7 x − 1) = ( 7 x ) − 2( 7 x )( 1) + ( 1)
2
( x )
2
= 7 − 14 x + 1
= 49x
− 14 x + 1
2 2 2
53. ( 4 x + 5) = ( 4 x ) + 2( 4 x )( 5) + ( 5)
2
( x )
2
= 4 + 40 x + 25
= 16x
+ 40 x + 25
2 2 2
55. ( x + 1) = ( x ) + 2( x )( 1) + ( 1)
57.
= x + 2 x + 1
1 1
+
5 1/ 2 1/ 5 2 5
x x = x ⋅ x = x
5 2
+
10 10
= x = x
=
10 7
x
7 /10
59. 5 4 5 3 5 4 3 5 7
x x = x ⋅ x = x
5 5 2 5 5 5 2
= x ⋅ x = x x
= x
5 2
x
61. −5 x ( 4 2x − 4) = −5 x 4 2x − 5 x ( −4)
4 4
5 2 20
= − x x + x
4 1/ 2 1/ 4
5 2 20
= − x ⋅ x + x
1 1
+
4 2 4
= − 5 2 x + 20
4 3/ 4
5 2 20
= − x +
4 4 3
5 2 20
= − x +
4 3
= − 5 2x
+ 20
3 3 3
63. ( x + 1) = ( x ) + 2( x )( 1) + ( 1)
x
x
x
2 2 2
3 2 3
= x + 2 x + 1
2 2 2
3 3 3
65. ( x − 2) = ( x ) − 2( x )( 2) + ( 2)
3 2 3
= x − 4 x + 4
2 2 2
4 3 4 4 3 3
67. ( x + x ) = ( x ) + 2( x )( x ) + ( x )
69. ( 2 x − 6)( 3 3 x + 1)
4 2 1/ 4 1/ 3 3 2
= x + 2x ⋅ x + x
1 1
+
2 / 4 4 3 3 2
= x + 2x + x
1/ 2 7 /12 3 2
= x + 2x + x
12 7 3 2
= x + 2 x + x
3 3
2 3 6 3 2 1 6 1
= x ⋅ x − ⋅ x + x ⋅ − ⋅
1/ 2 1/ 3 3
= 6x ⋅ x − 18 x + 2 x − 6
1 1
+
2 3 3
6 18 2 6
= x − x + x −
5/ 6 3
= 6x − 18 x + 2 x − 6
6 5 3
= 6 x − 18 x + 2 x − 6
6 5 3
= 6 x + 2 x −18 x − 6
x
4 4 4
71. ( 3 x + 5)( 3 x − 5) = ( 3 x ) − ( 5)
( x )
2 4
2 2
= 3 − 25
2
= 9 x − 25
279

Homework 9.3
SSM: Intermediate Algebra
73. a. ( )
( )
2 100 200
f 100 = = ≈ 2.49
32.2 32.2
It takes roughly 2.49 seconds for an object
to fall 100 feet.
b. ( )
( )
2 1483 2966
f 1483 = = ≈ 9.6
32.2 32.2
The parachutist was in freefall for roughly
9.6 seconds. This is an underestimate
because the chute would slow down his
descent.
c. f is an increasing function. This makes sense
because the higher an object is, the longer it
should take to fall to the ground.
b.
c.
d.
k
n
1/ k 1/ n
x x = x ⋅ x
3 4
= x
= x
= x
=
1 1
+
k n
n k
+
n⋅k
n⋅k
n+
k
n⋅k
n⋅ k n+
k
x
x x → k = 3, n = 4
3⋅ 4 3+
4 12 7
3 4
x x = x = x
The results are the same.
5⋅ 7 5+
7 35 12
x x = x = x
5 7
75.
3
1/ 2
1 1 3 2 1
− −
2 3 6 6 6 6
x x x x
x x
= = = = =
1/ 3
x x
77. Begin by letting the two expressions be denoted
by W and Y. We know the following from the
problem statement:
79. a.
W + Y = 9 x + 7
W − Y = x + 3
Adding the two equations together will eliminate
the Y terms and yield:
2W = 9 x + 7 + x + 3
2W
= 10 x + 10
W = 5 x + 5
Plug this into one of the original equations:
5 x + 5 + Y = 9 x + 7
Y = 9 x + 7 − 5 x − 5
Y = 4 x + 2
Thus, the two expressions are 5 x + 5 and
4 x + 2 .
3 4
1/ 3 1/ 4
x x = x ⋅ x
= x
= x
= x
=
1 1
+
3 4
4 3
+
12 12
7 /12
12 7
x
81. Answers will vary. See Key Points.
Homework 9.3
1.
3.
5.
2 2 3
= ⋅
3 3 3
=
=
2 3
( 3) 2
2 3
3
8 8 x
= ⋅
x x x
=
8
x
( x ) 2
8 x
=
x
3 3 5x
= ⋅
5x 5x 5x
=
3 5x
( 5x
) 2
3 5x
=
5x
280

SSM: Intermediate Algebra Homework 9.3
7.
5 5 x
= ⋅
2 x 2 x x
=
2
=
5⋅
x
( x ) 2
5x
2x
15.
2 2
=
x x
2
= ⋅
x
=
2 ⋅ x
( x ) 2
x
x
9.
4 4 2x
= ⋅
3 2x 3 2x 2x
4 2x
=
3 2
( x ) 2
4 2x
=
3 ⋅ 2x
2 2x
=
3x
17.
=
2x
x
x x
=
3 3
x 3
= ⋅
3 3
=
3⋅
x
( 3) 2
11.
4 4
=
x x
2
= ⋅
x
=
2
x
( x ) 2
2 x
=
x
x
x
19.
=
3x
3
3 3 x − 4
= ⋅
x − 4 x − 4 x − 4
=
3 x − 4
( x − 4) 2
3 x − 4
=
x − 4
13.
7 7
=
2 2
7 2
= ⋅
2 2
=
=
7 ⋅2
( 2) 2
14
2
21.
3
2 2 25
= ⋅
5 5 25
3 3 3
=
3
3
3
3
2 25
5⋅
25
2 25
=
3
125
=
2 25
5
281

Homework 9.3
SSM: Intermediate Algebra
23.
25.
3 3
3
3
3
3
3
5 5 4
= ⋅
3
16 16 4
=
5 4
16⋅
4
5 4
=
3
64
=
5 4
4
3 2
4 4 x
= ⋅
5 x 5 x x
3 3 3 2
29.
7 7 4x
= ⋅
4
4x
4x
4x
4 3 4 3
=
4
4 3
4
4
7 4x
4x
⋅4x
4
4
7 4x
=
4 4
16x
=
4
7 4x
16
7 4x
=
2x
4 4
x
27.
3 2
4 x
=
5 x ⋅ x
4
=
5
3 2
3 2
x
3 3
x
3 2
4 x
=
5x
6 6 4x
= ⋅
3
2x
2x
4x
3 2 3 2
=
3
3 2
3
6 4x
2x
⋅4x
31.
3 3
x x x
= ⋅
x x x
x
=
1/ 3 1/ 2
⋅ x
( x )
1 1
+
3 2
x
=
x
2 3
+
6 6
x
=
x
=
5/ 6
x
x
2
3
6 4x
=
3 3
8x
3
3
3
6 4x
=
3 3 3
8 x
6 4x
=
2x
3 4x
=
x
33.
5
=
6 5
x
x
5
2 2
=
3
x
5 3
x
5
2
= ⋅
x
=
5 2
x
5 3 5 2
5 2
2⋅
x
5 3 2
x
⋅ x
x
=
5 2
2x
5 5
x
=
5 2
2x
x
282

SSM: Intermediate Algebra Homework 9.3
35.
4
4
4 4
=
9x
9x
2 4 2
4
4 2
=
4
⋅
9x
9x
9x
4 2 4 2
41.
2 2 4 − 7
= ⋅
4 + 7 4 + 7 4 − 7
2⋅ 4 − 2⋅
7
=
( 4) 2
− ( 7 ) 2
=
=
4 2
4 ⋅9x
4 2 2
9x
⋅9x
4 2
36x
4 4
81x
8 − 2 7
=
16 − 7
=
8 − 2 7
9
37.
39.
5
=
6
=
4 2 2
6 x
3x
2/ 4 2/ 4
x
3x
1/ 2 1/ 2
6 x
=
3x
=
6x
3x
5
3 3
=
4x
4x
4 5 4
5 5
3 8x
= ⋅
5 4 5
4x
8x
=
=
=
5
5 4
5
5 5
5
3⋅8x
4x
⋅8x
24x
32x
24x
2x
1 1 5 − 3
= ⋅
5 + 3 5 + 3 5 − 3
=
1⋅5 −1⋅
3
( 5) 2
− ( 3) 2
43.
45.
47.
1 1 x + 7
= ⋅
x − 7 x − 7 x + 7
=
1⋅
x + 1⋅7
( x ) − ( 7)
x + 7
=
x − 49
2 2
x x x + 1
= ⋅
x −1 x − 1 x + 1
=
x ⋅ x + x ⋅1
( x ) − ( 1)
x + x
=
x −1
2 2
3 x 3 x 4 x + 5
= ⋅
4 x − 5 4 x − 5 4 x + 5
3 x ⋅ 4 x + 3 x ⋅5
=
( 4 x ) − ( 5)
2
( x ) +
2
2
( x ) −
2 2
12 15 x
=
4 25
12x
+ 15 x
=
16x
− 25
5 − 3
=
25 − 3
=
5 − 3
22
283

Homework 9.3
SSM: Intermediate Algebra
49.
2 x 2 x 3 x + 7
= ⋅
3 x − 7 3 x − 7 3 x + 7
2 x ⋅ 3 x + 2 x ⋅7
=
( 3 x ) − ( 7)
2
( x ) + x
2
2
( x ) −
2 2
6 14
=
3 49
57.
6 x + 5 6 x + 5 3 x + 7
= ⋅
3 x − 7 3 x − 7 3 x + 7
6 x ⋅ 3 x + 6 x ⋅ 7 + 5 ⋅ 3 x + 5 ⋅ 7
=
( )
2
( 3 x ) − ( 7)
( x )
2 2
18 x + 6 7 ⋅ x + 3 5⋅ x + 5⋅7
=
2
2
3 − 7
6x
+ 14 x
=
9x
− 49
18x + 6 7x + 3 5x
+ 35
=
9x
− 7
51.
53.
x − 5 x − 5 x − 5
= ⋅
x + 5 x + 5 x − 5
=
( x ) − 2( x )( 5) + ( 5)
2 2
( x ) − ( 5)
2 2
x − 10 x + 25
=
x − 25
x + 3 x + 3 4 + x
= ⋅
4 − x 4 − x 4 + x
=
x ⋅ 4 + x ⋅ x + 3⋅ 4 + 3⋅
x
2
( 4) − ( x ) 2
4 x + x + 12 + 3 x
=
16 − x
x + 7 x + 12
=
16 − x
59. Student 1 did the work correctly. Student 2’s
error was to square the entire expression. This
changes the value of the expression.
61. Answers may vary. The student did not
rationalize the denominator correctly. For the
radicand in the denominator to be a perfect cube,
2
x needs to be multiplied by x to yield
3 2 3 3
3 x ⋅ x = x .
3 2
5 5 x
= ⋅
x x x
3 3 3 2
=
5
5
=
3 2
x
3 2
x ⋅ x
3 2
5 x
=
3 3
x
3 2
x
x
55.
2 x + 5 2 x + 5 3 x + 1
= ⋅
3 x −1 3 x − 1 3 x + 1
2 x ⋅ 3 x + 2 x ⋅ 1+ 5⋅ 3 x + 5⋅1
=
( )
2
( 3 x ) −( 1)
( x )
2 2
6 x + 2 x + 15 x + 5
=
2
2
3 −1
63.
x x x
= ⋅
3 3 x
2
x
=
3 x
x
=
3 x
6x
+ 17 x + 5
=
9x
−1
284

SSM: Intermediate Algebra Homework 9.4
65.
67.
x + 2 − x x + 2 − x x + 2 + x
= ⋅
2 2 x + 2 + x
=
( x + 2) −( x )
2( x + 2 + x )
x + 2 − x
=
2 2
2 2
( x + + x )
2
=
2 2
=
1 3 1 x 3
− ⋅ −
x x x x x
=
2
+
1 2 x 1
x x ⋅ +
x x x
=
=
x 3
−
x x
2 x
+
1
x x
x −3
x
2 x + 1
x
( x + + x )
1
x + 2 +
x − 3 2 x + 1
= ÷
x x
x − 3 x
= ⋅
x 2 x + 1
x − 3
=
2 x + 1
x − 3 2 x −1
= ⋅
2 x + 1 2 x −1
=
x
( 2 x ) − ( 1)
2 2
( )
x ⋅ 2 x − x ⋅1− 3⋅ 2 x − 3 −1
2x − x − 6 x + 3
=
4x
−1
2x
− 7 x + 3
=
4x
−1
69. x 2 + 3 5 = 9 5
x
x
6 5
x =
2
2 = 9 5 − 3 5
2 = 6 5
6 5 2
= ⋅
2 2
=
6 10
2
= 3 10
71. Answers may vary.
1. Determine the conjugate of the denominator.
2. Multiply the original fraction by the fraction
conjugate
conjugate
3. Find the product of the fractions and
simplify.
Homework 9.4
1. y = 2
x y
0 0
1 2
4 4
9 6
16 8
x
3. y = − x
x y
0 0
1 −1
4 −2
9 −3
16 −4
8
4
−4
−8
y
y
4 8
4
8
x
x
285

Homework 9.4
SSM: Intermediate Algebra
5. y = x + 3
13. y = − x + 2
x y
0 3
1 4
4 5
9 6
16 7
8
4
y
4 8
x
x y
−2 0
−1 −1
2 −2
7 −3
14 −4
−4
−8
y
4
8
x
7. y = 2 x − 5
x y
0 −5
1 −3
4 −1
9 1
16 3
9. y = − 3 x + 4
x y
0 4
1 1
4 −2
9 −5
16 −8
y
4
−4
y
4
−4
2
4
8
8
x
x
15. 1
y = x − 4
2
x y
4 0
1
5
2
8 1
3
13
2
20 2
17. y = x + 3 + 2
x y
−3 2
−2 3
1 4
6 5
13 6
y
8
4
4 8
y
8
4
x
11. y = x − 2
x y
2 0
3 1
6 2
11 3
18 4
y
8
4
4 8
x
19. y = − 2 x + 3 − 4
x y
−3 −4
−2 −6
1 −8
6 −10
13 −12
−4
−4
y
−4
−8
4
4
x
x
286

SSM: Intermediate Algebra Homework 9.4
21. y = 4 x − 1 + 3
29. y = x + 2
x y
1 3
2 7
5 11
10 15
17 19
y
16
8
x y
0 2
1 3
4 4
9 5
16 6
y
8
4
23. x + y = 4
4 8
x
4 8
Domain: x ≥ 0
Range: y ≥ 2
x
y = − x + 4
x y
0 4
1 3
4 2
9 1
16 0
y
4
−4
4
8
x
31. y = − 2 x + 5 + 4
x y
−5 4
−4 2
−1 0
4 −2
11 −4
−4
y
4
−4
4
x
25. 2 y − 6 x = 8
2y
= 6 x + 8
y = 3 x + 4
33. a. a < 0, h = 0, and k > 0
Domain: x ≥ − 5
Range: y ≤ 4
x y
0 4
1 7
4 10
9 13
16 16
27. y = − 2
x
y
16
8
4 8
x
b. a > 0, h < 0, and k < 0
c. a > 0, h < 0, and k > 0
d. a < 0, h > 0, and k = 0
35. Answers may vary. One example: For the family
of curves y = a x − h + k , k = 3 , and h = 6 .
1 1
Let a = −4, −3, −2, −1, − , ,1,2,3, and 4 .
2 2
x y
0 0
1 −2
4 −4
9 −6
16 −8
y
4
−4
−8
Domain: x ≥ 0
Range: y ≤ 0
8
x
a < , f has a maximum point at ( , )
a > 0 , f has a minimum point at ( h,
k ) .
37. If 0
39. ( )
f 4 = 7 4 − 3
h k . If
= 7 ⋅ 2 − 3
= 11
287

Homework 9.4
SSM: Intermediate Algebra
41. ( )
f 9c = 7 9c
− 3
= 7 ⋅3 c − 3
= 21 c − 3
43. h( x) = ( x + 3) + ( x − 5)
= x + 3 + x − 5
= 2 x − 2
45. h( x) = ( x − 5) − ( x − 3)
47. h( x)
= x − x − 5 − 3
= −8
=
( x + 3)
( x − 5)
x + 3 x + 5
= ⋅
x − 5 x + 5
=
x ⋅ x + x ⋅ 5 + 3⋅ x + 3⋅5
( x ) − ( 5)
2 2
x + 5 x + 3 x + 15
=
x − 25
x + 8 x + 15
=
x − 25
49. h( x) = ( 5 x − 9) + ( 4 x + 1)
= 5 x − 9 + 4 x + 1
= 9 x − 8
51. h( x) = ( 4 x + 1) − ( 5 x − 9)
= 4 x + 1− 5 x + 9
= − x + 10
53. h( x)
5 x − 9
=
4 x + 1
5 x − 9 4 x −1
= ⋅
4 x + 1 4 x −1
5 x ⋅ 4 x − 5 x ⋅1− 9⋅ 4 x + 9⋅1
=
( 4 x ) − ( 1)
2 2
20x − 5 x − 36 x + 9
=
16x
−1
20x
− 41 x + 9
=
16x
−1
55. h( x) = ( 2 x − 3 5) + ( 2 x + 3 5)
= 2 x − 3 5 + 2 x + 3 5
= 4 x
57. h( x) = ( 2 x + 3 5) −( 2 x − 3 5)
59. h( x)
= 2 x + 3 5 − 2 x + 3 5
= 6 5
2 x − 3 5
=
2 x + 3 5
2 x − 3 5 2 x − 3 5
= ⋅
2 x + 3 5 2 x − 3 5
=
( 2 x ) − 2( 2 x )( 3 5) + ( 3 5)
2 2
( 2 x ) − ( 3 5)
2 2
4x
− 12 5x
+ 45
=
4x
− 45
61. h( x) = ( x + 1 − 2) + ( x + 1 + 2)
= x + 1 − 2 + x + 1 + 2
= 2 x + 1
63. h( x) = ( x + 1 + 2) − ( x + 1 − 2)
= x + 1 + 2 − x + 1 + 2
= 4
288

SSM: Intermediate Algebra Homework 9.5
65. h( x)
=
x + 1 − 2
x + 1 + 2
x + 1 − 2 x + 1 − 2
= ⋅
x + 1 + 2 x + 1 − 2
=
( x + 1) − 2( x + 1)( 2) + ( 2)
2 2
( x + 1) − ( 2)
x + 1− 4 x + 1 + 4
=
x + 1−
4
x − 4 x + 1 + 5
=
x − 3
2 2
67. a. The square root model would fit best
because it continues to increase slowly. The
quadratic function would reach a peak and then
start to decrease.
79. Answers may vary. The graph of f can be found
by translating the graph of y = a x horizontally
by h units (left if h < 0 and right if h > 0 ), and
vertically by k units (up if k > 0 and down if
k < 0 ).
Homework 9.5
1.
x = 5
( ) 2 2
x = 5
x = 25
Check x = 25
( ) =
?
25 5
5 = 5 true
The solution is 25.
b. S ( 20)
= 3.9 20 + 280 ≈ 297.44
The average test score in 2002 would be
about 297.
c. S ( 23)
= 3.9 23 + 280 ≈ 298.70
The average test score in 2005 would be
about 299.
69. f ( − 6)
= 0
71. f ( 0)
= 2.4
73. x = − 6
75. x = 3
77. ( ) ( )
f x = 2 x + 3 + 2; g x = − 2 x + 3+
2
The graph looks like a parabola that opens to the
right. The relation is not a function because it
would fail the vertical line test.
3.
5.
7.
x = −2
( x ) = ( −2)
2 2
x = 4
Check x = 4
( ) ?
4 =− 2
?
2 =− 2 false
There are no real solutions.
3 x − 1 = 5
3 x = 6
x = 2
( ) 2 2
x = 2
x = 4
Check x = 4
( )
?
3 4 − 1=
5
5 = 5 true
The solution is 4.
4 − 5 x = 2 x −10
− 7 x = −14
x = 2
( ) 2 2
x = 2
x = 4
Check x = 4
289

Homework 9.5
SSM: Intermediate Algebra
9.
11.
13.
?
( ) ( )
4 − 5 4 = 2 4 −10
− 6 = −6 true
The solution is 4.
3 7x
− 24 = −9 7x
12 7x
= 24
7x
= 2
( ) 2 2
7x
= 2
Check
7x
= 4
4
x =
7
4
x =
7
4
?
4
⎛ ⎞ ⎛ ⎞
3 7⎜ 24 9 7
7
⎟ − =− ⎜
7
⎟
⎝ ⎠ ⎝ ⎠
− 18 = −18 true
The solution is 4 7 .
x − 1 = 2
( ) 2 2
x − 1 = 2
x − 1 = 4
x = 5
Check x = 5
( )
?
5 − 1=
2
2 = 2 true
The solution is 5.
5x
− 7 = −8
( 5x
− 7 ) = ( −8)
Check
2 2
5x
− 7 = 64
5x
= 71
x =
x =
71
5
?
71
5
71
?
⎛ ⎞
5⎜
⎟ − 7 =− 8
⎝ 5 ⎠
8=−
8 false
There are no real solutions.
15.
17.
19.
10 6x
+ 3 = 100
6x
+ 3 = 10
( ) 2 2
6x
+ 3 = 10
6x
+ 3 = 100
6x
= 97
97
x =
6
97
Check x =
6
97
?
⎛ ⎞
10 6⎜
⎟ + 3 = 100
⎝ 6 ⎠
100 = 100 true
The solution is 97 6 .
3x
+ 1 = 2x
+ 6
( 3x
+ 1) = ( 2x
+ 6)
2 2
3x
+ 1 = 2x
+ 6
x = 5
Check x = 5
?
( ) ( )
3 5 + 1 = 2 5 + 6
4 = 4 true
The solution is 5.
2 1− x − 5 = 0
2 1− x = 5
( 2 1− x ) = ( 5)
( x)
2 2
4 1− = 5
4 − 4x
= 5
− 4x
= 1
1
x = −
4
1
Check x = −
4
1
?
⎛ ⎞
2 1− ⎜ − ⎟ − 5 = 0
⎝ 4 ⎠
0 = 0 true
1
The solution is − .
4
290

SSM: Intermediate Algebra Homework 9.5
21.
−4.91 3.18x
− 7.14 = −2.19
2.19
3.18x
− 7.14 =
4.91
( 3.18x
7.14 )
2 ⎛ 2.19 ⎞
− = ⎜ ⎟
⎝ 4.91 ⎠
⎛ 2.19 ⎞
3.18x
− 7.14 = ⎜ ⎟
⎝ 4.91 ⎠
Check x = 2.3078
⎛ 2.19 ⎞
3.18x
= 7.14 + ⎜ ⎟
⎝ 4.91 ⎠
( )
⎛ 2.19 ⎞
7.14 + ⎜ ⎟
4.91
x =
⎝ ⎠
3.18
x = 2.3078
−4.91 3.18 2.3078 − 7.14 =− 2.19
− 2.19 = −21.19 true
The solution is approximately 2.3078.
?
2
2
2
2
2
12x + 13 = 9x − 12x
+ 4
9x
− 24x
− 9 = 0
2
( x x )
3 3 − 8 − 3 = 0
( x )( x )
3 3 + 1 − 3 = 0
3x
+ 1 = 0 or x − 3 = 0
1
x = − or x = 3
3
1
Check x = −
3
1
?
1
⎛ ⎞ ⎛ ⎞
12⎜ − ⎟ + 13 + 2 = 3⎜ − ⎟
⎝ 3 ⎠ ⎝ 3 ⎠
Check x = 3
?
2
5=−1 false
( ) + + = ( )
12 3 13 2 3 3
9 = 9 true
The solution is 3.
?
23.
25.
x
( 3x
+ 3) = ( x − 5)
2
3x
+ 3 = x −5
( x )( x )
2 2
2
3x + 3 = x − 10x
+ 25
− 13x
+ 22 = 0
−11 − 2 = 0
x − 11 = 0 or x − 2 = 0
x = 11 or x = 2
Check x = 11
?
( ) ( )
3 11 + 3 = 11 − 5
6 = 6 true
Check x = 2
?
( ) ( )
3 2 + 3= 2 − 5
3=−
3 false
The solution is 11.
?
12x
+ 13 + 2 = 3x
12x
+ 13 = 3x
− 2
( 12x
+ 13) = ( 3x
− 2)
2 2
27.
29.
3x
+ 4 − x = 3
( 3x
+ 4) = ( x + 3)
x
2
3x
+ 4 = x + 3
2 2
3x + 4 = x + 6x
+ 9
+ 3x
+ 5 = 0
2
2
( ) − ( )( )
2( 1)
− 3 ± 3 4 1 5
x =
− 3 ± −11
=
2
There are no real solutions.
2
x − 5x + 1 = x − 3
2
2
( x − 5x + 1) = ( x − 3)
2 2
x − 5x + 1 = x − 6x
+ 9
Check x = 8
x = 8
?
2
( ) ( ) ( )
8 − 5 8 + 1 = 8 − 3
5 = 5 true
The solution is 8.
2
291

Homework 9.5
SSM: Intermediate Algebra
31.
2 + x = x + 12
( 2 + x ) = ( x + 12)
2 2
4 + 4 x + x = x + 12
35.
x −
2x
= −1
2x
= x + 1
( 2x
) = ( x + 1)
2 2
4 x = 8
2x = x + 2 x + 1
33.
( x )
x = 2
2
2
= 2
x = 4
Check x = 4
?
( ) ( )
2 + 4 = 4 + 12
4 = 4 true
The solution is 4.
x − 1 = 5 − x
( x − 1) = ( 5 − x )
2 2
x − 2 x + 1 = 5 − x
2
2
2x
− 4 = 2
x − 2 =
x
2
( x − 2) = ( x )
x − 4x + 4 = x
x
− 5x
+ 4 = 0
( x )( x )
− 4 − 1 = 0
x
x − 4 = 0 or x − 1 = 0
x = 4 or x = 1
Check x = 4
?
( 4) − 1=
5 − ( 4)
1 = 1 true
Check x = 1
?
( 1) − 1=
5 − ( 1)
0 = 2 false
The solution is 4.
?
2
37.
2
2
x − 1 = 2
( x − 1) 2
= ( 2 x ) 2
x − 2x + 1 = 4x
x
− 6x
+ 1 = 0
x
2
( 6) ( 6) 4( 1)( 1)
2( 1)
− − ± − −
x =
=
6 ± 32
2
= 3 ± 2 2
Check 3+ 2 2 ( x ≈ 5.83)
( ) ( ) ?
5.83 − 2 5.83 =−1
?
− 1.00 =−1 true
Check 3 − 2 2 ( x ≈ 0.172)
( ) ( ) ?
0.172 − 2 0.172 =−1
?
− 0.172=−1 false
The solution is 3 + 2 2 .
x + 1 = − x −7
( x + 1) = ( − x −7)
2 2
x + 1 = x + 14 x + 49
14 x = −48
7 x − 24
( 7 x ) = ( −24)
2 2
49x
= 576
576
x =
49
576
x = x ≈ 11.76
49
Check ( )
?
( ) ( )
11.76 + 1 =− 11.76 − 7
?
3.57 =−10.43 false
There are no real solutions.
292

SSM: Intermediate Algebra Homework 9.5
39.
x − 3 + x + 5 = 4
43.
2x
− 1 + 3x
− 2 = 2
x − 3 = 4 − x + 5
( x − 3) = ( 4 − x + 5)
2 2
2x
− 1 = 2 − 3x
− 2
( 2x
− 1) = ( 2 − 3x
− 2)
2 2
x − 3 = 16 − 8 x + 5 + x + 5
2x − 1 = 4 − 4 3x − 2 + 3x
− 2
8 x + 5 = 24
4 3x
− 2 = x + 3
41.
x + 5 = 3
( ) 2 2
x + 5 = 3
Check x = 4
x + 5 = 9
( ) ( )
x = 4
4 − 3 + 4 + 5 = 4
4 = 4 true
The solution is 4.
x − 4 = x + 6 + 2
( x − 4) = ( x + 6 + 2)
( x 6)
2 2
x − 4 = x + 6 + 4 x + 6 + 4
4 x + 6 = −14
7
x + 6 = −
2
2
2 ⎛ 7 ⎞
+ = ⎜ − ⎟
⎝ 2 ⎠
49
x + 6 =
4
25
x =
4
25
Check x =
4
25
?
25
⎛ ⎞ ⎛ ⎞
⎜ ⎟ − 4 = ⎜ ⎟ + 6 + 2
⎝ 4 ⎠ ⎝ 4 ⎠
?
3
?
11
= false
2 2
There are no real solutions.
45.
47.
( 4 3x
− 2) = ( x + 3)
x
2
( )
( x )( x )
2 2
16 3x − 2 = x + 6x
+ 9
48x − 32 = x + 6x
+ 9
− 42x
+ 41 = 0
− 41 − 1 = 0
x − 41 = 0 or x − 1 = 0
x = 41 or x = 1
Check x = 41
( ) ( )
2 41 − 1 + 3 41 − 2 = 2
( ) ( )
2 1 − 1 + 3 1 − 2 = 2
2
2
?
?
20=
2 false
2 = 2 true
The solution is 1.
x − 2 = 3
( x )
2
( x )
2
2
− 2 = 3
x − 2 = 9
x = 11
2
= 11
x = 121
Check x = 121
( )
?
121 − 2 = 3
3 = 3 true
The solution is 121.
1
+ x = 2
x
⎛ 1 ⎞
x ⋅ ⎜ + x ⎟ = x ⋅2
⎝ x ⎠
?
293

Homework 9.5
SSM: Intermediate Algebra
49.
2
1+ x = 2
2
( 1+ x) = ( 2 x )
1+ 2x + x = 4x
x
2
− 2x
+ 1 = 0
( x )( x )
−1 − 1 = 0
x − 1 = 0
x = 1
Check x = 1
1
+ ( 1) =
?
2
1
( )
2 = 2 true
The solution is 1.
1
= 3 − x + 2
x + 2
1
x + 2 ⋅ = x + 2 3 − x + 2
x + 2
x
( 3 x + 2) = ( x + 3)
x
2
( )
2 2
2
2
2
( )
1 = 3 x + 2 − x − 2
3 x + 2 = x + 3
9 x + 2 = x + 6x
+ 9
9x + 18 = x + 6x
+ 9
− 3x
− 9 = 0
( 3) (
2
3) 4( 1)( 9)
2( 1)
− − ± − − −
x =
3 ± 3 5
=
2
3 + 3 5
2
1
?
= 3 − ( 4.85)
+ 2
4.85 + 2
Check ( x ≈ 4.85)
( )
0.382 = 0.382 true
3 − 3 5
2
1
?
= 3 − ( − 1.8541)
+ 2
− 1.8541 + 2
Check ( x ≈ − 1.8541)
( )
2.62 = 2.62 true
The solutions are 3 ± 3 5
2
.
51.
53.
3 x + 4 − 7 x + 1
3 x − 7 x + 4 + 1
( 3 − 7) x + ( 4 + 1)
− 4 x + 5
3 x + 4 − 7 x + 1= −7
Check x = 9
− 4 x + 5 = −7
− 4 x = −12
x = 3
( ) 2 2
x = 3
( ) ( )
x = 9
3 9 + 4 − 7 9 + 1=−
7
− 7 =− 7 true
The solution is 9.
55. ( x )( x )
?
?
+ 3 + 1 = 3
x ⋅ x + 3⋅ x + x ⋅ 1+ 3⋅ 1 = 3
Check x = 0
( ( ) ) ( )
x + 3 x + x + 3 = 3
x + 4 x = 0
( 4 x ) = ( −x)
x
2
( ) ?
0 + 3 0 + 1 = 3
Check x = 16
( ( ) ) ( )
4
( )
x = −x
2 2
16x
= x
− 16x
= 0
x x − 16 = 0
x = 0 or x − 16 = 0
x = 0 or x = 16
3 = 3 true
( ) ?
16 + 3 16 + 1 = 3
The solution is 0.
?
35=
3 false
2
294

SSM: Intermediate Algebra Homework 9.5
57. ( x + 3)( x + 1)
x ⋅ x + 3⋅ x + x ⋅ 1+ 3⋅1
x + 3 x + x + 3
x + 4 x + 3
59. f ( x) = 5 x − 7
5 x − 7 = 0
x − 7 = 0
( ) 2 2
x − 7 = 0
x − 7 = 0
x = 7
Check x = 7
( )
?
5 7 − 7 = 0
0 = 0 true
7,0 .
The x-intercept is ( )
= − + −
61. h( x) 3 3x
4 15
3 − 3x
+ 4 − 15 = 0
3 − 3x
+ 4 = 15
− 3x
+ 4 = 5
( ) 2 2
− 3x
+ 4 = 5
− 3x
+ 4 = 25
− 3x
= 21
x = −7
Check x = − 7
( )
3 −3 − 7 + 4 − 15=
0
?
0 = 0 true
− 7,0 .
The x-intercept is ( )
63. f ( x) = 3x − 2 − x + 8
3x
− 2 − x + 8 = 0
3x
− 2 = x + 8
( 3x
− 2) = ( x + 8)
Check x = 5
( ) ( )
2 2
3x
− 2 = x + 8
2x
= 10
x = 5
3 5 − 2 − 5 + 8 = 0
?
0 = 0 true
The x-intercept is ( 5,0 ).
65. h( x) = 2 x + 4 + 3 x − 5
2 x + 4 + 3 x − 5 = 0
Check
2 x + 4 = −3 x − 5
( 2 x + 4) = ( −3 x − 5)
x =
2 2
( x + ) = ( x − )
4 4 9 5
4x
+ 16 = 9x
− 45
61
5
− 5x
= −61
x =
61
5
?
⎛ 61⎞ ⎛ 61⎞
2 ⎜ ⎟ + 4 + 3 ⎜ ⎟ − 5 = 0
⎝ 5 ⎠ ⎝ 5 ⎠
36
?
= 0 false
5
No real number solution. There are no
x-intercepts.
295

Homework 9.5
SSM: Intermediate Algebra
67. f ( x) = 3 x − 7
3 x − 7 = −1
3 x = 6
x = 2
( ) 2 2
x = 2
x = 4
Check x = 4
( )
3 4 − 7 =−1
?
− 1 = −1 true
When x = 4 , f ( x ) = − 1.
69. f ( x) = −2 x − 4 + 5
−2 x − 4 + 5 = 7
−2 x − 4 = 2
x − 4 = −1
( x − 4) = ( −1)
2 2
x − 4 = 1
x = 5
Check x = 5
( )
−2 5 − 4 − 5=
7
?
?
− 7 = 7 false
No real number solutions. There is no value of x
that would make f ( x ) = 7 .
71. a. Somewhere during the 1970’s average test
scores declined. During the 1980’s and
1990’s test scores have been improving to
near the 1970 level.
b.
3.9 t + 280 = 305
3.9 t = 25
( t )
t =
2
25
3.9
⎛ 25 ⎞
= ⎜ ⎟
⎝ 3.9 ⎠
2
2
⎛ 25 ⎞
t = ⎜ ≈ 41.09
3.9
⎟
⎝ ⎠
According to the model, average test scores
will return to 305 in 2023.
c.
3.9 t + 280 = 500
3.9 t = 220
( t )
t =
2
220
3.9
⎛ 220 ⎞
= ⎜ ⎟
⎝ 3.9 ⎠
2
2
⎛ 220 ⎞
t = ⎜ ⎟ ≈ 3182.12
⎝ 3.9 ⎠
According to the model, the average test
score will reach the maximum score of 500
in 5164. Model breakdown has occurred.
The model assumes that the average score
will continue to rise.
73. No. When the student squared both sides of the
equation, the work on the left side was incorrect.
The student did not square the factor 5 on the left
side.
( 5 2
x + 3 ) = 25 ( x + 3 )
75. y = 3 x − 4
y = − 2 x + 6
Since the left hand sides are equal, set the right
hand sides equal to each other and solve the
resulting equation.
3 x − 4 = − 2 x + 6
5 x = 10
x = 2
( ) 2 2
x = 2
x = 4
Substitute this value into either original equation
and solve for y.
y = 3 ( 4)
− 4 = 2
The solution is ( 4, 2 ) .
77. The left hand side was not squared properly in
the second line.
296

SSM: Intermediate Algebra Homework 9.6
Homework 9.6
1. ( 0,3 ) and ( 4,5 )
Substitute the point ( 0,3 ) into the equation
y = a x + b .
3 = a 0 + b
b = 3
Substitute the point ( )
y = a x + 3 and solve for a.
5 = a 4 + 3
2a
= 2
a = 1
The equation is y = x + 3 .
4,5 into the equation
3. ( 0, 2 ) and ( 9,6 )
Substitute the point ( 0, 2 ) into the equation
y = a x + b .
2 = a 0 + b
b = 2
Substitute the point ( )
y = a x + 2 and solve for a.
6 = a 9 + 2
3a
= 4
a =
4
3
The equation is
9,6 into the equation
4
y = x + 2 .
3
5. ( 0, 2 ) and ( 3,5 )
Substitute the point ( 0, 2 ) into the equation
y = a x + b .
2 = a 0 + b
b = 2
Substitute the point ( )
y = a x + 2 and solve for a.
a
5 = a 3 + 2
3 = 3
a =
=
3
3
3
3,5 into the equation
The equation is y = 3 ⋅ x + 2 or y = 3x
+ 2 .
7. ( 1,2 ) and ( 4,3 )
Substitute the points into the equation
y a x b = + .
2 = a 1 + b
3 = a 4 + b
Rewrite as:
a + b = 2
2a
+ b = 3
Solve the resulting system. Multiply the first
equation by − 1 and add to the second equation.
−a
− b = −2
2 a + b = 3
a = 1
Substitute the point ( )
y = x + b and solve for b.
2 = 1 + b
b = 1
The equation is y = x + 1 .
9. ( 2, 4 ) and ( 3,5 )
1,2 into the equation
Substitute the points into the equation
y a x b = + .
4 = a 2 + b
5 = a 3 + b
Rewrite as:
1.4142a
+ b = 4
1.7321a
+ b = 5
Solve the resulting system. Multiply the first
equation by − 1 and add to the second equation.
−1.4142a
− b = −4
1.7321 a + b = 5
0.3179a
= 1
a ≈ 3.15
2, 4 into the equation
Substitute the point ( )
y = 3.15 x + b and solve for b.
4 = 3.15 2 + b
b ≈ −0.45
The equation is roughly y = 3.15 x − 0.45 .
297

Homework 9.6
SSM: Intermediate Algebra
11. ( 2,6 ) and ( 5,4 )
Substitute the points into the equation
y a x b = + .
6 = a 2 + b
4 = a 5 + b
Rewrite as:
1.4142a
+ b = 6
2.2361a
+ b = 4
Solve the resulting system. Multiply the first
equation by − 1 and add to the second equation.
−1.4142a
− b = −6
2.2361a
+ b = 4
0.8129a
= −2
a ≈ −2.43
2,6 into the equation
Substitute the point ( )
y = − 2.43 x + b and solve for b.
6 = − 2.43 2 + b
b ≈ 9.44
The equation is roughly y = − 2.43 x + 9.44 .
13. Increase the value of b to shift the graph up.
15. a. Start by plotting the data.
( ) = 5.65 t + 10
f t
b. f ( 26) = 5.65 ( 26)
+ 10 ≈ 38.81
c.
In 2008, 38.8% of single-parent fathers will
have never been married.
5.65 t + 10 = 39
5.65 t = 29
( t )
t =
29
5.65
2 ⎛ 29 ⎞
= ⎜
5.65 ⎟
⎝ ⎠
t ≈ 26.35
In 2008, 39% of single-parent fathers will
have never been married.
d. ( 0,10 ) . In 1982, 10% of single-parent
fathers had never been married.
17. a. Start by plotting the data.
2
Answers may vary. A square root model
0,10
may be reasonable. Use the points ( )
and ( 18,34 ) .
Substitute the point ( 0,10 ) into the equation
y = a x + b .
10 = a 0 + b
b = 10
Substitute the point ( )
18,34 into the
equation y = a x + 10 and solve for a.
34 = a 18 + 10
24
a = ≈ 5.66
18
Answers may vary. A square root model
seems to be appropriate. Use the points
52.5,1.94 .
( 0,0 ) and ( )
Substitute the point ( 0,0 ) into the equation
y = a x + b .
0 = a 0 + b
b = 0
52.5,1.94 into the
Substitute the point ( )
equation y = a x and solve for a.
1.94 = a 52.5
a ≈ 0.27
S h =
( ) 0.27
h
298

SSM: Intermediate Algebra Homework 9.6
b. i. Graph all three functions.
19. a. Start by plotting the data.
The model S ( h ) appears to fit the best.
ii. S ( ) L( ) Q( )
iii.
0 = 0; 0 = 0.327; 0 = 0.165
S models the situation best near 0 since
it is the only model that passes through
the origin.
Zoom out.
Q is not possible since it indicates that
the falling time will reach 0 for larger
drop heights.
Answers may vary. A square root model
seems reasonable.
3,68 .
Use the points ( 0,0 ) and ( )
Substitute the point ( 0,0 ) in the equation
y = a x + b .
0 = a 0 + b
b = 0
Substitute the point ( )
y = a x and solve for a.
68 = a 3
a ≈ 39.26
S t =
( ) 39.26
t
3,68 into the equation
iv.
v.
S models the situation the best and has
no problems with larger h.
T =
=
2h
32.2
2
32.2
= 0.249 h
This is close to the model
( ) 0.27
S h
h
= h .
b. i. Both models seem to fit the data pretty
well but the square root model seems to
be a little better fit.
c.
0.27 h = 3
( h )
h =
3
0.27
2
2 ⎛ 3 ⎞
= ⎜ ⎟
⎝ 0.27 ⎠
h ≈ 123.46
According to the model, the height of the
cliff is roughly 123.46 feet.
d. S ( 1250)
= 0.27 1250 ≈ 9.55
It would take about 9.55 seconds for the
baseball to reach the ground if it were
dropped from the top of New York City’s
Empire State Building.
ii. As the length of release time increases,
the percent of criminals that were arrested
again cannot decrease since the
percentages are cumulative. The square
root function is an increasing function.
c. ( ) 4 39.26 4 78.52
S = =
About 78.5% of convicts released from state
prison have been arrested for another crime
after being out of prison for 4 years or less.
299

Homework 9.6
SSM: Intermediate Algebra
d.
100 = 39.26
( t )
t =
100
39.26
t
2
2 ⎛ 100 ⎞
= ⎜ ⎟
⎝ 39.26 ⎠
t ≈ 6.5
According to the model, after about 6.5
years, all convicts released from state prison
will have been arrested for a new crime.
This is not realistic since it is likely that
some released convicts will never be
arrested for another crime.
21. a. Start by plotting the data.
b. f ( 7) = 31.92 ( 7)
+ 9.16 ≈ 93.61
c.
93.6% of 7 th births occurred despite the use
of contraception.
100 = 31.92 n + 9.16
31.92 n = 90.84
( n )
n =
90.84
31.92
2 ⎛ 90.84 ⎞
= ⎜ ⎟
⎝ 31.92 ⎠
n ≈ 8.1
All 8 th births occurred despite the use of
contraception.
2
d. The higher the birth order, the higher the
percent of births that happened despite the
use of contraception.
A square root model my fit reasonably well.
4,73 .
Use the points ( 2,54.3 ) and ( )
Substitute the points into the equation
y a x b = + .
54.3 = a 2 + b
73 = a 4 + b
Rewrite as:
1.4142a
+ b = 54.3
2a
+ b = 73
Solve the system of equations. Multiply the
first equation by − 1 and add to the second
equation.
−1.4142a
− b = −54.3
2a
+ b = 73
0.5858a
= 18.7
a ≈ 31.92
2,54.3 into the
Substitute the point ( )
equation y = 31.92 x + b and solve for b.
54.3 = 31.92 2 + b
b = 54.3 − 31.92 2
b ≈ 9.16
( ) = 31.92 n + 9.16
f n
23. a. Start by plotting the data.
A square root function may fit reasonably
6,160 .
well. Use the points ( 2,85.3 ) and ( )
Substitute the points into the equation
y a x b = + .
85.3 = a 2 + b
160 = a 6 + b
Rewrite as:
1.41a
+ b = 85.3
2.45a
+ b = 160
Multiply the first equation by –1 and add to
the second equation.
−1.41a
− b = −85.3
2.45 a + b = 160
1.04a
= 74.7
a ≈ 71.83
2,85.3 into the equation
Substitute ( )
y = 71.83 x + b and solve for b.
85.3 = 71.83 2 + b
b = 85.3 − 71.83 2
b ≈ −16.28
300

SSM: Intermediate Algebra
Chapter 9 Review Exercises
( ) 71.83 n 16.28
f n
= − .
b. f is an increasing function. This suggests that
the greater the number of people living with
the child, the less likely the child is to try to
take control.
Chapter 9 Review Exercises
1.
2.
x
3/ 7 7 3
=
x = x
x
1/ 2
2/ 9 2
9
3. ( 2x
+ 1) = ( 2x
+ 1)
7 7/ 5
4. 5 ( 3 x + 4 ) = ( 3 x + 4 )
5. 16x
= 16
6.
7.
= 4
x
6 6
3
6
x
8x
= 4x
⋅2
=
4x
2
= 2x
2
7 6
3x = x ⋅3x
=
3
= x
x
8. 8 x
6 x
6/8
=
= x
=
3/ 4
4 3
x
6
3x
3x
9. 3 10 3 9
24x = 8x ⋅3x
=
3 9 3
8x
3x
3 3
= 2x
3x
27 25 2
5 5
10. ( 6x + 11) = ( 6x + 11) ⋅ ( 6x
+ 11)
( 6x
11) ( 6x
11)
5 25 5
2
= + +
5
( 6x
11) ( 6x
11)
5 2
= + +
11. 2 − 5 + 3 = ( 2− 5+
3)
12.
x x x x
3 3
3 3
= 0⋅
x
= 0
5 x − 2 x + 7 x + 4 x
= 5 x + 7 x − 2 x + 4 x
3
( 5 7) x ( 2 4)
= + + − +
3
= 12 x + 2
13. ( )
( 20 8) x ( 5 2)
x
3 3
5 4 x − x − 2 x + 8 x
3 3
= 20 x − 5 x − 2 x + 8 x
3
x
3 3
= 20 x + 8 x −5 x −2
x
= + + − −
= 28 x − 7
14. ( )
x
3 x x − 7 = 3 x ⋅ x − 3 x ⋅7
2
3
x
= 3 x ⋅ x − 21 x
= 3 x − 21
= 3x
− 21 x
15. ( 4 x − 3)( 2 x + 1)
= 4 x ⋅ 2 x − 3⋅ 2 x + 4 x ⋅1− 3⋅1
= 8 x ⋅ x − 6 x + 4 x − 3
2
= 8 x − 2 x − 3
= 8x
− 2 x − 3
16. 2( 3 x + 8)( 4 x + 2)
= 2( 3 x ⋅ 4 x + 8⋅ 4 x + 3 x ⋅ 2 + 8⋅
2)
= 2( 12 x ⋅ x + 32 x + 6 x + 16)
2
( x x )
= 2 12 + 38 + 16
= 24x
+ 76 x + 32
x
301

Chapter 9 Review Exercises
SSM: Intermediate Algebra
2 2
17. ( x + 1)( x − 1) = ( x ) −( 1)
= x −1
2 2
18. ( 3 x + 4)( 3 x − 4) = ( 3 x ) − ( 4)
= 9x
−16
2 2 2
3 3 3
19. ( 2 x + 5) = ( 2 x ) + 2( 2 x )( 5) + ( 5)
3 2 3
= 4 x + 20 x + 25
2 2 2
20. ( 5 x − 2) = ( 5 x ) − 2( 5 x )( 2) + ( 2)
21.
22.
23.
4
5
= 25x
− 20 x + 4
1/ 4 1/ 5
x x = x ⋅ x
= x
= x
= x
=
1 1
+
4 5
5 4
+
20 20
9/ 20
20 9
x
3 6 3 1/ 6
4
6
x =
=
1/ 6
( x )
= x
= x
=
x x
=
x x
= x
= x
= x
=
12
18
x
1 1
⋅
6 3
1/ 4
1/ 6
1/18
x
1 1
−
4 6
3 2
−
12 12
1/12
x
1/ 3
24.
x x 2
= ⋅
2 2 2
=
=
x
2
2⋅
2
x 2
2
25. 3 =
3 x x
3
= ⋅
x
26.
27.
=
=
3⋅
x
x ⋅ x
3x
x
x
x
3 2
5 5 x
= ⋅
x x x
3 3 3 2
5
=
5
5
=
3 2
x
3 2
x ⋅ x
3 2
5 x
=
3 3
x
3 2
x
x
5
7 7
=
27x
27x
2 5 2
5
5 3
=
7
⋅
9x
27x
9x
=
=
=
5 2 5 3
5 3
7 ⋅9x
5 2 3
27x
⋅9x
5 3
63x
5 5
243x
5 3
63x
3x
302

SSM: Intermediate Algebra
Chapter 9 Review Exercises
28.
29.
30.
5 5 3 − x
= ⋅
3 + x 3 + x 3 − x
=
5( 3 − x )
2
( 3) − ( x ) 2
15 − 5 x
=
9 − x
2 2 2 + 3 x
= ⋅
2 − 3 x 2 − 3 x 2 + 3 x
=
2( 2 + 3 x )
( 2) 2
− ( 3 x ) 2
4 + 6 x
=
4 − 9x
5 x − 4 5 x − 4 2 x − 3
= ⋅
2 x + 3 2 x + 3 2 x − 3
5 x ⋅ 2 x − 4⋅ 2 x − 5 x ⋅ 3 + 4⋅3
=
( 2 x ) − ( 3)
2 2
10 x ⋅ x − 8 x − 15 x + 12
=
4x
− 9
10x
− 23 x + 12
=
4x
− 9
31. h( x) = ( 3 x + 5) + ( 2 − 4 x )
= 3 x + 5 + 2 − 4 x
= − x + 7
32. h( x) = ( 3 x + 5) −( 2 − 4 x )
= 3 x + 5 − 2 + 4 x
= 7 x + 3
33. h( x) = ( 3 x + 5)( 2 − 4 x )
= 3 x ⋅ 2 + 5⋅ 2 − 3 x ⋅4 x − 5⋅4
x
= 6 x + 10 −12x − 20 x
= −12x
− 14 x + 10
34. h( x)
35.
36.
3 x + 5
=
2 − 4 x
3 x + 5 2 + 4
= ⋅
2 − 4 x 2 + 4
x
x
3 x ⋅ 2 + 5⋅ 2 + 3 x ⋅ 4 x + 5⋅4
x
=
( x + x + )
( − x)
( 2) 2
− ( 4 x ) 2
6 x + 10 + 12x + 20 x
=
4 −16x
12x
+ 26 x + 10
=
4 −16x
2 6 13 5
=
2 2 8
6x
+ 13 x + 5
=
2 − 8x
3 x + 4 = 13
3 x = 9
x = 3
( ) 2 2
x = 3
x = 9
Check x = 9
( )
3 9 + 4 = 13
13 = 13 true
The solution is 9.
2x
+ 1 + 4 = 7
( ) 2 2
2x
+ 1 = 3
?
2x
+ 1 = 3
2x
+ 1 = 9
2x
= 8
x = 4
Check x = 4
( )
2 4 + 1 + 4 = 7
7 = 7 true
The solution is 4.
?
303

Chapter 9 Review Exercises
SSM: Intermediate Algebra
37.
38.
( 4x
5)
x
2
4x
+ 5 = x
2
+ = x
4x
+ 5 = x
− 4x
− 5 = 0
( x )( x )
− 5 + 1 = 0
x − 5 = 0 or x + 1 = 0
x = 5 or x = −1
Check x = 5
?
( ) + = ( )
4 5 5 5
5 = 5 true
Check x = − 1
2
2
( − ) + = ( − )
4 1 5 1
1=−1 false
The solution is 5.
?
?
2x
− 4 = x − 2
( 2x
− 4) = ( x − 2)
2 2
40.
Check x = 9
?
( 9) + 6=
( 9)
9 = 9 true
Check x = 4
?
( 4) + 6 = ( 4)
8=
4 false
The solution is 9.
?
13x
+ 4 = 5x
− 20
( 13x
+ 4) = ( 5x
− 20)
2 2
13x
+ 4 = 5x
− 20
8x
= −24
x = −3
Check x = − 3
?
( ) ( )
13 − 3 + 4 = 5 −3 − 20
?
− 35 = −35
There is no real solution.
39.
x
2
( x )( x )
2x − 4 = x − 4x
+ 4
− 6x
+ 8 = 0
− 4 − 2 = 0
x − 4 = 0 or x − 2 = 0
x = 4 or x = 2
Check x = 4
?
( ) ( )
2 4 − 4 = 4 − 2
2 = 2 true
Check x = 2
?
( ) ( )
2 2 − 4 = 2 − 2
0 = 0 true
The solutions are 4 and 2.
x
2
x + 6 = x
2
( x ) = ( x − 6)
( x )( x )
x = x − 6
− 13x
+ 36 = 0
− 9 − 4 = 0
2 2
2
x = x − 12x
+ 36
x − 9 = 0 or x − 4 = 0
x = 9 or x = 4
41.
x
2x
− 1 = 1+ x + 3
( 2x
− 1) = ( 1+ x + 3)
2 2
2x − 1 = 1+ 2 x + 3 + x + 3
2 x + 3 = x −5
( 2 x + 3) = ( x − 5)
2
( )
( x )( x )
2 2
4 x + 3 = x − 10x
+ 25
2
2
4x + 12 = x − 10x
+ 25
− 14x
+ 13 = 0
−13 − 1 = 0
x − 13 = 0 or x − 1 = 0
x = 13 or x = 1
Check x = 13
?
( ) ( )
2 13 − 1 = 1+ 13 + 3
5 = 5 true
Check x = 1
?
( ) ( )
2 1 − 1= 1+ 1 + 3
1=
3 false
The solution is 13.
?
304

SSM: Intermediate Algebra
Chapter 9 Review Exercises
42.
x + 2 + x + 9 = 7
x + 2 = 7 − x + 9
( x + 2) = ( 7 − x + 9)
2 2
x + 2 = 49 − 14 x + 9 + x + 9
14 x + 9 = 56
x + 9 = 4
( ) 2 2
x + 9 = 4
Check x = 7
x + 9 = 16
( ) ( )
x = 7
7 + 2 + 7 + 9 = 7
7 = 7 true
The solution is 7.
43. y = − 2
x y
0 0
1 −2
4 −4
9 −6
16 −8
y
x
?
45. y = − x − 5 + 3
x y
5 3
6 2
9 1
14 0
21 −1
4
y
−4
46. y = 2 x + 4 − 1
x y
−4 −1
−3 1
0 3
5 5
12 7
y
4
8
1 6
x
4
8
x
−2
2
x
−4
−4
−8
44. y = 3 x + 1
x y
0 1
1 4
4 7
9 10
16 13
8
4
y
47. f ( x) = 4x − 7 − 2x
+ 1
4x
− 7 − 2x
+ 1 = 0
Check x = 4
4x
− 7 = 2x
+ 1
( 4x
− 7) = ( 2x
+ 1)
2 2
4x
− 7 = 2x
+ 1
2x
= 8
( ) ( )
x = 4
4 4 − 7 − 2 4 + 1 = 0
?
0 = 0 true
4 8
x
305

Chapter 9 Review Exercises
SSM: Intermediate Algebra
The x-intercept is ( 4,0 ) .
48. g ( x) 3 x 2 9
= − + +
− 3 x + 2 + 9 = 0
− 3 x + 2 = −9
x + 2 = 3
( ) 2 2
x + 2 = 3
x + 2 = 9
x = 7
Check x = 7
( )
− 3 7 + 2 + 9 = 0
?
0 = 0 true
7,0 .
The x-intercept is ( )
49. Increase b to raise the y-intercept and decrease a
to lower the rate of increase. Sketches may vary.
50. The equation is of the form y = a x + b .
The y-intercept is ( 0, 2 ) so b = 2 .
Substitute the point ( 8,9 ) into the equation
y = a x + 2 and solve for a.
a
9 = a 8 + 2
8 = 7
7
a = ≈ 2.475
8
The equation is roughly y = 2.475 x + 2 .
51. The equation is of the form y = a x + b .
The y-intercept is ( 0,3 ) so b = 3 .
Substitute the point ( 4,8 ) into the equation
y = a x + 3 and solve for a.
8 = a 4 + 3
2a
= 5
a =
5
2
The equation is
5
y = x + 3 .
2
52. The equation is of the form y = a x + b .
The y-intercept is ( 0,7 ) so b = 7 .
Substitute the point ( 9,3 ) into the equation
y = a x + 7 and solve for a.
3 = a 9 + 7
3a
= −4
4
a = −
3
The equation is
53. ( 2,5 ) and ( 3,6 )
4
y = − x + 7 .
3
Substitute the points into the equation
y a x b = + .
5 = a 2 + b
6 = a 3 + b
Rewrite as:
1.4142a
+ b = 5
1.7321a
+ b = 6
Solve the system of equations. Multiply the first
equation by − 1 and add to the second equation.
−1.4142a
− b = −5
1.7321a
+ b = 6
0.3179a
= 1
a ≈ 3.15
Substitute the point ( )
2,5 into the equation
y = 3.15 x + b and solve for b.
5 = 3.15 2 + b
b = 5 − 3.15 2
b ≈ 0.55
The equation is roughly y = 3.15 x + 0.55 .
306

SSM: Intermediate Algebra
Chapter 9 Test
54. ( 3,7 ) and ( 5,4 )
Substitute the points into the equation
y a x b = + .
7 = a 3 + b
4 = a 5 + b
Rewrite as:
1.7321a
+ b = 7
2.2361a
+ b = 4
Solve the system of equations. Multiply the first
equation by − 1 and add to the second equation.
−1.7321a
− b = −7
2.2361a
+ b = 4
0.504a
= −3
a ≈ −5.95
3,7 into the equation
Substitute the point ( )
y = − 5.95 x + b and solve for b.
7 = − 5.95 3 + b
b = 7 + 5.95 3
b ≈ 17.31
The equation is roughly y = − 5.95 x + 17.31.
55. a. Start by plotting the data.
A square root model, y = a x + b , seems
reasonable.
4, 28.8 .
Use the points ( 0,16 ) and ( )
The y-intercept is ( 0,16 ) so b = 16 .
Substitute the point ( 4, 28.8 ) into the
equation y = a x + 16 and solve for b.
28.8 = a 4 + 16
2a
= 12.8
a = 6.4
( ) = 6.4 t + 16
f t
b. ( 0,16 ) ; In 1998 there were 16 million U.S.
households with a Sony PlayStation.
c. 6.4 t + 16 = 36
6.4 t = 20
t = 3.125
t ≈ 9.77
In 2008, 36 million households in the United
States will have a Sony PlayStation.
d. f ( 5)
= 6.4 5 + 16 ≈ 30.31
The model predicts a slightly larger number
of households, but the values are fairly
close.
Chapter 9 Test
1.
9 8
32x = 16x ⋅2x
=
4
8
16x
2x
= 4x
2x
2. 3 64 22 3 64
21
x = x ⋅ x
=
3 21 3
= 4x
64x
7 3
4 4
3. ( 2x + 8) = ( 2x + 8) ⋅ ( 2x
+ 8)
4.
3
4 x 2x
=
5
6 x 3x
=
2
3
x
x
27 24 3
1/ 3
1/ 5
2
= x
3
2
= x
3
1 1
x3 − 5
( 2x
8) ( 2x
8)
4 24 4
3
= + +
4
( 2x
8) ( 2x
8)
6 3
= + +
5 3
−
15 15
2/15
15 2
2 x
=
3
307

Chapter 9 Test
SSM: Intermediate Algebra
5.
6.
x + 1 x + 1 2 x + 3
= ⋅
2 x − 3 2 x − 3 2 x + 3
=
x ⋅ 2 x + 1⋅ 2 x + x ⋅ 3 + 1⋅3
( 2 x ) − ( 3)
2 2
2x + 2 x + 3 x + 3
=
4x
− 9
2x
+ 5 x + 3
=
4x
− 9
4 4
3 12x + 2 x − 8 27x − 5 x
4
4 4
= 3 4⋅3x − 8 9⋅ 3x + 2 x − 5 x
4 4
= 3⋅2 3x − 8⋅ 3 3x + 2 x − 5 x
4 4
= 6 3x − 24 3x + 2 x − 5 x
= −18 3x
− 3
7. ( )
x
3 x 6 x − 5 = 3 x ⋅6 x − 3 x ⋅5
8. ( 2 + 4 x )( 3 − 5 x )
2
= 18 x −15
= 18x
−15
x
= 2⋅ 3 + 4 x ⋅3 − 2⋅5 x − 4 x ⋅5
x
= 6 + 12 x −10 x −20
x
= − 20x
+ 2 x + 6
9. ( 4 + 3 x )( 4 − 3 x ) = ( 4) 2
− ( 3 x ) 2
2
= 16 − 9x
2 2 2
5 5 5
10. ( 4 x − 3) = ( 4 x ) − 2( 4 x )( 3) + ( 3)
x
12. h( x) = ( 7 − 3 x ) + ( 4 + 5 x )
= 7 − 3 x + 4 + 5 x
= 2 x + 11
13. h( x) = ( 7 − 3 x ) − ( 4 + 5 x )
= 7 − 3 x − 4 − 5 x
= 3 − 8 x
14. h( x) = ( 7 − 3 x )( 4 + 5 x )
15. h( x)
= 7 ⋅4 − 3 x ⋅ 4 + 7 ⋅5 x − 3 x ⋅5
x
= 28 − 12 x + 35 x −15x
= − 15x
+ 23 x + 28
7 − 3 x
=
4 + 5 x
7 − 3 x 4 − 5
= ⋅
4 + 5 x 4 − 5
x
x
7 ⋅4 − 3 x ⋅4 − 7 ⋅ 5 x + 3 x ⋅5
x
=
( 4) 2
− ( 5 x ) 2
28 −12 x − 35 x + 15x
=
16 − 25x
15x
− 47 x + 28
=
16 − 25x
16. ( ) ( )
f 8 = 6 − 4 8 + 1
= 6 − 4 9
= 6 − 4⋅3
= −6
11.
n
k
x
x
x
=
x
= x
= x
= x
=
1/ n
1/ k
1 1
−
n k
k n
−
kn kn
k −n
kn
kn k −n
x
5 2 5
= 16 x − 24 x + 9
17.
− 2 = 6 − 4 x + 1
4 x + 1 = 8
x + 1 = 2
( ) 2 2
x + 1 = 2
x + 1 = 4
x = 3
Check x = 3
?
?
( )
− 2 = 6 − 4 3 + 1
− 2 =− 2 true
Therefore, 3
f x = − .
x = when ( ) 2
308

SSM: Intermediate Algebra
Chapter 9 Test
18. y = − 2 x + 3 + 1
x y
−3 1
−2 −1
1 −3
6 −5
13 −7
−4
4
−4
y
4
19. a. We need a < 0 and k ≥ 0 , or we need
a > 0 and k ≤ 0 . In either case, h can be
any real number.
b. f ( x)
= a x − h + k
a x − h + k = 0
a x − h = −k
( x h )
k
x − h = −
a
2 ⎛ k ⎞
− = ⎜ − ⎟
⎝ a ⎠
k
x − h =
a
k
x = h +
2
a
⎛
2
k
The x-intercept is h ,0 ⎞
⎜
+
2 a ⎟
.
⎝ ⎠
2
2
x
2
2
20.
21.
22.
2 x + 3 = 13
2 x = 10
x = 5
( ) 2 2
x = 5
x = 25
Check x = 25
( )
2 25 + 3=
13
13 = 13 true
The solution is 25.
3 5x
− 4 = 27
( ) 2 2
5x
− 4 = 9
?
5x
− 4 = 9
5x
− 4 = 81
5x
= 85
x = 17
Check x = 17
( )
3 5 17 − 4 = 27
27 = 27 true
The solution is 17.
3 − 2 x + 9 − x = 0
?
9 − x = 2 x − 3
( 9 − x ) = ( 2 x − 3)
2 2
( 12 x ) = ( 5x)
2
( x )
9 − x = 4x − 12 x + 9
12 x = 5x
2 2
144x
= 25x
25x
− 144x
= 0
x 25 − 144 = 0
x = 0 or 25x
− 144 = 0
144
x = 0 or x =
25
Check x = 0
( ) ( ) ?
3 − 2 0 + 9− 0 = 0
Check
144
x =
25
?
2
6=
0 false
309

Chapter 9 Test
SSM: Intermediate Algebra
144 144
?
⎛ ⎞ ⎛ ⎞
3 − 2 ⎜ ⎟ + 9 − ⎜ ⎟ = 0
⎝ 25 ⎠ ⎝ 25 ⎠
0 = 0 true
The solution is 144
25 .
23. f ( x) = 3 2x − 4 − 2 2x
+ 1
3 2x
− 4 − 2 2x
+ 1 = 0
Check x = 4
3 2x
− 4 = 2 2x
+ 1
( 3 2x
− 4) = ( 2 2x
+ 1)
2 2
( x − ) = ( x + )
9 2 4 4 2 1
18x
− 36 = 8x
+ 4
10x
= 40
( ) ( )
x = 4
3 2 4 − 4 − 2 2 4 + 1 = 0
The x-intercept is ( )
?
0 = 0 true
4,0 .
24. Decrease b to lower the y-intercept and increase
a to increase the rate of increase. Graphs may
vary.
25. Substitute the points ( 2, 4 ) and ( 5,6 ) into the
equation y = a x + b .
4 = a 2 + b
6 = a 5 + b
Rewrite as:
1.4142a
+ b = 4
2.2361a
+ b = 6
Solve the system of equations. Multiply the first
equation by − 1 and add to the second equation.
−1.4142a
− b = −4
2.2361a
+ b = 6
0.8219a
= 2
a ≈ 2.43
2, 4 into the equation
Substitute the point ( )
y = 2.43 x + b and solve for b.
4 = 2.43 2 + b
b = 4 − 2.43 2
b ≈ 0.56
The equation is y = 2.43 x + 0.56 .
26. a. Start by plotting the data.
Answers may vary. A square root model
seems reasonable.
60,43.4 .
Use the points ( 0, 20.5 ) and ( )
The y-intercept is ( 0, 20.5 ) so b = 20.5 .
Substitute the point ( 60,43.4 ) into the
equation y = a x + 20.5 and solve for a.
a
43.4 = a 60 + 20.5
60 = 22.9
a ≈ 2.96
( ) = 2.96 t + 20.5
f t
b. f ( 72)
= 2.96 72 + 20.5 ≈ 45.62
According to the model, the median height
of 6-year-old boys is about 45.6 inches.
c.
( t )
36 = 2.96 t + 20.5
2.96 t = 15.5
t =
15.5
2.96
2
2 ⎛ 15.5 ⎞
= ⎜ ⎟
⎝ 2.96 ⎠
t ≈ 27.42
The median height of 27-month-old boys is
3 feet.
d. The h-intercept is ( ) 0, 20.5 . The median
height of boys at birth is 20.5 inches.
310