# Chapter 12 Quadratic Functions by bfk20410

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```									    SSM: Elementary and Intermediate Algebra                                Chapter 12: Quadratic Functions

21.   h = -16t 2 + s0 , for t
7.   E = i2 r , for i                                16t 2 = s0 - h
E                                                   s -h
= i2                                        t2 = 0
r                                                     16
E                                                     s -h
=i
r                                             t= 0
16
9.   d = 16t 2 , for t                                       s0 - h
t=
d = t2                                                   4
16
23.           1
d =t ﬁ t = d                                  E=        mv 2 , for v
2
16          4
2E = mv 2
2E
11.   E = mc 2 , for c                                          = v2
E                                                   m
= c2
m                                                  2E
=v
E                                                  m
=c
m
25.              2
v 2 - v1
2
1 2                                        a=           , for v1
13.                                                           2d
V=     pr h , for r
3                                                   2

3V = pr2 h                                            2
2ad + v1 = v2 2
3V
= r2                                      2
v1 = v 2 - 2ad
2
ph
2
3V                                            v1 = v2 - 2ad
=r
ph
27.   v ¢ = c 2 - v 2 , for c
15.           2      2
d = L +W , for W                                ( v ¢) 2 = c 2 - v 2
d 2 = L2 +W 2                                   ( v ¢) 2 + v 2 = c 2
d 2 - L2 =W 2                                                 2
c = (v ¢ ) + v 2
d 2 - L 2 =W
29.   a.       P( n) = 2.4 n 2 + 9n - 3
17.    2     2      2
†          a + b = c , for b                                                        2
P( 6) = 2.4 ( 6) + 9( 6) - 3
b 2 = c2 - a 2
=137.4 ﬁ \$13,740
b = c2 - a 2
The profit would be \$13,740.
19.   d = L 2 +W 2 + H 2 , for H                      b.            P( n) = 2.4n 2 + 9n- 3
d 2 = L2 +W 2 + H 2                                           200= 2.4n 2 + 9n- 3
d 2 - L2 -W 2 = H 2                                           2.4n 2 + 9n- 203= 0
d 2 - L 2 -W 2 = H                                           -9± 92 - 4( 2.4 )(-203)
x=
2( 2.4 )

x = -9± 2029.8
4.8
x ª 8 or x ª -11
Eight tractors must be sold.

437
Chapter 12: Quadratic Functions                                         SSM: Elementary and Intermediate Algebra

Eight tractors must be sold.                        37.   a. m (t) = 0.05t 2 - 0.32t + 3.15
In 2003, t = 21.
31.   T = 6.2t 2 +12t + 32
( )
m( 21) = 0.05 212 - 0.32( 21) + 3.15
a. When the car is turned on, t = 0.                                      = 22.05- 6.72 + 3.15
T = 6.2( 0) 2 +12 ( 0) + 32 = 32                                       ª 18.5
The temperature is 32°F.                                         Veterinary bills for dogs in 2003 amounted
b. T = 6.2(1) 2 + 12(1) + 32 = 50.2
The temperature after 1 minute is 50.2°F.                     b. m (t) = 0.05t 2 - 0.32t + 3.15, m (t) = 12
25= 0.05t 2 - 0.32t + 3.15
c.    120 = 6.2t 2 +12t + 32
0 = 0.05t 2 - 0.32t - 21.85
0 = 6.2t 2 +12t - 88                                                            2
0.32 ± ( 0.32) - 4 (0.05)( -21.85)
-12 ± 12 2 - 4(6.2)( -88)                                  t=
t=                                                                               2( 0.05)
2(6.2)
tª
-12 ± 48.23                                                 = 0.32 ± 0.1024 + 4.37
0.1
12.4
t ª 2.92 or t ª -4.86                                           = 0.32± 4.4724
The radiator temperature will reach 120°F                               0.1
about 2.92 min. after the engine is started.                    ª 0.32 ± 2.1148050
0.1
tª  0.32+ 2.1148050 = 2.4348050 ª 24.3
0.1             0.1
33.   a.    C = -0.011500 )2 + 80(1500) + 20,000
(                                                   or
= 117, 500                                                  t ª 0.32- 2.1148050 = -1.794805 ª -17.9
0.1             0.1
Since 1£ t £ 28, the only solution is t ≈
24.3. Thus \$25 billion will be spent on
b. 150, 000 = -0.01s 2 + 80s + 20, 000                              veterinary bills for dogs approximately
0 = -0.01s2 + 80s - 130, 000                                     24.3 years after 1982 or in 2006.
-80 ± 802 - 4( -0.01)(-130, 000)
s=                                                   39.   Let x be the width of the playground. Then
2(-0.01)                                  the length is given by x + 5.
†                    -80 ± 34.64                                             Area = length ¥ width
sª
-0.02                                               500= x ( x + 5)
s ª 2268 or s ª 5732
Notice that s must fall between 1200 and                   x 2 + 5x - 500 = 0
4000. Therefore, Mr. Boyle can purchase                    ( x - 20)( x + 25) = 0
a 2268 sq ft house for \$150,000.
x = 20 or x = -25
†                         2                                                 Disregard the negative value. The width of
35.   a. 0 = -3. 3t - 2.3t + 62
the playground is 20 meters and the length is
-(-2 .3)± (-2.3)2 - 4(- 3. 3)(62)                          20 + 5 = 25 meters.
t=              2(- 3.3)
t = 2.3-6.6 7
± 28.

t ≈ –4.7 or t = 4
Since time must be positive, the car takes
4 seconds for the drop.

†          b. s = 6.74(4) + 2.3 = 29.26
The speed is 29.26 feet per second.
†

†
†
438
SSM: Elementary and Intermediate Algebra                                               Chapter 12: Quadratic Functions

41. Let r be the rate at which the present equipment                     mechanic.
drills.                                                                                   6      6
+       =1
d                                        x x +1
d         r        t=                                  6               6 ˆ
r                    x(x + 1)Ê ˆ + x(x + 1)Ê
Á ˜           Á      ˜ = x(x + 1)(1)
Ë x¯          Ë x + 1¯
present                                      64
64         r                                                   6(x +1) + 6x = x 2 + x
equipment                                     r
64                                        6x + 6 + 6x = x 2 + x
new                    64       r+1
equipment                                   r +1                                                  0 = x 2 - 11x - 6
They would have hit water in 3.2 hours less time                          -(-11) ± (-11)2 - 4(1)(-6)
x=
with the new equipment.                                                               2(1)
64      64                                              11 ± 121 + 24
=     - 3.2
r +1      r                                            =
2
64 ˆ              64
r(r + 1)ÊÁ       ˜ = r(r + 1)Ê ˆ - r(r + 1)(3.2)
Á ˜                                      11 ± 145
Ë r + 1¯            Ë r ¯                                  =
2
64r = 64(r + 1) - 3.2r( r + 1)
11 + 145               11- 145
64r = 64r + 64 - 3.2r 2 - 3.2r                      x=               or x =
2                     2
0 = 64 - 3.2r 2 - 3.2r                              ª 11.52               ª -0.52
3.2r2 + 3.2r - 64 = 0                                                Since the time must be positive, it takes Bonita
r2 + r - 20 = 0                                             about 11.52 hours and Pamela about 12.52 hours
to rebuild the engine.
( r + 5)( r - 4) = 0
r + 5 = 0 or r - 4 = 0                                          47. Let r be the speed of the plane in still air.
r = -5                r=4                                                                                   d
d         r      t=
Use the positive value. The present equipment                                                                    r
drills at a rate of 4 ft/hr.                                                                                   80
With wind           80      r + 30
43. Let x be Latoya’s rate going uphill so x + 2 is                                                              r + 30
d                                                                        80
her rating going downhill. Using = t gives                            Against wind        80      r – 30
r                                                                      r - 30
tuphill + tdownhill = 1.75
The total time is 1.3 hours
6      6                                       80 + 80 = 1.3
+       = 1.75
x x+2                                        r + 30 r - 30
6                 6 ˆ
x(x + 2)Ê ˆ + x(x + 2)Ê
Á ˜             Á      ˜ = x(x + 2)(1.75)                                  (           80
( r + 30)( r - 30) r 80 + r - 30 =1.3 )
Ë x¯            Ë x + 2¯                                                       + 30
6(x + 2) + 6x = 1.75x( x + 2)                                                 (
80( r - 3) + 80( r + 30) = 1.3 r 2 - 900 )
6x + 12 + 6x = 1.75x 2 + 3.5x                                                    2
80r - 240 + 80r + 240 = 1.3r -1170
0 = 1.75x 2 - 8.5x -12
160r = 1.3r2 -1170
- ( -8.5) ±   (-8.5) 2 - 4(1.75)(-12)                                               0 = 1.3r2 -160 r - 1170
x=
2(1.75)                                           - (-160) ± ( -160) 2 - 4(1.3)( -1170)
8 ± 156.25                                                      r=
=                                                                                      2(1.3)
3.5                                                            160 ± 25, 600 + 6084
8.5 ±12.5                                                        =
=                                                                                2.6
3.5
x = 6 or x ª -1.14                                                      160 ± 31,684
=
Since the time must be positive, Latoya’s uphill                                2.6
rate is 6 mph and her downhill rate is                                   160 ±178
=
x + 2 = 8 mph.                                                              2.6

45. Let x be the time of the experienced mechanic
then x + 1 is the time of the inexperienced

439
Chapter 12: Quadratic Functions                                             SSM: Elementary and Intermediate Algebra

160 +178               160 - 178                                  Including the 2.5 hours she spent in Plainview,
r=                 or r =
2.6                    2.6                                    the entire trip took Lisa 5.5 hours.
338                    -18                                             60 + 2.5+ 100 = 5.5
=                      =                                                  x          x -10
2.6                   2.6
= 130                  ª -6.92                                               60 + 100 = 3
Since speed must be positive, the speed of the                                  x x -10
plane in still air is 130 mph.
(
x(x -10) 60 + 100 = 3
x x -10        )
49. Let t be the number of hours for Chris to clean
60(x -10) +100x = 3(x 2 -10x)
alone. Then t + 0.5 is the number of hours for
John to clean alone.                                                   60x - 600+100x = 3x 2 - 30x
Rate of     Time         Part of Task                    -600+160x = 3x 2 - 30x
work      worked         completed
0= 3x 2 -190x + 600
1                             6                           0 = (x - 60)(3x -10)
Chris                         6
t                             t                      x - 60 = 0 or 3x - 10 = 0
10
1                           6                            x = 60              x=
John                          6                                                                     3
t + 0.5                     t + 0.5                   10
6       6                                        miles per hour is too slow for a car, so the
+         =1                              3
t t + 0.5                                   speed of the trip from Lubbock to Plainview was
6                   6 ˆ
t(t + 0.5)Ê ˆ + t(t + 0.5)Ê
Á ˜             Á        ˜ = t(t + 0.5)(1)                60 mph.
Ët¯             Ë t + 0.5¯
6(t + 0.5) + 6t = t(t + 0.5)             53.    Answers will vary.
6t + 3 + 6t = t 2 + 0.5t
55.    Let l = original length and w = original
12t + 3 = t 2 + 0.5t
width. A system of equations that describes
0 = t 2 -11.5t - 3                this situation is
11.5 ±    ( -11.5)2 - 4(1)(-3)                                   l ⋅ w = 18
t=
2(1)                                              ( l + 2)( w + 3) = 48
11.5 ± 132.25 + 12                                               If you solve for l in the first equation you
=
2                                                       get l = 18 . Substitute 18 into the l in the
w                 w
11.5 ± 144.25                                                    second equation. The result is an equation
=
2                                                         in only one variable which can be solved.
11.5 + 144.25            11.5 - 144.25
t=                   or t =                                          ( l + 2)( w + 3) = 48
2                          2
ª 11.76                  ª -0.26
Since the time must be positive, it takes                            ( 18 + 2)( w + 3) = 48
w
18+ 54 + 2w + 6 = 48
11.76 + 0.5 = 12.26 hours to clean alone.                                w
2w - 24+ 54 = 0
†                                                                                             w
51. Let x be the speed of the trip from Lubbock to
Plainview. Then x – 10 is the speed from
(
w 2w - 24 + 54 = 0
w            )
2
Plainview to Amarillo.                                                 2w - 24w + 54 = 0
2( w - 3) (w - 9) = 0
d        r            t
w = 3 or w = 9
60
first part           60       x
x                          If w = 3, then l = 18 = 6. One possible set
3
100                          of dimensions for the original rectangle is 6
second part        100     x – 10
x -10                         m by 3 m.

440

†

†                   †
SSM: Elementary and Intermediate Algebra                                                              Chapter 12: Quadratic Functions

4
If w = 9, then l = 18 = 2. Another possible                              60.   Ê x 3/4 y -2 ˆ                            4
9
set of dimensions for the original rectangle
Á 1/2 2 ˜ = x
Ë x y ¯
(
(3/4 )-(1/2) -2-2
y          )
is 2 m by 9 m.                                                                                                   4
(
= x 1/4 y -4   )
57.                     3
[
- 4( 5- 3) + 23      ]                                                                        = x1 y -16

[           ]                                                                            = x
3
- 4( 2) + 23
y16
-[ 4( 8) ] + 8
61.
-32+ 8                                                                  x 2 + 3x + 9 = x
-24                                                                   x 2 + 3x + 9 = x 2
3x + 9= 0
58.    IR+ Ir = E, for R
3x = -9 ﬁ x = -3
IR = E - Ir
Upon checking, this value does not satisfy
R = E - Ir                                                                     the equation. There is no real solution.
I

59.     2r - 2r + 64
r - 4 r + 4 r 2 -16
= 2r ⋅ r + 4 - 2r ⋅ r - 4 +      64
r - 4 r + 4 r + 4 r - 4 ( r + 4 )( r - 4)

=       2r 2 +8r - 2r 2 - 8r +                64
( r + 4 )( r - 4) ( r + 4)( r - 4) ( r + 4) (r - 4)

=
(
2r 2 + 8r - 2r 2 -8r + 64    )
( r + 4) ( r - 4)
=        16r + 64
( r + 4 )( r - 4)
16( r + 4 )                   16
=                            or
( r + 4 )( r - 4)               r- 4

†
Exercise Set 12.4

1. A given equation can be expressed as an equation in quadratic form if the equation can be written in the form
au 2 + bu + c = 0.
2
†      3. Let u = x 2 . Then 3x 4 - 5x 2 +1= 0            ﬁ 3 x2( )         - 5x 2 +1= 0 ﬁ 3u 2 - 5u+1= 0

-1 2
5. Let u = z -1 . Then z-2 - z -1 = 56         ﬁ    (z )       - z -1 = 56 ﬁ u 2 - u = 56

441
Chapter 12: Quadratic Functions                     SSM: Elementary and Intermediate Algebra

7.    x4 + x2 - 6                           15. 6( a + 2) 2 - 7( a + 2) - 5
Let u = x 2                               Let u = a + 2
2                                 6u 2 - 7u - 5
( x 2)       + x2 - 6
( 2u + 1)( 3u - 5)
u2 + u - 6
Back substitute a + 2 for u
( u + 3)( u - 2)
( 2(a + 2) + 1) (3( a + 2) - 5)
Back substitute x 2 for u
( 2a + 4 + 1) (3a + 6 - 5)
( x 2 + 3)( x 2 - 2)                       ( 2a + 5)( 3a + 1)
9.    x 4 + 5x 2 + 6                        17. a 2b 2 + 8ab + 15
2
Let u = x                                 Let u = ab
2
( x 2)       + 5x 2 + 6                    ( ab) 2 + 8ab + 15
u 2 + 5u + 6                               u 2 + 8u + 15
( u + 2)( u + 3)                           ( u + 3)( u + 5)
Back substitute ab for u
Back substitute x 2 for u
( ab + 3)( ab + 5)
( x 2 + 2)( x 2 + 3)
19. 3x 2 y 2 - 2xy - 5
11. 6a 4 + 5a 2 - 25
Let u = xy
Let u = a 2                                      2
3( xy) - 2xy - 5
2 2
( )
6a            + 5a 2 - 25                  3u 2 - 2u - 5
2
6u + 5u - 25                               ( 3u - 5)( u + 1)
( 2u + 5)( 3u - 5)                         Back substitute xy for u
Back substitute a 2 for u                  ( 3xy - 5) ( xy + 1)
( 2a   2
)(   2
+ 5 3a - 5   )                 2
21. 2a ( 5- a) - 7a (5 - a ) + 5(5 - a )
13. 4( x + 1) 2 + 8( x + 1) + 3                  Factor out ( 5- a)
Let u = x + 1                                 ( 5- a)( 2a 2 - 7a + 5)
†          4u 2 + 8u + 3                               ( 5- a)( 2a - 5)( a - 1)
( 2u + 3)( 2u + 1)
23. 2x 2 ( x - 3) + 7x ( x - 3) + 6( x - 3)
Back substitute x +1 for u
Factor out x – 3
( 2(x + 1) + 3)( 2(x + 1) + 1)
( 2x + 2 + 3)( 2x + 2 + 1)                  ( x - 3)( 2x 2 + 7x + 6)
( 2x + 5)( 2x + 3)                          ( x - 3)( 2x + 3)( x + 2)

†

442
SSM: Elementary and Intermediate Algebra                                           Chapter 12: Quadratic Functions

25. y 4 + 13y 2 + 30                                                   5 2    4 3    3 4
29. 5a b - 8a b + 3a b
Let u = y 2                                                    Factor out a 3b 2
2
( y 2)       + 13y 2 + 30                                              (
a 3b 2 5a 2 - 8ab + 3b 2   )
3 2
u 2 + 13u + 30                                                  a b ( 5a - 3b) (a - b)
( u + 3)( u + 10)
Back substitute y 2 for u
( y 2 + 3)( y 2 + 10)
27. x 2 ( x + 3) + 3x ( x + 3) + 2( x + 3)
Factor out x + 3
( x + 3)( x 2 + 3x + 2)
( x + 3)( x + 1) ( x + 2)

31. x 4 - 10x 2 + 9 = 0
2
(x2 )          - 10x 2 + 9 = 0
u2 - 10u + 9 = 0 ¨ Replace x 2 with u
(u - 9)(u -1) = 0
u-9=0               or u - 1 = 0
u=9                        u =1
2
x =9                      x 2 = 1 ¨ Replace u with x 2
x = ± 9 = ±3               x = ± 1 = ±1
The solutions are 3, –3, 1, and –1.

33. x 4 - 26x 2 + 25= 0
2
( )
x 2 - 26x 2 + 25= 0
†
u 2 - 26u+ 25= 0 ¨ Replace x 2 with u
( u- 25)( u-1) = 0
u- 25= 0                 u-1= 0
u = 25                   u =1
2
x = 25                  x 2 = 1 ¨ Replace u with x 2
x = ± 25 = ±5           x = ± 1 = ±1
The solutions are 5, –5, 1, and –1.

35. x 4 - 13x 2 + 36 = 0
2
(x2 )        - 13x 2 + 36 = 0
u2 - 13u + 36 = 0 ¨ Replace x 2 with u
(u - 9)(u - 4) = 0
u-9=0                   u-4=0
u=9                       u=4
x2 = 9                   x 2 = 4 ¨ Replace u with x 2
x = ± 9 = ±3              x = ± 4 = ±2
The solutions are 3, –3, 2, and –2.

443
Chapter 12: Quadratic Functions                                              SSM: Elementary and Intermediate Algebra

37. a 4 - 7a 2 +12 = 0                                                     u- 9 = 0       u+ 2 = 0
2
( )
a 2 - 7a 2 +12 = 0                                                       u =9
2
u = -2
z =9            z 2 = -2 ¨ Replace u with z 2
u 2 - 7u+12 = 0 ¨ Replace a 2 with u
z = ±3           z = ± -2 = ±i 2
( u - 4)( u - 3) = 0                                                The solutions are 3, - 3, i 2, and - i 2 .
u- 4 = 0              or u - 3= 0
u=4                      u =3                                45. -c 4 = 4c 2 - 5
2
a =4                      a 2 = 3 ¨ Replace u with a 2                c 4 + 4c 2 - 5= 0
2
a = ± 4 = ±2             a =± 3
(c )
2
+ 4c 2 - 5= 0

The solutions are 2, –2,          3 , and - 3 .                              u 2 + 4u - 5= 0 ¨ Replace c 2 with u
( u -1)( u + 5) = 0
39. 4x 4 - 5x 2 +1= 0
2                                                                u-1= 0            u + 5= 0
( )
4 x 2 - 5x 2 +1= 0                                                         u =1              u = -5
4u 2 - 5u +1= 0¨ Replace x 2 with u                              c2 = 1            c 2 = -5 ¨ Replace u with c 2
( 4u -1)( u-1) = 0                                                      c = ±1            c = ± -5 = ±i 5
4u -1= 0             or          u-1= 0                               The solutions are 1, -1, i 5, and -i 5 .
u=  1                         u =1
4
x2 = 1                        x 2 = 1¨ Replace u with x 2   47.     x = 2x - 6
4
2x - x - 6 = 0
x =± 1 =± 1                   x = ± 1 = ±1
4     2                                                                  2
( )
2 x1/2             - x 1/2 - 6 = 0
1    1
The solutions are          , - , 1, and –1.                                       2u 2 - u- 6 = 0 ¨ Replace x1/2 with u
2    2
(2u+ 3)( u- 2) = 0
4          2                                                         2u+ 3= 0                 or     u- 2 = 0
41. r - 8r = -15
r 4 - 8r 2 + 15 = 0                                                       u =- 3              or        u =2
†                2                                                                         2
)
( r2† - 8r 2 + 15 = 0                                                    x1/2 = - 3
2
or      x 1/2 = 2 ¨ Replace u with x1/2
u 2 - 8u + 15 = 0 ¨ Replace r2 with u
†
( u - 3)(u - 5) = 0                                                  x = 22 = 4
u-3=0              u -5 = 0                                            x 1/2 = - 3 has no solution since there is no value
2
u=3                  u=5
2                                                                    of x for which x 1/2 = - 3 .
r =3                r 2 = 5 ¨ Replace u with r 2                                               2
The solution is 4.
r=± 3               r=± 5
†
The solutions are            3, - 3, 5, and - 5 .                49. x + x = 6
x+ x -6 =0
2
†          4
43. z - 7z =18   2                                                         (x )
1/2
+ x 1/2 - 6 = 0

†4 - 7z 2 -18 = 0
z                                                                               u 2 + u- 6 = 0 ¨ Replace x 1/2 with u
2                                                                 (u + 3) (u - 2) = 0
†
(z )
2
- 7z 2 -18 = 0
u+ 3= 0 or                    u- 2 = 0
u 2 - 7u-18 = 0 ¨ Replace z 2 with u                                u = -3 or                     u =2
1/2
x = 2 ¨ Replace u with x 1/2
1/2
( u - 9)( u+ 2) = 0                                               x = -3 or

444

†
SSM: Elementary and Intermediate Algebra                                                      Chapter 12: Quadratic Functions

x = 22 = 4
x 1/2 = -3 has no solution since there is no value
of x for which x 1/2 = -3.
The solution is 4.

9x + 3 x = 2
51.
9x + 3 x - 2 = 0
2
( )
9 x1 /2       + 3x 1/2 - 2 = 0
9u2 + 3u - 2 = 0 ¨ Replace x 1/2 with u
(3u -1)(3u + 2) = 0
3u -1 = 0 or 3u + 2 = 0
1                    2
u=                 u= -
3                    3
1/ 2 = 1           1 /2 = - 2 ¨ Replace u with x 1/2
x                 x
3                    3
1
x=
9
1 /2      2                                                               2
x      = - has no solution since there is no value of x for which x1 /2 = - .
3                                                               3
1
The solution is .
9

53.        (x + 3)2 + 2(x + 3) = 24
(x + 3) 2 + 2(x + 3) - 24 = 0
u 2 + 2u - 24 = 0 ¨ Replace x + 3 with u
(u - 4)(u + 6) = 0
u - 4 = 0 or u + 6 = 0
u=4              u = -6
The solutions are 1 and –9.
x +3 = 4         x + 3 = -6 ¨ Replace u with x + 3
x =1             x = -9

55.                           6( a - 2) 2 = -19( a - 2) - 10
6( a - 2) 2 + 19( a - 2) + 10 = 0
6u 2 + 19u + 10 = 0       ¨ Replace a - 2 with u
( 3u + 2)( 2u + 5) = 0
3u + 2= 0        or     2u + 5= 0
u =- 2                  u =-5                                                        4      1
3                       2                                  The solutions are     and - .
a - 2= - 2              a- 2 = - 5    ¨ Replace u with a - 2                            3      2
3                       2
a= 4                    a =-1
3                         2

57.            (x 2 - 1) 2 - (x 2 - 1) - 6 = 0
u 2 - u - 6 = 0 ¨ Replace x 2 - 1 with u
( u + 2)( u - 3) = 0
u +2 = 0                or u - 3 = 0
u = -2                        u=3         445
2 - 1 = -2                     2 -1 = 3 ¨ Replace u with x 2 - 1
x                              x
x 2 = -1                       x2 = 4
x = -1 = ±i                   x = ± 4 = ±2
The solutions are i, –i, 2, and –2.
Chapter 12: Quadratic Functions                                        SSM: Elementary and Intermediate Algebra

u +2 = 0             or u - 3 = 0
u = -2                     u=3
2 - 1 = -2                  2 -1 = 3 ¨ Replace u with x 2 - 1
x                           x
x 2 = -1                    x2 = 4
x = -1 = ±i                x = ± 4 = ±2
The solutions are i, –i, 2, and –2.

59.      2(b+ 2) 2 + 5(b+ 2) - 3= 0
2u 2 + 5u - 3= 0 ¨ Replace b + 2 with u
(u + 3)(2u -1) = 0
u+ 3= 0 or 2u -1= 0
u = -3             u=1
2
b+ 2= -3           b+ 2 = 1     ¨ Replace u with b + 2
2
b= -5              b =- 3
2
3
The solutions are –5 and - .
2

61.      18(x 2 - 5)2 + 27(x 2 - 5) +10 = 0
18u 2 + 27u +10 = 0 ¨ Replace x 2 - 5 with u
( 3u + 2)(6u + 5) = 0
3u + 2 = 0                  or 6u + 5 = 0
2                               5
u=-                              u=-
3                               6
2           2                               5
x - 5 = - ¨ Replace u with x 2 - 5
2
x -5=-
3                               6
2   13                           2   25
x =                               x =
3                               6
13                              25
x=±                              x =±
3                               6
13      3                      5     6
=±          ⋅                    =±       ⋅
3      3                       6 6
39                            5 6
=±                               =±
3                               6
39       39 5 6              5 6
The solutions are           ,-       ,      , and -      .
3        3       6            6

63.             x -2 +10x -1 + 25= 0
2
(x )
-1
( )
+10 x -1 + 25= 0
2
u +10u + 25= 0 ¨ Replace x -1 with u
( u+ 5)( u+ 5) = 0
u+ 5= 0
u = -5
-1                                      The solution is - 1 .
x = -5       ¨ Replace u with x -1                         5
x=-1
5

446
SSM: Elementary and Intermediate Algebra                                        Chapter 12: Quadratic Functions

65.          6b-2 - 5b-1 +1= 0
2
( ) - 5(b ) +1= 0
6 b-1             -1

6u 2 - 5u+1= 0 ¨ Replace b-1 with u
( 2u-1)( 3u-1) = 0
2u-1= 0          or     3u-1= 0
u= 1                    u=1
2                       3                        The solutions are 2 and 3.
b-1 = 1                 b-1 = 1 ¨ Replace u with b-1
2                       3
b= 2                    b= 3

67.                            2b-2 = 7b-1 - 3
2b-2 - 7b-1 + 3= 0
2
( ) - 7( b ) + 3= 0
2 b-1             -1

2u 2 - 7u + 3= 0 ¨ Replace b-1 with u
( 2u -1)( u - 3) = 0
2u-1= 0 or u- 3= 0
u= 1              u=3
2
b-1 = 1          b-1 = 3 ¨ Replace u with b-1            The solutions are 2 and 1 .
2                                                                          3
b= 2              b= 1
3

69.                    x -2 + 9x -1 = 10
x -2 + 9x -1 -10= 0
-1 2
( x ) + 9( x ) -10= 0
-1

u 2 + 9u -10= 0 ¨ Replace x -1 with u
( u+10)( u -1) = 0
u+10 = 0               or u -1= 0
u = -10                  u =1                                                 1
-1                                                      The solutions are -      and 1.
x =1 ¨ Replace u with x -1
-1
x = -10                                                                        10
x =- 1                   x =1
10

71.                             x -2 = 4x -1 +12
x -2 - 4x -1 -12 = 0
-1 2
(x )           ( )
- 4 x -1 -12 = 0

u 2 - 4u-12 = 0 ¨ Replace x -1 with u
( u+ 2)( u - 6) = 0
u+ 2 = 0     or u - 6= 0
u = -2              u=6
-1†                                                      The solutions are - 1 and 1 .
x = 6 ¨ Replace u with x -1
-1
x = -2                                                                        2     6
x =-1               x= 1
2                6
447
Chapter 12: Quadratic Functions                                        SSM: Elementary and Intermediate Algebra

u+ 2 = 0          or u - 6= 0
u = -2                u=6
-1                                                          The solutions are - 1 and 1 .
x = 6 ¨ Replace u with x -1
-1
x = -2                                                                           2     6
x =-1                 x= 1
2                  6

73.              x 2/3 - 3x1/3 = -2
2
(x )
1/3
- 3x1/3 + 2= 0

u 2 - 3u + 2= 0 ¨ Replace x1/3 with u
( u -1)( u - 2) = 0
u-1= 0 or u - 2= 0
u =1                 u=2
1/3
x = 2 ¨ Replace u with x 1/3
1/3
x =1                                                         The solutions are 1 and 8.
x = 13               x = 23
=1                  =8

75.         b 2/3 +11b1/3 + 28= 0
2
( )
b1/3 +11b1/3 + 28= 0

u 2 +11u + 28= 0 ¨ Replace b1/3 with u
( u + 7)( u+ 4) = 0
u+ 7 = 0      or u+ 4 = 0
u = -7    or       u = -4
† b1/3 = -7    or    b1/3 = -4 ¨ Replace u with b1/3
3                  3
b = (-7) or        b= (-4)
= -343             = -64
The solutions are –343 and –64.

77.                        -2a - 5a1/2 + 3 = 0
2
-2 ( )
a1/ 2     - 5a1/2 + 3 = 0
-2u 2 - 5u + 3 = 0 ¨ Replace a1/2 with u
2u 2 + 5u - 3 = 0
( 2u - 1)(u + 3) = 0
2u-1= 0             or u+ 3= 0
u=  1                     u = -3
2
1/2
a =2                       1/2
a = -3 ¨ Replace u with a1/2
2
a= 1 =1
2  ()  4

1                          1
a1/ 2 = -3 has no solution since there is no value of a for which a 2 = -3. The solution is     .
4

448
SSM: Elementary and Intermediate Algebra                                           Chapter 12: Quadratic Functions

79.        c 2/5 - 3c1/5 + 2= 0
2
(c )
1/5
- 3c1/5 + 2= 0

u 2 - 3u + 2= 0 ¨ Replace c1/5 with u
( u - 2)( u-1) = 0
u- 2 = 0         u -1= 0
u =2              u =1
1/5
c =1 ¨ Replace u with c 1/5
1/5
c =2
c = 25            c =15
= 32              =1
The solutions are 32 and 1.

81.      f (x) = x - 5 x + 4 , f (x) = 0
2
( )
0 = x1 /2           - 5x 1/2 + 4
0 = u2 - 5u + 4 ¨ Replace x1 /2 with u
0 = (u - 1)(u - 4)
u - 1 = 0 or u - 4 = 0
u =1             u=4
x 1/2 = 1        x 1/2 = 4 ¨ Replace u with x 1/2
x =1             x = 16
The x-intercepts are (1, 0) and (16, 0).

83.      h( x ) = x +13 x + 36, h ( x ) = 0
2
( )
0 = x 1/2 +13x1/2 + 36

0 = u 2 +13u+ 36 ¨ Replace x 1/2 with u
0 = ( u+ 9)( u+ 4 )
u+ 9 = 0 or u+ 4 = 0
u = -9             u = -4
1/2
x = -9              1/2
x = -4 ¨ Replace u with x 1/2

There are no values of x for which x 1/2 = -9 or x1/2 = -4 . There are no x-intercepts.

85.      p( x) = 4x -2 -19x -1 - 5 , p( x) = 0
2
( )
0 = 4 x -1          - 19x -1 - 5
0 = 4u 2 - 19u - 5 ¨ Replace x -1 with u
0 = (4u +1)(u - 5)
4u +1 = 0      or u - 5 = 0
1
u=-                u=5
4
1
x -1 = -           x -1 = 5 ¨ Replace u with x -1
4                              449
1
x = -4             x=
5
Ê1 ˆ
The x-intercepts are (–4, 0) and Á , 0˜ .
Ë5 ¯
Chapter 12: Quadratic Functions                                              SSM: Elementary and Intermediate Algebra

4u +1 = 0            or u - 5 = 0
1
u=-                    u=5
4
1
x -1 = -                     x -1 = 5 ¨ Replace u with x -1
4
1
x = -4                         x=
5
Ê1 ˆ
The x-intercepts are (–4, 0) and Á , 0˜ .
Ë5 ¯

87.      f ( x ) = x 2/3 + x 1/3 - 6, f ( x ) = 0
2
( )
0 = x1 /3        + x 1/3 - 6
0 = u2 + u - 6 ¨ Replace x1 /3 with u
0 = (u + 3)(u - 2)
u+3=0          or u - 2 = 0
u = -3             u=2
x1/ 3 = -3        x1 /3 = 2 ¨ Replace u with x 1/3
x = -27            x=8
The x-intercepts are (–27, 0) and (8, 0).

89.                               2
(
g( x ) = x 2 - 3x    )          (            )
+ 2 x 2 - 3x - 24 , g( x ) = 0
0=  u2  + 2u - 24 ¨ Replace x 2 - 3x with u
0 = (u + 6)(u - 4)
u + 6 = 0 or           u-4 =0
2 - 3x + 6 = 0         2 - 3x - 4 = 0 ¨ Replace u with x 2 - 3x
x                      x
( x - 4)( x +1) = 0
x - 4 = 0 or x + 1 = 0
x =4         x = -1
There are no x-intercepts for x 2 - 3x + 6 = 0 since b 2 - 4ac = ( -3)2 - 4(1)( 6) = 9 - 24 = -15.
The x-intercepts are (4, 0) or (–1, 0).

91.
f ( x ) = x 4 - 20x + 64, f ( x ) = 0
2
( )
0 = x 2 - 20x 2 + 64

0 = u 2 - 20u + 64 ¨ Replace x 2 with u
0 = ( u-16)( u- 4 )
u-16 = 0              or u- 4 = 0
u =16                       u=4
2
x =16                       x 2 = 4 ¨ Replace u with x 2
x = ± 16 = ±4                x = ± 4 = ±2
The x-intercepts are ( 4, 0), (–4, 0), ( 2, 0), and (–2, 0).
450
SSM: Elementary and Intermediate Algebra                                                          Chapter 12: Quadratic Functions

The x-intercepts are ( 4, 0), (–4, 0), ( 2, 0), and (–2, 0).

93. When solving an equation of the form ax 4 + bx 2 + c = 0 , let u = x 2 .

95. When solving an equation of the form ax -2 +bx -1 + c = 0, let u = x -1 .

97. If the solutions are ±2 and ±4, the factors must be ( x - 2), ( x + 2) , ( x - 4) and ( x + 4) .
0 = ( x - 2 )( x + 2)( x - 4)( x + 4)
(
0 = x 2 - 4 x 2 -16  )(                )
0=     x4      -   20x 2      + 64

(        )(
99. If the solutions are ± 2 and ± 3 , the factors must be x + 2 , x - 2 , x + 3 , x - 3 .)(        )(      )
(               )( )(
0= x + 2 x - 2 x + 3 x - 3                   )(           )
0= ( x         - 2)( x - 3)
2              2

0= x 4 - 5x 2 + 6

101. No. An equation of the form ax 4 + bx 2 + c = 0 can have no imaginary solutions, two imaginary solutions, or
four imaginary solutions.

3 3
103.     a.                                - = 60 The LCD is x 2
x2 x
Ê 3 ˆ         3
x 2 Á 2 ˜ - x 2 Ê ˆ = x 2 (60)
Á ˜
Ëx ¯        Ë x¯
3 - 3x = 60x 2
0 = 60x 2 + 3x - 3
(
0 = 3 20x 2 + x - 1    )
0 = 3( 5x - 1)( 4x +1)
5x -1 = 0 or 4x + 1 = 0
1                    1
x=                  x =-
5                    4
1         1
The solutions are      and -
5         4
3    3
b.                              - = 60
x2 x
3x -2 - 3x -1 = 60
2
†
( )
3 x -1             - 3x -1 - 60 = 0
3u2 - 3u - 60 = 0 ¨ Replace x -1 with u
(              )
3 u2 - u - 20 = 0
3(u - 5)(u + 4) = 0

451
Chapter 12: Quadratic Functions                                   SSM: Elementary and Intermediate Algebra

u-5=0       or u + 4 = 0
u=5              u = -4
-1 = 5           -1 = -4 ¨ Replace u with x -1
x                x
1                  1
x=               x =-
5                  4
1       1
The solutions are and - .
5       4

8
112. 15( r + 2 ) + 22 = -
r +2
8
15( r + 2 )(r + 2) + 22( r + 2) = -      (r + 2 )
r+2
15( r + 2)2   + 22( r + 2) = -8
15(r + 2 )2 + 22(r + 2 ) + 8 = 0
15u 2 + 22u + 8 = 0 ¨ Replace r + 2 with u
(5u + 4)(3u + 2) = 0
5u + 4 = 0       or 3u + 2 = 0
4                     2
u= -                   u= -
5                     3
4                     2
r+2= -                 r + 2 = - ¨ Replace u with r + 2
5                     3
14                    8
r= -                   r=-
5                    3
14         8
The solutions are -       and - .
5         3

107. 4 - ( x - 2 )-1 = 3( x - 2) -2
4 - u -1 = 3u -2 ¨ Replace x - 2 by u
1    3
4- = 2
u u
2Ê      1ˆ        Ê 3ˆ
u Á4 - ˜ = u2Á 2 ˜
Ë     u¯        Ëu ¯
2 -u=3
4u
2 - u -3 = 0
4u
( 4u + 3)(u -1) = 0
4u + 3 = 0        or u - 1 = 0
3
u=-                   u =1
4
3
x -2 = -              x - 2 = 1 ¨ Replace u by x - 2
4
5
x=                    x =3
4
5
The solutions are and 3.
4

452
SSM: Elementary and Intermediate Algebra                                               Chapter 12: Quadratic Functions

109. x 6 - 9x 3 + 8 = 0
2
(x3 )       - 9x 3 + 8 = 0
u2 - 9u + 8 = 0 ¨ Replace x 3 with u
(u - 8)( u - 1) = 0
u - 8 = 0 or u -1 = 0
u=8               u =1
x3 = 8            x 3 = 1 ¨ Replace u with x 3
x =2              x =1
The solutions are 2 and 1.

2
(
111. x 2 + 2x - 2       )      (           )
- 7 x 2 + 2x - 2 + 6 = 0
- 7u + 6 = 0 ¨ Replace x 2 + 2x - 2 with u
u2
(u - 6)(u -1) = 0
u-6=0        or             u - 1= 0
u=6                        u=1
2 + 2x - 2 = 6              2 + 2x - 2 = 1 ¨ Replace u with x 2 + 2x - 2
x                          x
2 + 2x - 8 = 0
x                          x 2 + 2x - 3 = 0
( x + 4)( x - 2 ) = 0       ( x + 3)( x -1 ) = 0
x +4 = 0        or x - 2 = 0       x + 3 = 0 or x - 1= 0
x = -4                x=2           x = -3          x=1
The solutions are –4, 2, –3, and 1.

113. 2n4 - 6n2 - 3 = 0                                       115.
(         )
4- 3-2 =4 - 9 - 8
( )
2                                                        5 4 3    5 12 12
2( )n2       - 6n2 - 3 = 0
2u 2 - 6u - 3 = 0 ¨ Replace n2 with u
= 4- 1
5 12         ( )
48 - 5
6±      (-6)2 - 4( 2)( -3)                                    =
60 60
u=
2( 2)                                              = 43
6 ± 60                                                          60
=
4                                           116.   6( x + 4) - 4 (3x + 3) = 6
6 ± 2 15
=                                                          6x + 24 -12x -12 = 6
4
-6x +12 = 6
3 ± 15
=                                                                       -6x = -6
2
3 ± 15                                                                x =1
n2 =              ¨ Replace u with n2
2                                                  D : all real numbers
117.
3 ± 15
n=   ±                                                     R:       {y   y ≥ 0}
2

453

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