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
                                                           2ad = v2 - v12
                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
                                                                               to about $18.5 billion.
           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
                  The cost is about $117,500.
                                                                               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
               Chris about 11.76 hours and John about
                                                                                    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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