PDF Solutions Manual for College Algebra 12th Edition by Sullivan

INSTRUCTOR’S SOLUTIONS MANUAL

TIM BRITT

Jackson State Community College

C OLLEGE A LGEBRA

TWELFTH EDITION

Michael Sullivan

Chicago State University

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3.4

Table of Contents

Chapter 4 Linear and Quadratic Functions

4.1

4.4

4.5

Chapter 5 Polynomial and Rational Functions

5.1

5

5

5.6

5.7

Chapter 6

6.1

6.5

6.6

6.7

6.8

6.9

Chapter 7 Analytic Geometry

Chapter 8 Systems of Equations and Inequalities

8.1

8.2

8.3

Chapter 9

9.1

Chapter 10 Counting and Probability

Appendix Graphing Utilities

Chapter R Review

Section R.1

1. rational

2. 4563430331 

3. Distributive

4. c

5. a

6. b

7. True

8. False;TheZero-ProductPropertystatesthatifa productequals0,thenatleastoneofthefactors mustequal0.

9. False;6istheGreatestCommonFactorof12 and18.TheLeastCommonMultipleisthe smallestvaluethatbothnumberswilldivide evenly.TheLCMfor12and18is36.

10. True

11.   1,3,4,5,92,4,6,7,8 1,2,3,4,5,6,7,8,9 AB

12.   1,3,4,5,91,3,4,6 1,3,4,5,6,9

13.  1,3,4,5,92,4,6,7,84 AB

14.  1,3,4,5,91,3,4,61,3,4 AC

15.

ABC 

16.

    ABC

     ()

1,3,4,5,92,4,6,7,81,3,4,6 41,3,4,6 1,3,4,6

17.  0,2,6,7,8 A 

18.  0,2,5,7,8,9 C 

19.

AB 

   1,3,4,5,92,4,6,7,8 40,1,2,3,5,6,7,8,9

BC 

20.    2,4,6,7,81,3,4,6 1,2,3,4,6,7,80,5,9

AB 

21.   0,2,6,7,80,1,3,5,9 0,1,2,3,5,6,7,8,9

BC 

22.   0,1,3,5,90,2,5,7,8,9 0,5,9

23. a.  2,5

b.  6,2,5

c.  1 6,,1.333...,2,5 2

d.  

e.  1 6,,1.333...,,2,5 2 



 

  ()

1,3,4,5,92,4,6,7,81,3,4,6 1,2,3,4,5,6,7,8,91,3,4,6 1,3,4,6

  

24. a.  1

b.  0,1

c.  5,2.060606...2.06,1.25,0,1 3 

d.  5

e.  5,2.060606...2.06,1.25,0,1,5 3 

25. a.  1

b.  0,1

c.  1110,1,,,234

d. None

e.  1110,1,,,234

26. a. None

b.  1

c.  1.3,1.2,1.1,1

d. None

e.  1.3,1.2,1.1,1

27. a. None

b. None

c. None

d.  2,,21,1 2 

e.  2,,21,1 2 

28. a. None

b. None

c.  110.3 2 

d.  2,2   e.  1 2,2,10.3 2  

29. a. 18.953 b. 18.952

30. a. 25.861 b. 25.861

31. a. 28.653 b. 28.653

32. a. 99.052 b. 99.052

33. a. 0.063 b. 0.062

34. a. 0.054 b. 0.053

35. a. 9.999 b. 9.998

36. a. 1.001 b. 1.000

37. a. 0.429 b. 0.428

38. a. 0.556 b. 0.555

39. a. 34.733 b. 34.733

40. a. 16.200 b. 16.200

41. 325 

42. 5210 

43. 234 x 

44. 322 y 

45. 312 y 

46. 246 x 

47. 26 x 

48. 26 y 

49. 6 2 x 

50. 26 x 

51. 942527 

52. 643235 

53. 6436126 

54. 842880 

55. 185218108 

56. 1001021002080 

57. 4112113 333  

58. 21413 222 

21100374253

73. 3215823 452020

74. 418311 3266

75. 74493281 875656

76. 81516135151 921818

77. 5110313 18123636

2864046 1594545

79. 5825643913 241512012040

80. 32945 14214242

81. 32981 20156060

82. 6312153 35147070

83. 5 1852759359 1118119211 27

5 2153557557 2212732

85. 141741721 1 3721212121

86. 24122222222 35635323532315 25210210212 351515151515 43434 53535

87. 3323363623 2 4814848428 12312315 8888

88. 51351351351 3 62162322322 51514 2 2222

64624 xx

42184 xx

2 44 xxxx

2 43412 xxxx

2121323 333 3636332 3231 2 3322 x

2 2 24428 68 xxxxx

2 2 5155 65 xxxxx

xxxxx

2 2 927271863 21163

98.  2 2 3153155 3145   xxxxx xx

99.  2 2 822816 1016 xxxxx xx  

100.  2 2 42248 68 xxxxx xx  

101. 2 22 3(5)360 315360 1560 4     xxkxx xxkxx xkx k

102. 2 222 222 222 222 ()(3)412 33412 (3)3412 (3)3412 (2)3412 24 2 

 

    xkxkxx xkxkxkxx xxkkkxx xxkkkxx xxkkxx k k

103.   2323 23 5 5 xxxx x x x  

104. 23421214  sincemultiplicationcomesbeforeadditioninthe orderofoperationsforrealnumbers.

2345420  sinceoperationsinsideparenthesescomebefore multiplicationintheorderofoperationsforreal numbers.

105.  23421224 

23246848 

106. 4371 257    ,but 43453220626132.6 251010105  

107. Subtractionisnotcommutative;for example:231132 

108. Subtractionisnotassociative;for example:  52124521  .

109. Divisionisnotcommutative;forexample: 23 32  .

110. Divisionisnotassociative;for example:  1222623  ,but

122212112

111. TheSymmetricPropertyimpliesthatif2= x, then x =2.

112. Fromthe principleofsubstitution, if5 x  ,then

113. Therearenorealnumbersthatarebothrational andirrational,sinceanirrationalnumber,by definition,isanumberthatcannotbeexpressed astheratiooftwointegers;thatis,notarational number

Everyrealnumberiseitherarationalnumberor anirrationalnumber,sincethedecimalformofa realnumbereitherinvolvesaninfinitely repeatingpatternofdigitsoraninfinite,nonrepeatingstringofdigits.

114. Thesumofanirrationalnumberandarational numbermustbeirrational.Otherwise,the irrationalnumberwouldthenbethedifferenceof tworationalnumbers,andthereforewouldhave toberational.

115. Answerswillvary.

116. Since1day=24hours,wecompute 12997541.5416 24 

Nowweonlyneedtoconsiderthedecimalpart oftheanswerintermsofa24hourday.Thatis, 

0.54162413  hours.Soitmustbe13hours laterthan12noon,whichmakesthetime1a.m. CST.

117. Answerswillvary.

Section R.2

1. variable

2. origin

3. strict

4. base;exponent(orpower)

5. 31.234567810 

6. d

7. a

8. b

9. True

10. False;theabsolutevalueofarealnumberis nonnegative.00  whichisnotapositive number.

11. False;anumberinscientificnotationis expressedastheproductofanumber,x, 110 x  or101 x  ,andapowerof10.

12. True 13.

15. 10 2  16. 56  17. 12 18. 35 2  19. 3.14  20. 21.41 

21. 10.5 2 

22. 10.33 3 

23. 20.67 3 

24. 10.25 4 

25. 0 x 

26. 0 z 

27. 2 x 

28. 5 y 

29. 1 x 

30. 2 x 

31. Graphonthenumberline:2 x 

32. Graphonthenumberline:4 x 

33. Graphonthenumberline:1 x 

34. Graphonthenumberline:7 x 

35. (,)(0,1)1011dCDd

36. (,)(0,3)3033dCAd

37. (,)(1,3)3122dDEd

38. (,)(0,3)3033dCEd

39. (,)(3,3)3(3)66dAEd

40. (,)(1,1)1122dDBd

41. 2223264xy

42. 33(2)3633 xy

43. 525(2)(3)230228 xy 

44. 22(2)(2)(3)462 xxy 

45. 2(2)4 24 2355 x xy 

46. 2311 2355 xy xy   

47. 323(2)2(3)6600 22355 xy y

48. 2(2)343 237 333 x y 

49. 3(2)11 xy

50. 3(2)55 xy

51. 32325 xy

52. 32321 xy

53. 331 33 x x 

54. 221 22 y y 

55. 454(3)5(2) 1210 22 22 xy   

56. 323(3)2(2)9455 xy

57. 454(3)5(2) 1210 1210 2 2 xy

58. 323322 3322 94 13 xy

59. 21 x x

Part(c)mustbeexcluded.Thevalue0 x  must beexcludedfromthedomainbecauseitcauses divisionby0.

60. 21 x x 

Part(c)mustbeexcluded.Thevalue0 x  must beexcludedfromthedomainbecauseitcauses divisionby0.

61. 2(3)(3) 9 xx xxx  

Part(a),3 x  ,mustbeexcludedbecauseit causesthedenominatortobe0.

62. 29 x x 

Noneofthegivenvaluesareexcluded.The domainisallrealnumbers.

63. 2 21 x x 

Noneofthegivenvaluesareexcluded.The domainisallrealnumbers.

64. 33 2(1)(1) 1 xx xxx  

Parts(b)and(d)mustbeexcluded.Thevalues 1,and1xx mustbeexcludedfromthe domainbecausetheycausedivisionby0.

65. 22 3 510510 (1)(1) xxxx xxxxx   

Parts(b),(c),and(d)mustbeexcluded.The values0,1,and1 xxx mustbeexcluded fromthedomainbecausetheycausedivisionby 0.

66. 22 32 9191 (1) xxxx xxxx   

Part(c)mustbeexcluded.Thevalue0 x  must

beexcludedfromthedomainbecauseitcauses divisionby0.

67. 4 5 x 5 x  mustbeexludedbecauseitmakesthe denominatorequal0.

Domain5 xx 

68. 6 4 x  4 x  mustbeexcludedsineitmakesthe denominatorequal0.

 Domain4 xx 

69. 4 x x  4 x  mustbeexcludedsineitmakesthe denominatorequal0.

 Domain4 xx 

70. 2 6 x x 6 x  mustbeexcludedsineitmakesthe denominatorequal0.

Domain6 xx 

71. 555 (32)(3232)(0)0C 999 CF

72. 555 (32)(21232)(180)100C 999 CF

73. 555 (32)(7732)(45)25C 999 CF

74. 55(32)(432)99 5(36) 9 20C CF  

75. 2 (9)(9)(9)81 

76. 22 4(4)16

77. 2 2 411 416 

78. 2 2 411 416

79. 64642 2 333311 39

80. 23231 44444

90.

313 13 2228

2224

33311333 3 y xyxyxy x

91. 25 235411 34  xyy xyxy xyx

92. 2 211231 23 1 xy xyxy xyxy 

93. 253533 37272 315732 221 2 2 (4)()16 (3)27 16 27 16 27 16 27

yxzyxz xyzxyz xyz xyz xz y 94. 21211 344 24111 621 62 4()4 28 4 8 1 2 1 2 xyzxyz xyxy xyz xyz xyz 

  95. 22 2 33266 13222 3 2339 2

122 2 24 1 x xy y

131 333 22 y xy x

222221415 xy

222221414 xy

 2 222124 xy

 2 222111 xy 103. 222xx 104.

2 2 xx 105.

222221415 xy 106. 2221213 xyxy 107. 211 2 y x 

113. 6 (8.2)304,006.671 

114. 5 (3.7)693.440  115. 3 (6.1)0.004 

116. 5 (2.2)0.019  117. 6 (2.8)481.890 

118. 6 (2.8)481.890

119. 4 (8.11)0.000

120. 4 (8.11)0.000

121. 2 454.24.54210 

122. 132.143.21410 

123. 0.0131.3102  124. 0.004214.21103 

125. 432,1553.215510 

126. 4 21,2102.12110 

127. 0.0004234.23104 

128. 0.05145.14102 

129. 4 6.151061,500 

130. 3 9.7109700 

131. 3 1.214100.001214 

132. 4 9.88100.000988 

133. 8 1.110110,000,000 

134. 2 4.11210411.2 

135. 2 8.1100.081 

136. 1 6.453100.6453 

137. Alw 

138.  2 Plw 

139. Cd  

140. 1 2 Abh 

141. 32 4 Ax 

142. 3 Px 

143. 43 3 Vr  

144. 42Sr  

145. 3 Vx 

146. 62Sx 

147. a. If1000, x  40002 40002(1000) 40002000 $6000 Cx  



Thecostofproducing1000watchesis $6000.

b. If2000, x  40002 40002(2000) 40004000 $8000 Cx

Thecostofproducing2000watchesis $8000.

148. 210801202560325$98  Hisbalanceattheendofthemonthwas$98.

149. Wewantthedifferencebetween x and4tobeat least6units.Sincewedon’tcarewhetherthe valuefor x islargerorsmallerthan4,wetake theabsolutevalueofthedifference.Wewantthe inequalitytobenon-strictsincewearedealing withan‘atleast’situation.Thus,wehave 46 x 

150. Wewantthedifferencebetween x and2tobe morethan5units.Sincewedon’tcarewhether thevaluefor x islargerorsmallerthan2,we taketheabsolutevalueofthedifference.We wanttheinequalitytobestrictsinceweare dealingwitha‘morethan’situation.Thus,we have 25 x 

151. a. 110108110225 x  108voltsisacceptable.

b. 110104110665 x  104voltsis not acceptable.

152. a. 220214220668 x  214voltsisacceptable.

b. 22020922011118 x  209voltsis not acceptable.

153. a. 32.9993 0.001 0.0010.01 x    Aradiusof2.999centimetersisacceptable.

b. 32.893 0.11 0.110.01 x     Aradiusof2.89centimetersis not acceptable.

154. a. 98.69798.6 1.6 1.61.5 x    97˚Fisunhealthy.

b. 98.610098.6 1.4 1.41.5 x    100˚Fis not unhealthy.

155. ThedistancefromEarthtotheMoonisabout 8 410400,000,000  meters.

156. TheheightofMt.Everestisabout 388488.84810  meters.

157. Thewavelengthofvisiblelightisabout 7 5100.0000005  meters.

158. Thediameterofanatomisabout 10 1100.0000000001  meters.

159. Thediameterisabout2 0.04034.0310  inches.

160. Thetiniestmotorislessthan5 0.00004410  millimeterstall.

161.  5112 2 1.86106102.4103.6510

186,000606024365   1012 586.5696105.86569610  Thereareabout12 5.910  milesinonelightyear.

162. 7 2 5 93,000,0009.310510 186,0001.8610 500seconds8min.20sec.   

Ittakesabout8minutes20secondsforabeam oflighttoreachEarthfromtheSun.

163. 10.333333...0.333 3  1 3islargerbyapproximately0.0003333...

164. 2 30.666666...0.666  2 3islargerbyapproximately0.000666...

165.

613613 1920 5.24106.5105.246.51010 34.06103.40610

166. 44 6 1010 5 1.62101.62100.3610 4.5 4.51010

167. No.Foranypositivenumber a,thevalue2 a is smallerandthereforecloserto0.

168. Wearegiventhat2110 x  .Thisimpliesthat 110 x  .Since103.162 x  and 3.142 x   ,thenumbercouldbe3.15or3.16 (whicharebetween1and10asrequired).The numbercouldalsobe3.14sincenumberssuchas 3.146whichliebetween  and10would equal3.14whentruncatedtotwodecimalplaces.

169. Answerswillvary.

170. Answerswillvary. 5<8isatruestatementbecause5isfurtherto theleftthan8onarealnumberline.

Section R.3

1. right;hypotenuse

2. 1 2 Abh  3. 2 Cr  

4. similar

5. c 6. b

7. True.

8. True.222 68366410010 

9. False;thesurfaceareaofasphereofradius r is givenby2 4 Vr  

10. True.Thelengthsofthecorrespondingsidesare equal.

11. True.Twocorrespondinganglesareequal.

12. False.Thesidesarenotproportional.

13. 222 22 5,12, 512 25144 16913 ab cab c 









14. 222 22 6,8, 68 3664 10010 ab cab c 









15. 222 22 10,24, 1024 100576 67626 ab cab c 









16. 222 22 4,3, 43 169 255 ab cab c  







17. 222 22 7,24, 724 49576 62525 ab cab c 

18. 222 22 14,48, 1448 1962304 250050 ab cab c 

19. 222 534 25916 2525

Thegiventriangleisarighttriangle.The hypotenuseis5.

20. 222 1068 1003664 100100  

Thegiventriangleisarighttriangle.The hypotenuseis10.

21. 222 645 361625 3641false  

Thegiventriangleisnotarighttriangle.

22. 222 322 944 98false 

Thegiventriangleisnotarighttriangle.

23. 222 25724 62549576 625625 

Thegiventriangleisarighttriangle.The hypotenuseis25.

24. 222 261024 676100576 676676   

Thegiventriangleisarighttriangle.The hypotenuseis26.

25. 222 634 36916 3625false   

Thegiventriangleisnotarighttriangle.

26. 222 754 492516 4941false    Thegiventriangleisnotarighttriangle.

27. 67422 in Alw

28. 94362 cm Alw

29. 112 22(14)(4)28in

Abh

30. 112 22(4)(9)18cmAbh

31. 222 (5)25m 22(5)10m Ar Cr 



32. 222 (2)4ft 22(2)4ft Ar Cr





33. 6852403 ft Vlwh



 Slwlhwh

 

 2 222 268265285 966080

236ft

34. 9482883 in Vlwh

Slwlhwh    

 2 222 294298248 7214464 280in

35. 333 222 445005cm 333 445100cm

 Vr Sr



36. 333 222 44336f 33 44336ft



Vrt Sr



37. 223 (9)(8)648in Vrh

2 2 2 22 29298 162144 306in

38. 223 (8)(9)576in Vrh

2 2 2 22 28289 128144 272in

39. Thediameterofthecircleis2,soitsradiusis1. 22(1)squareunits Ar

40. Thediameterofthecircleis2,soitsradiusis1. 22 2(1)4squareunits A 

41. Thediameterofthecircleisthelengthofthe diagonalofthesquare.

222 22 44 8 822 222 22 d d d r 

Theareaofthecircleis:

2222squareunits Ar

42. Thediameterofthecircleisthelengthofthe diagonalofthesquare.

Theareais:

22 2224squareunits A

Section R.3: Geometry Essentials

43. Sincethetrianglesaresimilar,thelengthsof correspondingsidesareproportional.Therefore, weget 8 42 82 4 4 x x x 

Inaddition,correspondinganglesmusthavethe sameanglemeasure.Therefore,wehave 90 A  ,60 B  ,and30 C 

44. Sincethetrianglesaresimilar,thelengthsof correspondingsidesareproportional.Therefore, weget 6 1216 616 12 8 x x x    Inaddition,correspondinganglesmusthavethe sameanglemeasure.Therefore,wehave 30 A

,75 B  ,and75 C 

45. Sincethetrianglesaresimilar,thelengthsof correspondingsidesareproportional.Therefore, weget 30 2045 3045 20 135 or67.5 2 x x xx 

Inaddition,correspondinganglesmusthavethe sameanglemeasure.Therefore,wehave 60 A

46. Sincethetrianglesaresimilar,thelengthsof correspondingsidesareproportional.Therefore, weget 8 1050 850 10 40 x x x 

Inaddition,correspondinganglesmusthavethe sameanglemeasure.Therefore,wehave 50 A  ,125 B  ,and5 C  .

47. Thetotaldistancetraveledis4timesthe circumferenceofthewheel. TotalDistance44()416 64201.1inches16.8feet

48. Thedistancetraveledinonerevolutionisthe circumferenceofthedisk4  Thenumberofrevolutions= dist.traveled2051.6revolutions circumference4

49. Areaoftheborder=areaofEFGH–areaof ABCD222 1061003664ft

50. FG=4feet;BG=4feetandBC=10feet,so CG=6feet.TheareaofthetriangleCGFis: 12 2(4)(6)12ft

51. Areaofthewindow=areaoftherectangle+ areaofthesemicircle. 122 (6)(4)224230.28ft 2

Perimeterofthewindow=2heights+width+ one-halfthecircumference. 1 2(6)4(4)1242 2 16222.28feet

52. Areaofthedeck=areaofthepoolanddeck–areaofthepool. 22 22 (13)(10)169100 69ft216.77ft

Theamountoffenceisthecircumferenceofthe circlewithradius13feet. 2(13)26ft81.68ft

53. WecanformsimilartrianglesusingtheGreat Pyramid’sheight/shadowandThales’ height/shadow: h 126114 240 {{ 2 3

Thisallowsustowrite 2 2403 2240 160

54. Let x =theapproximatedistancefromSanJuan toHamiltonand y =theapproximatedistance fromHamiltontoFortLauderdale.Usingsimilar triangles,weget 1046

5853.5 104653.5 58 964.8 x x x 

1046 5857 104657 58 1028.0 y y y

TheapproximatedistancebetweenSanJuanand Hamiltonis965milesandtheapproximate distancebetweenHamiltonandFortLauderdale is1028miles.

55. Convert20feettomiles,andsolvethe PythagoreanTheoremtofindthedistance:

201milefeet20feet0.003788miles 5280feet (39600.003788)396030 5.477miles

222sq.miles

56. Convert6feettomiles,andsolvethe PythagoreanTheoremtofindthedistance:

61mile feet6feet0.001136miles 5280feet (39600.001136)39609 3miles

222sq.miles

57. Convert100feettomiles,andsolvethe PythagoreanTheoremtofindthedistance: 1001milefeet100feet0.018939miles 5280feet 

222sq.miles (39600.018939)3960150 12.2miles

Convert150feettomiles,andsolvethe PythagoreanTheoremtofindthedistance: 1501milefeet150feet0.028409miles 5280feet 

222sq.miles (39600.028409)3960225 15.0miles

58. Given0,0andmnmn  , if2222 ,2and amnbmncmn  ,then

and  2 2224224 2 cmnmmnn  222,and abcabc representthesides ofarighttriangle.

Ifyoudoubletheradiusthevolumeis8times theoriginalvolume.

63. Let l= lengthoftherectangle and w =widthoftherectangle. Noticethat 22 ()() [()()][()()] (2)(2)44 lwlw lwlwlwlw lwlwA

Since2()0 lw ,thelargestareawilloccur when l–w =0or l=w;thatis,whenthe rectangleisasquare.But 1000222() 5002 250 lwlw lwl lw

Vrh So,

59. 2 2 3 (10)(4.5) 450ft

3 33 1ft7.48052galso 450ft7.48052gal/ft10,575gal

60. 3 2 2 10000(5.61458)56145.8ft 56145.8(25) 56145.828.6ft 625

61. 2 2 2 2 2 (2) 4 44

Ar Ar r rA

Ifyoudoubletheradius,theareaisfourtimes theoriginalarea.

Thelargestpossibleareais225062500  sqft. Acircularpoolwithcircumference=1000feet yieldstheequation:500 21000rr  

Theareaenclosedbythecircularpoolis: 22 22 50050079577.47ft Ar



Thus,acircularpoolwillenclosethemostarea.

64. Considerthediagramshowingthelighthouseat pointL,relativetothecenterofEarth,usingthe radiusofEarthas3960miles.LetPrefertothe furthestpointonthehorizonfromwhichthe lightisvisible.Notealsothat 362362feetmiles. 5280 

ApplythePythagoreanTheoremto CPL :

2 22 1 39603960362 5280 d 

2 22 1 22 1 362 5280 362 5280 39603960 3960396023.30mi. d d  

Therefore,thelightfromthelighthousecanbe seenatpointPonthehorizon,wherepointPis approximately23.30milesawayfromthe lighthouse.Brochureinformationisslightly overstated.

Verifytheshipinformation:

LetSrefertotheship’slocation,andlet x equal theheight,infeet,oftheship.

Weneed1240 dd

Since123.30miles d  weneed 24023.30=16.70miles. d 

ApplythePythagoreanTheoremto CPS :

396016.73960

222 22 22

396016.73960

396016.73960

Theshipwouldhavetobeatleast186feettallto seethelighthousefrom40milesaway.

Verifytheairplaneinformation:

LetArefertotheairplane’slocation.The distancefromtheplanetopointPis2 d Wewanttoshowthat12120 dd Assumethealtitudeoftheairplaneis 10,000feet=10000miles. 5280

ApplythePythagoreanTheoremto CPA :

2 22 2 3960396010000 5280 d

2 22 2 2 2 2 10000 39603960 5280 10000 39603960 5280 122.49miles. d d

    61Therefo re,1223.30122.49145.79120. dd

Thebrochureinformationisslightlyunderstated. Notethataplaneatanaltitudeof6233feet couldseethelighthousefrom120milesaway.

Section R.4

False;monomialscannothavenegativedegrees.

False;thedividend=(quotient)(divisor)+ remainder

9. 23 x Monomial;Variable: x ; Coefficient:2;Degree:3

10. 42 x Monomial;Variable: x ;Coefficient: –4;Degree:2

11. 81 8 x x  Notamonomial;whenwrittenin theform k ax ,thevariablehasanegative exponent.

12. 23 x Notamonomial;whenwritteninthe form k ax ,thevariablehasanegativeexponent.

13. 22 xy Monomial;Variables:,xy ; Coefficient:–2;Degree:3

14. 523 xy Monomial;Variables:,xy ; Coefficient:5;Degree:5

15. 81 8 x xy y  Notamonomial;whenwritten intheform nm axy ,theexponentonthevariable y isnegative.

16. 2 23 3 22 x xy y  Notamonomial;when writtenintheform nm axy ,theexponentonthe variable y isnegative.

17. 22 xy  Notamonomial;theexpression containsmorethanoneterm.Thisexpressionis abinomial.

18. 2 34 x  Notamonomial;theexpression containsmorethanoneterm.Thisexpressionis abinomial.

19. 2 35 x Polynomial;Degree:2

20. 14 x Polynomial;Degree:1

21. 5Polynomial;Degree:0

22. –π Polynomial;Degree:0

23. 325 x x Notapolynomial;thevariableinthe denominatorresultsinanexponentthatisnota nonnegativeinteger.

Section R.4: Polynomials

24. 32 x  Notapolynomial;thevariableinthe denominatorresultsinanexponentthatisnota nonnegativeinteger.

25. 3 22 y Polynomial;Degree:3

26. 102zz  Polynomial;Degree:2

27. 2 3 5 1 x x  Notapolynomial;thepolynomialin thedenominatorhasadegreegreaterthan0.

28. 3 2 321 1 xx xx

  Notapolynomial;the polynomialinthedenominatorhasadegree greaterthan0.

29. 22 22 2 (68)(347) (3)(64)(87) 4215

 xxxx xxxx xx







30. 322 322 32 (32)(44) (3)(4)(24) 446 xxxx xxxx xxx





31. 322 322 (2510)(243) 2510243 xxxxx xxxxx 

 322 32 (22)(54)(103) 497 xxxxx xxx  

32. 232 232 (34)(35) 3435 xxxxx xxxxx   322 32 (3)(3)(45) 449 xxxxx xxx 



33.  53432 542 653 653 xxxxxx xxxx 



34.  5232 532 108326 103106 xxxx xxx 



35. 22 22 2 (64)3(25) 646315 7311

 xxxx xxxx xx





Chapter R: Review

36. 22 22 2 2(1)(52) 22252 73 xxxx xxxx xx

37. 3232 3232 32 6(3)4(23) 6618812 21818 xxxx xxxx xx

38. 323 323 8(431)6(482) 32248244812 xxxx xxxx

32 824484 xxx 

39.

40.

222 222 22351 22351 xxxxx xxxxx

2 246 xx 

222 222 1452 1452 xxxx xxxx 

2 26 xx

41.  22 22 2 75343 73521124 11359    yyy yyy yy

42.  323 323 8141 884444 yyyy yyyy 

 32 44412 yyy 

43. 22432 (25)25  xxxxxx

44. 23532 4(2)448 xxxxxx 

45. 2352 2(45)810 xxxx

46. 343 5(34)1520 xxxx 

47. 2 22 (1)(24) (24)1(24) xxx xxxxx   322 32 2424 324 xxxxx xxx  

48. 2 22 (23)(1) 2(1)3(1) xxx xxxxx   322 32 222333 23 xxxxx xxx  

49. 2 2 (2)(4)428 68 xxxxx xx  

50. 2 2 (3)(5)5315 815 xxxxx xx  

51. 2 2 (27)(5)271035 21735   xxxxx xx

52. 2 2 (31)(21)6321 651 xxxxx xx 



53. 2 2 (4)(2)248 28 xxxxx xx  

54. 2 2 (4)(2)248 28 xxxxx xx  

55. 2 2 (6)(3)6318 918   xxxxx xx

56. 2 2 (5)(1)55 65 xxxxx xx  

57. 2 2 (23)(2)2436 26 xxxxx xx  

58. 2 2 (24)(31)62124 6104 xxxxx xx  

59. 2 2 (34)(2)3468 3108   xxxxx xx

60. 2 2 (31)(1)331 341 xxxxx xx  

61. 2 2 (5)(27)210735 21735   xxxxx xx

62. 2 2 (23)(3)6293 239 xxxxx xx  

63. 22 22 (2)()22 2 xyxyxxyxyy xxyy 

64. 22 22 (23)()2233 23 xyxyxxyxyy xxyy

65. 22 22 (23)(32)6496 6136 xyxyxxyxyy xxyy  

66. 22 22 (3)(2)263 273 xyxyxxyxyy xxyy 

67. 222 (7)(7)749 xxxx 

68. 222 (1)(1)11 xxxx 

69. 222 (23)(23)(2)349 xxxx 

70. 222 (32)(32)(3)294 xxxx 

71. 2222 (4)244816 xxxxx 

72. 2222 (5)2551025 xxxxx 

73. 2222 (4)244816 xxxxx 

74. 2222 (5)2551025 xxxxx 

75. 222 (34)(34)(3)4916 xxxx 

76. 222 (53)(53)(5)3259 xxxx 

77. 222 2 (23)(2)2(2)(3)3 4129 xxx xx  

78. 222 2 (34)(3)2(3)(4)4 92416 xxx xx  

79.  2 222 ()()() xyxyxyxy 

80.  2 222(3)(3)()39 xyxyxyxy 

81.  2 222(3)(3)(3)9 xyxyxyxy 

82.  2 222(34)(34)(3)4916 xyxyxyxy 

83. 222()2 xyxxyy 

84. 222()2 xyxxyy 

85.    222 22 (2)222 44 xyxxyy xxyy  

86.  222 22 (23)22233 4129 xyxxyy xxyy  

87. 33223 32 (2)32322 6128 xxxx xxx 

88. 33223 32 (1)31311 331 xxxx xxx  

89. 33223 32 (21)(2)3(2)(1)3(2)11 81261 xxxx xxx 

90. 33223 32 (32)(3)3(3)(2)3(3)22 2754368 xxxx xxx 

91. 2 32 32 2 2 41123 2431 48 11 1122 231 2346 45 xx xxxx xx xx xx x x 

2 322 32 Check: (2)(41123)(45) 411238224645 431 xxx xxxxx xxx    Thequotientis241123 xx ;theremainder is–45.

Chapter R: Review

92. 2 32 32 2 2 3715 232 36 7 714 152 1530 32 xx xxxx xx xx xx x x      2 322 32

Check:

(2)(3715)(32) 37156143032 32 xxx xxxxx xxx  

Thequotientis23715 xx ;theremainderis –32.

93. 232 3 2 2 43 431 4 31 3 1 x xxxx x xx x x   

Check: ()(43)(1)431 xxxxxx  Thequotientis43 x ;theremainderis1 x 

232

94. 232 3 2 2 31 32 3 2 2 x xxxx x xx x x 

232 Check: ()(31)(2)32 xxxxxx 

Thequotientis31 x ;theremainderis2 x

95. 2 2432 42 2 2 513 25031 510 131 1326 27 x xxxxx xx xx x x  

 

Check:



  22 422 42 251327 510132627 531 xxx xxxx xxx 



Thequotientis2513 x ;theremainderis 27 x 

96. 2 2432 42 2 2 511 2502 510 112 1122 20 x xxxxx xx xx x x 

Check:   22 422 42 251120 510112220 52 xxx xxxx xxx  



Thequotientis2511 x ;theremainderis 20 x 

97. 2 35432 52 2 2 2140031 42 1 x xxxxxx xx xx 

Check:  322 52252 2121 421431 xxxx xxxxxxx 



Thequotientis2 2 x ;theremainderis 21xx 

Thequotientis221

22 432322

42 Check: (1)(1)(22) 122 1 xxxxx xxxxxxxx x xx 

. 105. 22 323 32 2 22 23 23 00 0 xaxa xaxxxa xax ax axax axa axa

22 322223 33 Check: ()()0 xaxaxa xaxaxaxaxa xa

Thequotientis22 xaxa  ;theremainderis0.

106. 432234 54325 54 4 423 0000 xaxaxaxa xaxxxxxa xax ax axax

23 2332 32 324 45 45 0 ax axax ax axax axa axa

Thequotientis432234 xaxaxaxa  ;the remainderis0.

108. Theproducts()()  xyxy and()()  zwzw willeachresultinabinomialthatisthe differenceofsquares.Theproductofthose resultingbinomialswillhave4terms.

109. Whenwemultiplypolynomials   1 px and  2 px ,eachtermof  1 px willbemultiplied byeachtermof  2 px .Sowhenthehighestpoweredtermof   1 px multipliesbythehighest poweredtermof  2 px ,theexponentsonthe variablesinthosetermswilladdaccordingtothe basicrulesofexponents.Therefore,thehighest poweredtermoftheproductpolynomialwill havedegreeequaltothesumofthedegreesof   1 px and  2 px .

110. Whenweaddtwopolynomials   1 px and  2 px ,wherethedegreeof  1 px  thedegree of  2 px ,eachtermof  1 px willbeaddedto eachtermof   2 px .Sinceonlythetermswith equaldegreeswillcombineviaaddition,the degreeofthesumpolynomialwillbethedegree ofthehighestpoweredtermoverall,thatis,the degreeofthepolynomialthathadthehigher degree.

111. Whenweaddtwopolynomials  1 px and  2 px ,wherethedegreeof  1 px =thedegree of  2 px ,thenewpolynomialwillhavedegree  thedegreeof   1 px and  2 px .

112. Answerswillvary.

113. Answerswillvary.

Section R.5

1.  322 xxx

2. prime

3. c 4. b

5. d

6. c

7. True;24 x  isprimeoverthesetofreal numbers.

8. False;  322 3264322 xxxxx 

9. 363(2) xx

10. 7147(2) xx

11. 22(1)axaax

12. (1)axaax

13. 322(1)xxxxxx 

14. 322(1)xxxxxx 

15. 2 222(1) xxxx 16. 2 333(1) xxxx

17. 22 36123(24) xyxyxyxyxy 

18. 2232 60487212(546) xyxyxyxyxyx 

19. 22211(1)(1)xxxx 

20. 22242(2)(2)xxxx 

21. 222 41(2)1(21)(21) xxxx 

22. 222 91(3)1(31)(31) xxxx 

23. 222164(4)(4)xxxx 

24. 222255(5)(5)xxxx 

25. 2 254(52)(52) xxx 

26. 22 3699419(21)(21) xxxx 

27. 22 21(1)xxx

28. 22 44(2)xxx

29. 22 44(2)xxx

30. 22 21(1)xxx

31. 22 1025(5)xxx

32. 22 1025(5)xxx

33. 22 441(21) xxx

34. 22 961(31) xxx

35. 22 1681(41) xxx

36. 22 25101(51) xxx

37. 3332273(3)(39)xxxxx 

38. 33321255(5)(525)xxxxx 

39. 3332273(3)(39)xxxxx 

40.   333 2 2 2783(2) (32)(964) 23469 xx xxx xxx 

 

41. 333 2 827(2)3 (23)(469) xx xxx  

42.   333 2 2 64274(3) (43)(16129) 3491216 xx xxx xxx







43. 256(2)(3) xxxx 

44. 268(2)(4) xxxx 

45. 276(6)(1) xxxx 

46. 298(8)(1) xxxx 

47. 2710(2)(5) xxxx 

48. 21110(10)(1) xxxx 

49. 21016(2)(8) xxxx 

50. 21716(16)(1) xxxx 

51. 278(1)(8) xxxx 

52. 228(2)(4) xxxx 

53. 278(8)(1) xxxx 

54. 228(4)(2) xxxx 

55. 2 24362(2)3(2) (2)(23) xxxxxx xx  

56. 2 33223(1)2(1) (1)(32) xxxxxx xx  

57. 2 51535(3)1(3) (3)(51)   xxxxxx xx

58. 2 3623(2)1(2) (2)(31) xxxxxx xx  

Section R.5: Factoring Polynomials

59. 2 6218283(27)4(27) (27)(34)   xxxxxx xx

2 9632332132 3231 xxxxxx xx  

60.  

61. 2 341(31)(1) xxxx 

62. 2 231(21)(1) xxxx 

63. 2 297(27)(1)  zzzz

64. 2 651(31)(21) zzzz 

65. 2 568(54)(2)  xxxx

66. 2 3108(34)(2) xxxx 

67. 2 568(54)(2)  xxxx

68. 2 3108(34)(2) xxxx 

69. 2 5228(52)(4)  xxxx

70. 2 3148(32)(4) xxxx 

71. 2 5188(52)(4)  xxxx

72. 2 3108(32)(4) xxxx 

73. Since b is10thenweneedhalfof10squaredto bethelastterminourtrinomial.Thus

12 2 22 (10)5;(5)25 1025(5)xxx

74. Since b is14thenweneedhalfof14squaredto bethelastterminourtrinomial.Thus

12 2 22 (14)7;(7)49 1449(7)ppp

75. Since b is-6thenweneedhalfof-6squaredto bethelastterminourtrinomial.Thus

12 2 22 (6)3;(3)9 69(3)yyy  

76. Since b is-4thenweneedhalfof-4squaredto bethelastterminourtrinomial.Thus 12 2 22 (4)2;(2)4 44(2)xxx

77. Since b is12thenweneedhalfof12squared tobethelastterminourtrinomial.Thus 2 11111 224416 22 111 2164 ();() ()xxx

78. Since b is13thenweneedhalfof13squaredto bethelastterminourtrinomial.Thus 2 11111 236636 22 111 3366 ();() ()xxx  

79. 236(6)(6) xxx

80. 29(3)(3) xxx

81.  22 282(14)21212 xxxx 

82.  22 3273(19)31313 xxxx 

83. 21110(1)(10) xxxx 

84. 254(4)(1) xxxx 

85.  2102173 xxxx 

86. 268(2)(4) xxxx 

87. 22 4832428 xxxx 

88. 22 31215345 xxxx 

89. 2416xx isprimeovertherealsbecause therearenofactorsof16whosesumis4.

90. 22 1236(6)xxx

91. 22 152(215)(5)(3) xxxxxx 

92. 22 146(614) xxxx  isprimeoverthe integersbecausetherearenofactorsof–14 whosesumis–6.

93. 22 312363(412) 3(6)(2) xxxx xx

94. 322820(820) (10)(2) xxxxxx xxx

95. 43222 2 1130(1130) (5)(6) yyyyyy yyy

96. 322 318483(616) 3(2)(8) yyyyyy yyy 

97. 22 4129(23) xxx

98. 22 9124(32) xxx

99.

22 6822341 2311 xxxx xx

100.

22 8622431 2411 xxxx xx

101.  2 42222 2 819(9)(9) (3)(3)(9) xxxx xxx

102.  2 42222 2 11(1)(1) (1)(1)(1) xxxx xxx 

103. 6332 22 222 21(1) (1)(1) (1)(1) xxx xxx xxx 

104. 6332 22 222 21(1) (1)(1) (1)(1) xxx xxx xxx   



105. 75525(1)(1)(1)xxxxxxx 

106. 855352(1)(1)(1)xxxxxxxx 

107.  22 1624943 xxx

108.  22 9241634 xxx

109. 22 51616(16165) (45)(41) xxxx xx  

110. 22 51116(16115) (165)(1) xxxx xx  

111. 2 41615(25)(23) yyyy 

112. 2 994(34)(31) yyyy 

113. 2442 22 2 189(981) (91)(1) (31)(31)(1) xxxx xx xxx   



114. 2442 22 2 41482(472) 2(41)(2) 2(21)(21)(2) xxxx xx xxx





115. (3)6(3)(3)(6) xxxxx 

116. 5(37)(37)(37)(5) xxxxx 

117.

118.

  2 (2)5(2)(2)(2)5 (2)(3) xxxx xx  

  2 (1)2(1)(1)(1)2 (1)(3) xxxx xx 



119.   



 

 3 33 2 2 2 3227 323 323323329 359124969 35937 x x xxx xxxx xxx

120.

121.

3 33 2 2 2 511 511 511511511 525101511 525153 x x xxx xxxx xxx

2 2 3102545 3545 5354 53154 5311

122.

123.

2 2 76953 7353 3735 37215 3716 xxx xx xx xx xx

322 2 22(2)12 (2)(1) (2)(1)(1) xxxxxx xx xxx

124.

322 2 33(3)13 (3)(1) (3)(1)(1) xxxxxx xx xxx

125.  433 3 2 1(1)11 (1)(1) (1)(1)(1) xxxxxx xx xxxx





126.  433 3 2 22 1(1)11 (1)(1) (1)(1)(1) (1)(1) xxxxxx xx xxxx xxx

127. 



  2 234232343 23434233 2343469 234913 xxx xxx xxx xx

128.     2 521562212 21521564 211052024 213019 xxx xxx xxx

129. 

130.

 2 2252225 225 235 xxxxxx xxx xx

232 2 2 38383838 2498 329 xxxxxx xxx xx

132.



xxx



 

324 3 3 3 3 3 451521 251215 251225 25133 23511 6511 xxxx xxxx xxxx xxx xxx xxx

2 432434 43438 43438 43123 34341 xxx xxx xxx xx xx

134.  

2 23 2 2 2 3342343 334342 334342 33454 xxxx xxxx xxxx

135.

322 2 2 2 235321353212 635212135 635212135 6352154 xxxx xxxx

136. 

223 2 2 2 345451452515 24551651545 245513062025 245515031 xxxx xxxx xxxx xxx 

137. 4242(1)xxxx 

138. 54352 2682(134) xxxxxx 

139. 212 22 22 (1)(1)[(1)(1)] (1) (1) xxxxxxxx xxxx xx   

140. 1222 2 (3)4(3)(3)[(3)4] (3)(53) xxxxxxxx xxx  

141. Thepossiblefactorizationsare  2 1454xxxx  or  2 2244xxxx  ,noneofwhich equals24 x  .

142. Thepossibilefactorizationsare

 22 121xxx  ,neitherofwhichequals 21xx

143. Answerswillvary.

144. Answerswillvary.

Section R.6

1. quotient;divisor;remainder 2. 32051

3. d 4. a

5. True 6. True

7. 217510 21010 1550

Quotient:255 xx

Remainder:0

8. 11231 114 1145

Quotient:24 xx Remainder:5

9. 33213 93396 3113299

Quotient:231132 xx Remainder:99

10. 24211 82042 4102143

Quotient:241021 xx  Remainder:43

11. 3104010 391545138 1351546138

Quotient:432351546

Remainder:138 xxxx 

12. 210102 241020 1251022

Quotient:322510xxx Remainder:22

13. 14030105 441122 4411227

Quotient:54324422 Remainder:7 xxxxx 

14. 11050010 11666 1166616

Quotient:432666xxxx Remainder:–16

15. 1.10.100.20

0.110.1210.3531

0.10.110.3210.3531

Quotient:20.10.110.321 xx Remainder:–0.3531

16. 2.10.100.2

0.210.441

0.10.210.241

Quotient:0.10.21 x Remainder:0.241

17. 21000032 2481632 1248160

Quotient:43224816xxxx Remainder:0

18. 1100001 11111 111110

Quotient:4321 xxxx

Remainder:0

19. 24384 8104 4528

Remainder=8 ≠ 0.Therefore,2 x isnota factorof324384 xxx

20. 34508 1251153 41751161

Remainder=161 ≠ 0.Therefore,3 x  isnota factorof32458 xx 

21. 3260721 60021 20070

Remainder=0.Therefore,3 x isafactorof 43 26721 xxx

22. 2401504 81624 48120

Remainder=0.Therefore,2 x isafactorof 42 4154 xx

23. 2500430024 10204061224 3102036120

Remainder=0.Therefore,3 x  isafactorof 63 54324 xx

24. 320180109 6180039 2600130

Remainder=0.Therefore,2  x isafactorof 642 2189 xxx

25. 410161019 4160416 140143

Remainder=1 ≠ 0.Therefore,4  x isnota factorof5321619xxx

26. 4101601016 41600416 1400140

Remainder=0.Therefore,4 x  isafactor 6421616xxx

27. 131062 3 1002 30060

Remainder=0;therefore1 3 x isafactorof 43 362 xxx

28. 131031 3 1001 30032

Remainder=20  ;therefore1 3 x  isnota factorof43331 xxx

29. 21235 2822 141117 32 2 23517 411 22 xxx xx xx 

 1411179 abcd

30. 23 23 2 1322 3 130 hhhhh hhhh hh 322 3  xxhxh isthequotientand0isthe remainder.

31. 234 233 23 1334 254 12540 yyyyy yyyy yyy

Yes,  xy isafactorof 432234 334  xxyxyxyy .

32. Answerswillvary.

Section R.7

1. lowestterms

2. LeastCommonMultiple

3. d

4. a

5. True; 113 33 115 55 355(3) 353(5)       x xx x xx xxx xxx

6. False;    322 433 3 2623 64232 2332 xxxx xxxx LCMxxx   

7. 2 3(3) 393 9(3)(3)3 xx xxxx    

8. 24(2)48 122412(2)3 xxxxx xx    

9. 2(2) 2 363(2)3 xxxxx xx 

10. 2 22 3(58) 152458 33 xx xxx xxx   

11. 22 2 24244 1266(21)21 xxx xxxxx 

12.  2 2 (2)2 442 4(2)(2)2 xx xxx xxxx    

15. 2 2 (6)(2) 4126 44(2)(2)2

16. 2 2 (1) 2(2)(1)22 xxxxxx xxxxxx

xxxx

254 20 42 54 14 5

18. 2(21)(3) 253(3)3 121(21) xxxx xx xx

19. 222 363(2) 545(2)(2) 3 5(2) x xxx xxxxx xx

20. 2 333 261022(35)4(35) xxx xxxx

23 2 22 2 2 464 162 44416 (4)(4)2 224416 244 2416 4 xx xx xxxx xxx xxxxx xxx xxx

26.

329 627223 541855 229 xx xxxx x

xxxx xxxx xxxx xxxx x x

22 22 625 45215 2355 5153 235 513 xxx xxxx xxxx xxxx xxx xxx

27. 2 2 6 4624 3939 4 24 62(2) (2)(2)3(3) 4 (2)(3) x xxx xx x x xx xxx x xx

32. 3 22 3 3 339 3 99 9 x xxx x xx x 

3 3 3 2 3 2 39 333 9 33 9 3 9 3 xx xxx x xx x x x x

33. 2 222 222 2 2 2 712 71271212 1271212 12 (3)(4)(4)(3) (3)(4)(4)(3) (3) (3) xx xxxxxx xxxxxx xx xxxx xxxx x x

 34. 2 222 222 2 76 67656 56656 56 (6)(1)(2)(3) (3)(2)(6)(1) (1)(2) (2)(1) xx xxxxxx xxxxxx xx xxxx xxxx xx xx



    35. 2 222 222 2 576 23557621320 1514323515143 21320 (53)(2)(25)(4) (1)(25)(53)(31) (2)(4) (1)(31)

  xx xxxxxx xxxxxx xx xxxx xxxx xx xx

    

 

36.

43. 3524(35)(24) 212121 3524 21 9 21 xxxx

44. 541(54)(1) 343434 541 34 45 34 xxxx

666 11111

47. 737(1)3(3) 31(3)(1)(1)(3) 7739 (1)(3) 416 (1)(3) 4(4) (1)(3)

48. 252(5)5(5) 55(5)(5)(5)(5) 210525 (5)(5) 335 (5)(5) 335 (5)(5) xx xxxxxx xx xx x xx x xx

49. 22 2 23(1)(23)(1) 11(1)(1)(1)(1) 23 (1)(1) 323 (1)(1) xxxxxx xxxxxx xxxx xx xx xx

50.

22 2 323(3)2(4) 43(4)(3)(4)(3) 3928 (4)(3) 5 (4)(3) 51 (4)(3) xxxxxx xxxxxx xxxx xx xx xx xx xx

51. 22 22 34(3)(2)(4)(2) 22(2)(2)(2)(2) 56(68) (2)(2) 5668 (2)(2) 112(112) or (2)(2)(2)(2) xxxxxx xxxxxx xxxx xx xxxx xx xx xxxx

52.

55.  2422 xxx

2212 xxxx  Therefore,

56.  21234 xxxx 

281644 xxxx  Therefore,  LCM342 xx  .

57.   32111 xxxxxxx 

21 xxxx

Therefore,  LCM11 xxx  .

58.   22 32739333 xxxx   2 215253 xxxx 

Therefore,  LCM32533 xxx 

59.

322 44441 2121 xxxxxx xxx     322 3 221 xxxx x 

Therefore,  32LCM21 xx  .

60. 3 x

 2 32 33 9933 xxxx xxxxxxx   Therefore,  LCM33 xxx  .

61.   32111 xxxxxxx 



 3222 32 2211 111 xxxxxxxx xxxx  

Therefore,  22 LCM111 xxxxx 

62.  22 442xxx   322 3 22 2 xxxx x   Therefore,  23LCM2 xx 

63. 22 22 76224 (6)(1)(6)(4) (4)(1) (6)(1)(4)(6)(4)(1) 45 (6)(4)(1)(6)(4)(1) xx xxxx xx xxxx xxxx xxxxxx xxxxx xxxxxx 

64. 2 22 1 3524 1 (3)(3)(8) (8)1 (3)(8)(3)(8) 8171 (3)(8)(3)(8) xx xxx xx xxx xxx xxxx xxxxx xxxx     

     

65. 22 42 46 42 (2)(2)(3)(2) x xxx x xxxx

2 2 2 4(3)2(2) (2)(2)(3)(3)(2)(2) 41224 (2)(2)(3) 4104 (2)(2)(3) 2(252) (2)(2)(3) xxx xxxxxx xxx xxx xx xxx xx xxx

69. 22 22 2 423 228 423 (2)(1)(4)(2) (4)(4)(23)(1) (2)(1)(4)(4)(2)(1) 816(253) (2)(1)(4) 313 (2)(1)(4) xx xxxx xx xxxx xxxx xxxxxx xxxx xxx xx xxx 

66. 22 2 2

68.

2 3434 1(1)21(1) 3(1)4 (1)(1)(1) 334 (1) 344 (1) xxxx xxxxx xxx xxx xxx x xx x

22 22 22 22 32 1111 3121 11 3322 11 51 11 xxxx xx xx xx xx x

22 26 2121xxxx

70. 22 232 87(1) xx xxx 2 2 22 2 2 2 232 (1)(7)(1) (23)(1)(2)(7) (1)(7)(1)(1)(7) 23(514) (1)(7) 611 (1)(7) xx xxx xxxx xxxxx xxxx xx xx xx

71. 232 123 xxxxx 

2 2 22 2 32 2 32 2 123 11 112131 11 12233 11 253 11 243 11 xxxxx xxxxxx xxx xxxxx xxx xxxx xxx xxx xxx

73. 111111() ()() 1 () () 1 () xxh hxhxhxhxxxh xxh hxxh h hxxh xxh

74. 22 111 () hxhx

22 2222 222 22 2 22 22 22 22 111() ()() 1(2) () 2 () (2) () 2 () 2 () xxh hxhxxxh xxxhh hxxh xhh hxxh hxh hxxh xh xxh xh xxh

78. 2 11 1111 1 21211 1 11 (1) xxx xxxx xxxx xxxx x xx x x

79. 43 21 1 xx xx x 

22 2 (4)(1)(3)(2) (2)(1)(1)(2) 1 54(56) (2)(1) 1 1021 (2)(1)1 2(51) (2)(1) xxxx xxxx x xxxx xx x x xxx x xx

80. 2 12 3 xx xx x 

22 (2)(2)(1) (1)(2)(2)(1) 3 44() (2)(1) 3 541 (2)(1)3 54 (2)(1)(3) 54 (2)(1)(3) xxxx xxxx x xxxx xx x x xxx x xxx x xxx

81. 21 21 23 1 xx xx xx xx

2 22 22 2 2 2 (2)(1)(1)(2) (2)(1)(1)(2) (23)(1) (1)()(1) 22 (2)(1) (23) (1) 24 (2)(1) 3 (1) 2(2)( (2)(1) xxxx xxxx xxx xxxx xxxx xx xxx xx x xx xx xx xxx xx

2 2 2 2 2 1) (3) 2(2) (2)(3) 2(2) (2)(3) xx xx xxx xx xxx

82. 22 25 3 (1) 33 xx xx xx xx

  22 22 3232 2 2 2 (25)(3)() (3)(3) (3)(3)(1) (3)(3)(3)(3) 215 (3) 3(53) (3)(3) 15 (3) 453 (3)(3) 15( (3) xxxx xxxx xxxx xxxx xxx xx xxxxx xx xx xx xx xx xxx xx

2 2 2 3)(3) 453 (15)(3) (453) x xx xxx xxx  

 83. 1111 111 1 1 1 1 1 1 x xx x x xx x x

84. 111111 1111 111 1111 1 1 xx xxx xx xx xx x x

2 7121 236 (3)7(2)121 3714121 (7)314121 712 5

11 111 1111 121 11 2,1,1 x xx xx xxx xx abc

111111 121121 111 21132 2121 3,2,1 x xx x x xxx xx abc

1121111 132132 1121 11

322153 3232 5,3,2 x xx x x xxx xx abc

Ifwecontinuethisprocess,thevaluesof a, b and c producethefollowingsequences: :1,2,3,5,8,13,21,.... :1,1,2,3,5,8,13,21,..... :0,1,1,2,3,5,8,13,21,.....

a b c Ineachcasewehavea FibonacciSequence, wherethenextvalueinthelistisobtainedfrom thesumoftheprevious2valuesinthelist.

100. Answerswillvary.

Answerswillvary.

21. 33433882 xxxxx

22. 3335322 19264343

xxxxx 23. 444 24381333

24. 44544 4816323 xxxxx

44 41283232 4 xyxyxy  26. 5 5105252 5 xyxyxy 

27. 97 44842 3 xy xyxy xy 

28. 23 33 42333 3111 81273 27 xy xyxx x 

29. 648  xx

30. 542933 xxxxx 

31.

 444491223 4 234 16223 32 xyxxy xyx 

32.

33314103243 3 432 3 405(2) 25 xyxxyy xyxy

33. 222 1557525353  xxxxxxxx

34. 34252010010 xxxx 

 22233 32333 5959 59581533153 

 44433 34233 310310 310331003003

3622612643123 

51.

445444 44 44 3221622 222 22or22 xxxxx xxx xxxx

32 8350423252 22152 2152 xxxxx xxx xx

52.

39425920 920 xyyxyy xy

53.

4433 3 3333 3 333 3 33 163252 823252 223252 2352 52or52 xyxxyxy xxyxxyyxy xxyxxyyxy xxyxy xyxyxyxy

2233 3 8258852 852 5 xyxyxyxyxyxy xy xy

55. 1122 2222

56. 22323 3333

57. 33515 5555

58. 333266 822222224

59.

60.

3352 525252 352 252 352536 or 2323

2272 727272 272 74 2721422 or 33

61. 2525235

62. 3131233 233233233 623333953 1293

63. 5521 212121 525525 21

64. 3354 545454 35123512 51611 3512 11

65. 33 333 55454 2242

66. 33 333 22323 9933

2 2 2 2 22 xhxxhxxhx xhxxhxxhx xhxxhx xhx xhxxhx xhx xhxxh h

22 22 22 2 2 22 2 xhxhxhxh xhxhxhxh xhxhxhxh xhxh xhxhxh xhxh xxh h

111111111 22111 11110 21112111 5 111

615615615 1515615 6159 901531015 93 3105105

80. 3 3/43 4 161628 

3/2 3/233 41111 428 4 

3/2 3/233 161111 16464 16

3/2333 33 9933 882222 2727272 8221621622

2 2/32 3 272739 8824

1/31/31/3 36362 xyxyxy

3/43/43/421/33/421/3

1/41/4 21/42 433/21/4

1/41/2

33/21/41/41/2 5/43/4 5/4 3/4 1616 16 2 8 8 xyxy xyxy xy xy xy xy x y

3/23/23/211/33/211/3

3/23/23/2

33/21/2

3/23/2

1/32/31/32/3 1/32/3 2222

2/32/32/32/3

2/31/32/34/3 2/32/3

2/32/32/31/34/32/3

2/312/3 xyxyxyxy xyxy xyxy xy xy xyxy

1/41/21/21/2 221/41/422

3/43/4 223/4

1/41/4

3/23/4

1/413/21/413/4 1/2 1/41/2 1/4 xyxyxyxy xyxy xyxy xy xy y xy x

33/23/21/23/2 31 3 44 4 2 8 8 xyxy xyxy xy xy xy xy xy 

1/21/2 1/2 1/21/2 1/2 1/2 1/2 211 21 (1)(1) 21 (1) 22 (1) 32 (1) xxxx x xx xx x xx x x x   

102. 1/21/2 1/2 1/21/2 1/21/2 112 22 1231 22 xxxx x xx xxx xx

1/32/3 1 11,1 3 xxxx

106.

33 2233 8121 3,2,8 22481 xx xx xx

22

22

24281 8812 24281 88+12 24281 6482 24281 656 24281 xxxx xx xx xx xx xx xx xx x xx

112.

21/2 2 21/2 2 1 1 ,1or1 x x x xx x 

1/21/2222 21/2 2 1/21/2222 21/22 11 1 111 1 xxx x x xxx xx

22 21/22 22 21/22 221/2 11 1 11 1 1 1 xx xx xx xx xx

1/21/2222 2 44 4 xxx x  

21/22 21/2 2 1/21/2 222 21/2 2 1/21/2 222 21/22 22 1/223/222 4 4 4 44 4 4 441 44 414 4 44 x x x x xxx x x xxx xx xx x xx

 

118.

1/223/21/2 1/22 1/22 1/2 688 23()44 234 2(34)(1) xxxxx xxxx xxx xxx

1/23/2 1/2 1/2 6238 23(23)4 2109 xxx xxx xx

119.

4/31/322 1/3 222 1/3 222 1/3 22 34442 4348 43128 41112 xxxx xxx xxx xx

120.

4/31/3 2 1/3 1/3 234434 234342 23454 xxxx xxxx xxx 

121.

1/33/24/31/2 4352333523 xxxx 

1/31/2 1/31/2 1/31/2 3523423335 3523812915 35231727 xxxx xxxx xxx   

3 where 2 x 

136. 161222 3222 24.44seconds T 

137. 314313 334312 134431

Thequotientis2(34)43 xx Theremainderis1

138. 1219137 126727 1387720

Yes,12  isafactorof329137xxx

139. Answersmayvary.Onepossibilityfollows:If 5 a  ,then  225255 aa  Sinceweusetheprincipalsquareroot,whichis alwaysnon-negative, 2if0 if0 aa a aa 

whichisthedefinitionof a ,so 2 aa 

Section R.8: nth Roots; Rational Exponents

Chapter 1

Equations and Inequalities

Section 1.1

1. Distributive

2. Zero-Product

3.  4 xx 

4. False.Multiplyingbothsidesofanequationby zerowillnotresultinanequivalentequation.

5. identity

6. linear;first-degree

7. False.Thesolutionis 8 3

8. True 9. b

10. d 11. 721 721

3 x x x 

Thesolutionsetis{3}. 12. 624 624 66 4 x x x

Thesolutionsetis{4}. 13. 3150 31515015 315 315 33 5 x x x x x

Thesolutionsetis{5}. 14. 6180 61818018 618 618 66 3 x x x x x

Thesolutionsetis{3}. 15. 230 23303 23 23 22 3 2 x x x x x 

Thesolutionsetis3 2

16. 340 34404 34 34 33 4 3 x x x x x 

Thesolutionsetis 4 3

17. 17 420 17 44 420 287 205 

x x x

Thesolutionsetis7 5

18. 29 32 29 66 32 427 427 44 27 4 x x x x x 

Thesolutionsetis27. 4

19.

20.

Thesolutionsetis{2}.

Thesolutionsetis{3}.

21. 263 26636 29 29 39 39 33 3 tt tt tt tttt t t t

Thesolutionsetis{3}.

22. 5618 566186 524 524 624 624 66 4 yy yy yy yyyy y y y

Thesolutionsetis{4}.

23. 629 66296 23 2232 33 33 33 1 xx xx xx xxxx x x x

Thesolutionsetis{1}.

24. 322 32323 21 21 1 1 11 1 xx xx xx xxxx x x x

 Thesolutionsetis{1}.

25. 3247 323473 244 24444 24 24 22 2 nn nn nn nnnn n n n        Thesolutionsetis{2}.

Chapter 1: Equations and Inequalities

26. 6231 626316 235 23353 55 55 55 1 mm mm mm mmmm m m m

 Thesolutionsetis{1}.

30. 7(21)10 72110 8210 828108 22 22 22 1 x x x x x x x

Thesolutionsetis{1}.

27. 3(53)8(1) 15988 9815888 1515815

xx xx xxxx x x Thesolutionsetis{23}.

28. 3(2)21 xx  6321 636216 327 32272 57 57 55 7 5 xx xx xx xxxx x x x

Thesolutionsetis7. 5

29. 8(32)310 xxx  832310 52310 5223102 538 53383 28 28 22 4 xxx xx xx xx xxxx x x x

Thesolutionsetis{4}.

31. 311 2 222 xx  311 222 222 341 34414 33 33 43 43 44 3 4 xx xx xx xx xxxx x x x

Thesolutionsetis3 4 

32. 12 2 33 12 332 33 62 2622 36 36 33 2 xx xx xx xxxx x x x

Thesolutionsetis{2}.

33. 13 5 24 13 454 24 2203 220232 20 20 xx xx xx xxxx x x

Thesolutionsetis{20}.

34.

1 16 2 1 2126 2 212 22122 10 10 11 10 x x x x x x x

Thesolutionsetis{10}.

35.

Thesolutionsetis{2}.

36. 114 233 p  114 66 233 328 32383 25 25 22 5 2 p p p p p p

Thesolutionsetis5. 2

37. 0.20.90.5 0.20.50.90.50.5 0.30.9 0.30.9 0.30.3 3

Thesolutionsetis{-3}.

38. 0.91 0.91 0.11 0.11 0.10.1 10 tt tttt t t t

Thesolutionsetis{10}.

39. 12 2 37 xx 

12 21212 37 713242 773642 101342 1013134213 1029 1029 1010 29 10 xx xx xx x x x x x

Thesolutionsetis29 10 

40. 21 163 3 x x    21 31633 3 21489 2499 249292 497 497 77 7 x x xx xx xxxx x x x 

Thesolutionsetis{7}.

Chapter 1: Equations and Inequalities

41. 5111 (3)2(23) 8416 10(3)324(23)11 10303281211

42. 122 3515 (1)3(4) 5(1)456(4)2 55456242 540626

Thesolutionsetis

Since y =2doesnotcauseadenominatorto equalzero,thesolutionsetis{2}. 44. 45 5 2 yy  45 252 2 8105 810858

Since3 10 y  doesnotcauseadenominatorto equalzero,thesolutionsetis310

123 24 123 44 24 283 28232 8 x xx x xx xxxx x

Since x =8doesnotcauseanydenominatorto equalzero,thesolutionsetis{8}. 46. 311 36 311 66 36 182 18222 183 183 33 6 x xx x xx xxxx x x x

Since6 x  doesnotcauseadenominatorto equalzero,thesolutionsetis{6}.

47. 2 (7)(1)(1) xxx  22 2222 6721 6721 6721 677217 628 62282 48 48 44 2 xxxx xxxxxx xx xx xx xxxx x x x

Thesolutionsetis{2}.

48. 2 (2)(3)(3) xxx  22 2222 669 669 669 66696 615 66156 715 715 77 15 7 xxxx xxxxxx xx xx xx xxxx x x x

Thesolutionsetis15 7

49. 22 2222 (23)(21)(4) 23274 2322742 374 37747 44 44 44 1 xxxx xxxx xxxxxx xx xxxx x x x

Thesolutionsetis{1}.

Section 1.1: Linear Equations

xxxx xxxxxx xx xxxx x x x

50. (12)(21)(2) xxxx  22 2222 2252 222522 52 5525 62 621 663

Thesolutionsetis 1 3 

51. 23 33 3333 312 312 312 312 312 4 33 

  ppp ppp ppppp p p p

Thesolutionsetis{4}.

52. 23 33 3333 (4)8 48 48 48 48 44 2 www www wwwww w w w

  

Thesolutionsetis{2}.

53. 2 3 22 x xx     2 322 22 322 362 462 46626 48 48 44 2 x xx xx xx xx x x x x x

     Since x =2causesadenominatortoequalzero, wemustdiscardit.Thereforetheoriginal equationhasnosolution.

Chapter 1: Equations and Inequalities

54.

Since x =–3causesadenominatortoequalzero, wemustdiscardit.Thereforetheoriginal equationhasnosolution.

Since x =2causesadenominatortoequalzero,wemustdiscardit.Thereforetheoriginalequationhasnosolution.

56.

Since x =3causesadenominatortoequalzero,wemustdiscardit.Thereforetheoriginalequationhasnosolution.

Since x =–6doesnotcauseanydenominatorto equalzero,thesolutionsetis{6}.

Since x =41doesnotcauseanydenominatorto equalzero,thesolutionis{41}.

Since x =–2doesnotcauseanydenominatorto equalzero,thesolutionsetis{2}.

63.

Since1 x  doesnotcauseanydenominatortoequalzero,thesolutionsetis{1}.

Since3

Since6 x  doesnotcauseanydenominatortoequalzero,thesolutionsetis{14}.

Chapter 1: Equations and Inequalities

69.

Thesolutionsetisapproximately{0.41}.

Thesolutionsetisapproximately{0.94}.

73. ,0axbca axbbcb axbc axbc aa bc x a

74. 1,0 111 1 1 11 axba axb axb axb aa bb x aa

75. ,0,0, () () xxcabab ab xx ababc ab bxaxabc abxabc abxabc abab abc x ab

76. ,0 ab cc xx ab xxc xx abcx abcx cc ab x c

Chapter 1: Equations and Inequalities

77. 2166,if4

4216(4)6 421646 42162 412 412 44 3 xaaxax aaa aaa aa a a a 

78. 242,for2 22242(2) 22244 2224 42 4 2 2

bb bb bb b b b

79. 12 111 RRR  1212 12 1221 1221 1221 2121 12 21 111 () () RRRRRR RRR RRRRRR RRRRR RRRRR RRRR RR R RR

80. (1)APrt

APPrt

APPrt APPrt PtPt AP r Pt

82. PVnRT PVnRT nRnR PV T nR

1 Sa r  (1)(1) 1 a Srr r SSra SSrSaS SraS SraS SS Sa r S

0 0 0 00 vgtv vvgt vvgt gg

85. AmountinbondsAmountinCDsTotal 300020,000 xx

300020,000 2300020,000 223,000 11,500 xx x x x

$11,500willbeinvestedinbondsand$8500 willbeinvestedinCD's.

86. Sean'sAmountGeorge'sAmountTotal 300010,000 xx

Seanwillreceive$6500andJorgewillreceive $3500.

87. DollarsHoursMoney perhourworkedearned Regular4040 wage

Sandra’sregularhourlywageis$17.50.

88. DollarsHoursMoney perhourworkedearned Regular4040 wage Overtime1.566(1.5) wage Sunday244(2) wage xx xx

Leigh’sregularhourlywageis$19.00.

89. Let x representthescoreonthefinalexam. 8083716195 80 7 3902 80 7 3902560 2170 85

Camilaneedsascoreof85onthefinalexam.

90. Let x representthescoreonthefinalexam. Note:sincethefinalexamcountsfortwo-thirds oftheoverallgrade,theaverageofthefourtest scorescountforone-thirdoftheoverallgrade. ForaB,theaveragescoremustbe80.

1868084902 80 343 13402 80 343 852 80 33 852 3380 33 852240 2155 77.5 x x x x x x x

Alineedsascoreof78toearnaB. ForanA,theaveragescoremustbe90.

1868084902 90 343 13402 90 343 852 90 33 852 3390 33 852270 2185 92.5 x x

Alineedsascoreof93toearnanA.

91. Let x representtheoriginalpriceofthephone. Then0.12x representsthereductionintheprice ofthephone. Thenewpriceofthephoneis$572. originalpricereductionnewprice

 xx x x Theoriginalpriceofthephonewas$650. Theamountofthereduction(i.e.,thesavings)is 0.12($650)=$78.

650

92. Let x representtheoriginalpriceofthecar. Then0.15x representsthereductionintheprice ofthecar.

Thenewpriceofthecaris$8000. listpricereductionnewprice

Thelistpriceofthecarwas$21,176.47. Theamountofthereduction(i.e.,thesavings)is 0.15($21176.47)$3176.47 

93. Let x representthepricethetheaterpaysforthe candy.

Then2.75 x representsthemarkuponthecandy. Thesellingpriceofthecandyis$4.50. suppierpricemarkupsellingprice

 xx x x Thetheaterpaid$1.20forthecandy.

94. Let x representsellingpriceforthenewcar. Thedealer’scostis0.85($24,000)$20,400.  Themarkupis$300. sellingprice=dealer’scost+markup 20,400300$20,700 x At$300overthedealer’scost,thepriceofthe careis$20,700.

95. TicketsPriceperMoney soldticketearned Adults7.507.50 Children52004.504.50(5200) xx xx

7.504.50520029,961 7.5023,4004.5029,961 3.0023,40029,961 3.006561 2187 xx xx x x x     

Therewere2187adultpatrons.

96. Let p representtheoriginalpricefortheboots. Then,0.30p representsthediscountedamount. originalpricediscountclearanceprice 0.30399 0.70399 570 pp p p 

Thebootsoriginallycost$570.

97. Let w representthewidthoftherectangle. Then8 w  isthelength. Perimeterisgivenbytheformula22. Plw  2(8)260 216260 41660 444 11 ww ww w w w     

Now,11+8=19. Thewidthoftherectangleis11feetandthe lengthis19feet.

98. Let w representthewidthoftherectangle. Then2 w isthelength. Perimeterisgivenbytheformula22. Plw  2(2)242 4242 642 7 ww ww w w    

Now,2(7)=14. Thewidthoftherectangleis7metersandthe lengthis14meters.

99. WewillletBbethecaloriesfrombreakfast,L thecaloriesfromlunchandDthecaloriesfrom dinner.Sowehavethefollowingequations: 125 2300 2025 BL DL BLD   

Nowwesubstitutethefirsttwointothelastone andsolveforL.

2025(125)(2300) 20254175 22004 550 LLL L L L

NowwesubstituteLintothefirsttwoequations togetBandD. 550125675 2(550)300800 B D

SoHerscheltookin675caloriesfrombreakfast, 550caloriesfromlunchand800caloriesfrom dinner.

100. WewillletBbethecaloriesfrombreakfast,L thecaloriesfromlunch,Dthecaloriesfrom dinnerandSthecaloriesfromsnacks.Sowe havethefollowingequations:

Nowwesubstitutethefirstfourintothelastone andsolveforB. 14800.5(200)(200)700 14803.5620 21003.5 600 BBBB B B B

NowwesubstituteBtogetS. 120600120480 SB

Judypays$10.80andTompays$7.20.

102. Anisoscelestrianglehasthreeequalsides. Therefore:410240318  xxx .Solve eachsetseparately:

Section 1.1: Linear Equations

xx x x xx x xx x x Since22isthelargestofthenumbersthenthe largestperimeteris:

422102224032218266

MultiplybothsidesbytheLCD80toclear fractions. 608488012064 1087258 10716 16 107

104. Ifahexagonisinscribedinacirclethenthesides ofthehexagonareequaltotheradiusofthe circle.LettheP=6rbetheperimeterofthe hexagon.Letrbetheradiusofthecircle. 610 510 2

Thusr=2inchesistheradiusofthecirclewhere theperimeterofthehexagonis10inchesmore thantheradius.

105. Tomovefromstep(6)tostep(7),wedivided bothsidesoftheequationbytheexpression 2 x .Fromstep(1),however,weknow x =2, sothismeanswedividedbothsidesofthe equationbyzero.

106– 107. Answerswillvary.

Chapter 1: Equations and Inequalities

Section 1.2 1.

25661 xxxx  2.

3.  5,3 3 4. True

5.

6. discriminant;negative

7. False;aquadraticequationmayhavenoreal solutions.

8. False;If2  xp then x couldalsobenegative.

9. b

10. d

11.

290 90 xx xx

0or90 0or9 xx xx

Thesolutionsetis{0,9}.

12. 240 (4)0 xx xx   0or40 0or4 xx xx   Thesolutionsetis{–4,0}.

13. 2250 (5)(5)0 x xx  

50or50 5or5 xx xx   Thesolutionsetis{–5,5}.

14. 290 (3)(3)0 x xx   30or30 3or3 xx xx   Thesolutionsetis{–3,3}.

15. 260 (3)(2)0 zz zz

  30or20 3or2 zz zz   Thesolutionsetis{–3,2}.

16. 2760 (6)(1)0 vv vv   60or10 6or1 vv vv   Thesolutionsetis{–6,–1}

17. 2 2530 (21)(3)0 xx xx   210or30 1or3 2 xx xx 



Thesolutionsetis  1,3 2

18. 2 3520 (32)(1)0 xx xx

  320or10 2 or1 3 xx xx  

Thesolutionsetis  1,2 3

19. 2 2 51800 5(36)0 5(6)(6)0

   w w ww 60or60 6or6 

 ww ww Thesolutionsetis{–6,6}.

20. 2 2 2500 2(25)0 2(5)(5)0 y y yy  

50or50 5or5 yy yy

Thesolutionsetis{–5,5}.

21. 

2 3100 3100 (2)50

Section 1.2: Quadratic Equations

25. 2 2 2 6(1)5 665 6560 (32)(23)0 pp pp pp pp 

 320or230 23 or 32 pp pp  

Thesolutionsetis23 , 32 

xx xx

xx xx xx 20or50 2or5

Thesolutionsetis 5,2.

22. 2 (4)12 4120 (6)(2)0 xx xx xx

60or20 6or2 xx xx

Thesolutionsetis 6,2.

23. 2 2 2 4912 41290 (23)0 230 3 2 xx xx x x x     

Thesolutionsetis  3 2

24. 2 2 2 251640 2540160 (54)0 540 4 5 xx xx x x x



Thesolutionsetis  4 5 .

26. 2 2 2(24)30 4830 (21)(23)0 uu uu uu 

210or230 13 or 22 uu uu

Thesolutionsetis13 , 22

27.

2 2 6 65 6 65 656 6560 (32)(23)0 x x xxx x xx xx xx 

320or230 23 or 32 xx xx

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis23 , 32 

28. 2 2 12 7 12 7 127 7120 (3)(4)0 x x xxx x xx xx xx

30or40 3or4 xx xx

Chapter 1: Equations and Inequalities

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis{3,4}.

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis

32. 236 36 6 x x x 

Thesolutionsetis  6,6

33.  2 14 14 12 12or12 3or1 x x x xx xx 

Thesolutionsetis  1,3

34.  2 21 21 21 21or21 1or3 x x x xx xx

Thesolutionsetis  3,1

35. 12416 3 1416 3 144 3 1144or44 33 110or8 33 0or24

     h h h hh hh hh

Thesolutionsetis  24,0

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis

31. 225 25 5 x x x 

Thesolutionsetis  5,5

36.  2 324 324 322 322or322 34or30 4or0 3

z z z zz zz zz

Thesolutionsetis  0,4 3

37.

Section 1.2: Quadratic Equations

40. 2210 33 xx

Thesolutionsetis7,3.

38.

44 x x xx

The13 solutionsetis,. 44

The1 solutionsetis1,. 3

41. 21 30 2 xx 2 2 2 2 110 36 11 36 1111 336636 17 636 17 636 17 66 17 6 xx xx xx x x x x

 The1717 solutionsetis,. 66

42. 2 2310 xx 2 2 2 310 22 31 22 3919 216216 xx xx xx

Chapter 1: Equations and Inequalities

43. 2420xx 2 1,2,13 (2)(2)4(1)(13)2452 2(1)2 2562214114 22

 abc x

Thesolutionsetis114,114. 

44. 2420xx 2 1,4,2 444(1)(2)4168 2(1)2 48422 22 22 abc x 

 Thesolutionsetis22,22. 

45. 2410xx 2 1,4,1 (4)(4)4(1)(1)4164 2(1)2 42042525 22 abc x 

  

 

 Thesolutionsetis25,25. 

46. 2610xx 2 1,6,1 664(1)(1)6364 2(1)2 632642 322 22

abc x 

  

 

 Thesolutionsetis322,322. 

47. 2 2530 xx 2 2,5,3 (5)(5)4(2)(3) 2(2) 525245151 444 5151 or 44 64 or 44 3or1 2 abc x xx xx xx



 





The3 solutionsetis1,. 2

48. 2 2530 xx 2 2,5,3 554(2)(3) 2(2) 525245151 444 5151 or 44 46 or 44 13 or 2 abc x xx xx xx



 

 

 





The3solutionsetis,1. 2



49. 2 420 yy 2 4,1,2 (1)(1)4(4)(2) 2(4) 1132131 88 abc y 

 Norealsolution.

50. 2 410 tt 2 4,1,1 114(4)(1) 2(4) 1116115 88 abc t 



 Norealsolution.

51. 2 2 985 9850   xx xx 2 9,8,5 884(9)(5) 2(9) 8641808244 1818 8261461 189

abc x

Thesolutionsetis461461,.99

52. 2 2 212 2210 xx xx   2 2,2,1 224(2)(1)248 2(2)4 21222313 442 abc x

Thesolutionsetis1313,.22

Section 1.2: Quadratic Equations

53. 2 2 49 490 (49)0 xx xx xx    0or490 09 or 4 xx xx  

Thesolutionsetis  9 0,. 4

54. 2 2 54 045 0(45) xx xx xx    0or450 05 or 4 xx xx 

Thesolutionsetis  5 0,. 4

55. 2 9610 tt 2 9,6,1 (6)(6)4(9)(1) 2(9) 63636601 18183 abc t     

Thesolutionsetis   1 . 3

56. 2 4690 uu 2 4,6,9 (6)(6)4(4)(9) 2(4) 6361446108 88 abc u      Norealsolution.

57.  2 2 2 3110 442 311 440 442 320 xx xx xx 

 3,1,2abc

Chapter 1: Equations and Inequalities

 2 11432 23 112412515 666 1515 or 66 64 or 66 12 or 3 x xx xx xx

The2solutionsetis,1. 3

58. 2230 3 xx

2 2 2 3330 3 2390 2,3,9 xx xx abc

2 33429 22 397238139 444 3939 or 44 126 or 44 33 or 2 x xx xx xx

The3solutionsetis,3. 2

59. 2 2 2 2 51 33 5133 33 531 5310 xx xx xx xx

2 5,3,1 33451 25 3920329 1010 abc x 

Thesolutionsetis329329 , 1010 

.

60. 2 2 2 2 31 55 3155 55 351 3510 xx xx xx xx 

2 3,5,1 55431 23 52512537 66 abc x 

Thesolutionsetis537537 , 66   

.

61. 2 2(2)3 2430 xx xx   2 2,4,3 444(2)(3)41624 2(2)4 4404210210 442 abc x       

Thesolutionsetis210210,.22

62. 2 3(2)1 3610 xx xx 

2 3,6,1 664(3)(1)63612 2(3)6 648643323 663

63.  2 22 2 2 11 40 11 40 410 xx xx xx

equalzero,sothesolutionsetis

64. 2 83 20 

Section 1.2: Quadratic Equations

65. 2 22 2 314 2 31(2)4(2) 2 3()(2)48 3248 092 x xx x xxxx xx xxxxx xxxx xx

1,9,2abc 2 (9)(9)4(1)(2) 2(1) 9818973 22 x  

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis 973973 , 22 

66. 2 22 2 214 3 21(3)4(3) 3 2()(3)412 23412 02133 x xx x xxxx xx xxxxx xxxx xx 

 2,13,3abc  2 (13)(13)4(2)(3) 2(2) 131692413145 44 x  

 

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis 1314513145 , 44   

67. 24.12.20xx   2 1,4.1,2.2 4.14.1412.2 21 4.116.818.84.18.01 22 3.47or0.63 abc x xx     



Thesolutionsetis  0.63,3.47

Chapter 1: Equations and Inequalities

68. 23.91.80xx

2 1,3.9,1.8 3.93.9411.8 21 3.915.217.23.98.01 22 0.53or3.37 abc x xx

Thesolutionsetis  3.37,0.53.

69. 2330xx

Thesolutionsetis  2.80,1.07

70. 2220xx

2 1,2,2 22412 21 228210 22 0.87or2.29 abc x xx

Thesolutionsetis  2.29,0.87.

71. 20xx

 2 2 ,1, 114 2 114 2 1.17or0.85 abc x xx

Thesolutionsetis  0.85,1.17.

72. 220xx 

  2 2 ,,2 42 2 8 2 0.44or1.44 abc x xx



  Thesolutionsetis  1.44,0.44

73. 2 2670 xx  22 2,6,7 4(6)4(2)7365620 abc bac

  Sincethe240, bac theequationhasnoreal solution.

74. 2470xx

 22 1,4,7 4(4)4(1)7162812 abc bac 

 Sincethe240, bac theequationhasnoreal solution.

75. 2 930250 xx



 22 9,30,25 4(30)4(9)259009000 abc bac

 Since240, bac theequationhasone repeatedrealsolution.

76. 2 252040 xx

 22 25,20,4 4(20)4(25)44004000 abc bac

  Since240, bac theequationhasone repeatedrealsolution.

77. 2 3580 xx



 22 3,5,8 4(5)4(3)82596121 abc bac

 Since240, bac theequationhastwo unequalrealsolutions.

78. 2 2370 xx



 22 2,3,7 4(3)4(2)795665 abc bac

 Since240, bac theequationhastwo unequalrealsolutions.

79. 2 2 50 5 5 x x x     Thesolutionsetis5,5.

80. 2 2 60 6 6 x x x     Thesolutionsetis6,6.

81.  2 16810 41410 410 1 4 xx xx x x    

Thesolutionsetis  1 . 4

82.  2 91240 32320 320 2 3 xx xx x x     Thesolutionsetis  2 . 3

83.  2 1019150 53250 xx xx   530or250 35 or 52 xx xx  

Thesolutionsetis  35 ,.52

84.  2 67200 34250 xx xx 

340or250 45 or 32 xx xx  

Thesolutionsetis  54 ,.23

Section 1.2: Quadratic Equations

85.  2 2 26 062 03221 zz zz zz 

  320or210 21 or 32 zz zz  

Thesolutionsetis  12 , 23

86.  2 2 26 062 03221 yy yy yy 

 320or210 21 or 32 yy yy  

Thesolutionsetis  21 , 32

87.

 2 2 2 2 21 2 1 20 2 1 2220 2 22210 xx xx xx xx  

 

 2,22,1abc 

 2 (22)(22)4(2)1 2(2) 22882216 44 22422 42 x  

   

The2222 solutionsetis,22    

88.

 2 2 2 2 121 2 1210 2 1 22120 2 2220 xx xx xx xx 

 

  1,22,2abc 

Chapter 1: Equations and Inequalities

The1515 solutionsetis,. 22

91. 2 271 212

xx xxxx xx xxxx xxxx xxxx xxxx xxx xx xx

Thevalue1 x  causesadenominatortoequalzero,sowedisregardit.Thus,thesolutionsetis{5}.

92. 2 3147 212 xx xxxx

 2 2 2 3147 21(2)(1) 3147(2)(1)(2)(1)21(2)(1) 3(1)(2)47 33247 32247 3520 (31)(2)0 xx

Thevalue2 x  causesadenominatortoequalzero,sowedisregardit.Thus,thesolutionsetis

93. Sincethisisarighttrianglethenwecanusethe PythagoreanTheorem.So 222 222 2 2 (23)(25)(7) 4129420251449 129674 01865 0(5)(13) xxx xxxxxx xxx xx xx

50or130 5or13 xx xx

Thismeansthereare2possiblethatmeetthese requirements.Substituting x intothegivensides gives:

When x =5:5m,12m,13m When x =13:20m,21m,29m Thusthereare2solutions.

94. Sincethisisarighttrianglethenwecanusethe PythagoreanTheorem.So 222 222 2 2 (45)(313) 164025978169 6381440 2(31972)0 2(38)(9)0 xxx xxxxx xx xx xx

Thismeansthereare2possiblesolutionsthat meettheserequirements.Substituting x intothe givensidesgives:

When x =9:41m,40m,9m

When x = 8 3atleastonesideofthetriangle hasanegativemeasurementwhichisimpossible. Thusthereisonly1trianglepossible

95. Let w representthewidthofwindow. Then2 lw representsthelengthofthe window.

Sincetheareais143squarefeet,wehave: 2 (2)143 21430 (13)(11)0 ww ww ww

Discardthenegativesolutionsincewidthcannot benegative.Thewidthoftherectangular windowis11feetandthelengthis13feet.

96. Let w representthewidthofwindow. Then1 lw representsthelengthofthe window.

Sincetheareais306squarecentimeters,we have:(1)306 ww  23060 (18)(17)0 ww ww

Discardthenegativesolutionsincewidthcannot

Chapter 1: Equations and Inequalities

benegative.Thewidthoftherectangular windowis17centimetersandthelengthis18 centimeters.

97. Let l representthelengthoftherectangle. Let w representthewidthoftherectangle. Theperimeteris26metersandtheareais40 squaremeters. 2226 13so13

Thedimensionsare5metersby8meters.

98. Let r representtheradiusofthecircle. Sincethefieldisasquarewitharea1250square feet,thelengthofasideofthesquareis

1250252  feet.Thelengthofthediagonal is2r

UsethePythagoreanTheoremtosolvefor r :

Theshortestradiussettingforthesprinkleris25 feet.

99. Let x =lengthofsideoforiginalsheetinfeet.

Lengthofbox:2 x feet

Widthofbox:2 x feet

Heightofbox:1foot

Discard0 x  sincethatisnotafeasiblelength fortheoriginalsheet.Therefore,theoriginal sheetshouldmeasure4feetoneachside.

100. Let x =widthoforiginalsheetinfeet.

Lengthofsheet:2 x

Lengthofbox:22 x feet

Widthofbox:2 x feet

Heightofbox:1foot

2 2 2 42221 4264 026 03 03 Vlwh xx xx xx xx xx

0or3xx

Discard0 x  sincethatisnotafeasiblelength fortheoriginalsheet.Therefore,theoriginal sheetis3feetwideand6feetlong.

101. a. Whentheballstrikestheground,the distancefromthegroundwillbe0. Therefore,wesolve

2 2 2 9680160 1680960 560 610 tt tt tt tt 

6or1tt

Discardthenegativesolutionsincethetime offlightmustbepositive.Theballwill strikethegroundafter6seconds.

b. Whentheballpassesthetopofthebuilding, itwillbe96feetfromtheground.Therefore, wesolve

2 2 2 96801696 16800 50 50 tt tt tt tt  

0or5tt Theballisatthetopofthebuildingattime 0 t  whenitisthrown.Itwillpassthetop ofthebuildingonthewaydownafter5 seconds.

102. a. Tofindwhentheobjectwillbe15meters abovetheground,wesolve 2 2 4.92015

Theobjectwillbe15metersabovethe groundafterabout0.99seconds(ontheway up)andabout3.09seconds(ontheway down).

b. Theobjectwillstrikethegroundwhenthe distancefromthegroundis0.Therefore,we solve

Theobjectwillstrikethegroundafterabout 4.08seconds.

c. 2 2 4.920100 4.9201000 tt tt 

Thereisnorealsolution.Theobjectnever reachesaheightof100meters.

103. Let x representthenumberofcentimetersthe lengthandwidthshouldbereduced. 12 x =thenewlength,7 x =thenewwidth. Thenewvolumeis90%oftheoldvolume.

2 2 2 (12)(7)(3)0.9(12)(7)(3) 357252226.8 35725.20 198.40 xx xx xx xx

Section 1.2: Quadratic Equations

2 (19)(19)4(1)(8.4)19327.4 2(1)2

0.45or18.55 x xx

Since18.55exceedsthedimensions,itis discarded.Thedimensionsofthenewchocolate barare:11.55cmby6.55cmby3cm.

104. Let x representthenumberofcentimetersthe lengthandwidthshouldbereduced. 12 x =thenewlength,7 x =thenewwidth. Thenewvolumeis80%oftheoldvolume.

(12)(7)(3)0.8(12)(7)(3)

357252201.6 35750.40 1916.80 xx xx xx xx

2 (19)(19)4(1)(16.8)19293.8 2(1)2

0.93or18.07 x xx

Since18.07exceedsthedimensions,itis discarded.Thedimensionsofthenewchocolate barare:11.07cmby6.07cmby3cm.

105. Let x representthewidthoftheborder measuredinfeet.Theradiusofthepoolis5 feet.Then5 x  representstheradiusofthe circle,includingboththepoolandtheborder. Thetotalareaofthepoolandborderis

2 (5) T Ax Theareaofthepoolis2(5)25 PA  Theareaoftheborderis

2 (5)25

BTP AAAx Sincetheconcreteis3inchesor0.25feetthick, thevolumeoftheconcreteintheborderis

 2 0.250.25(5)25 B Ax Solvingthevolumeequation:

 

  2 2 2 0.25(5)2527 102525108 101080 x xx xx

  

 2 2 10(10)4()(108) 2()

2.71or12.71 x xx

31.42100432

 

6.28

 Discardthenegativesolution.Thewidthofthe borderisroughly2.71feet.

Chapter 1: Equations and Inequalities

106. Let x representthewidthoftheborder measuredinfeet.Theradiusofthepoolis5 feet.Then5 x  representstheradiusofthe circle,includingboththepoolandtheborder. Thetotalareaofthepoolandborderis

2 (5) T Ax

Theareaofthepoolis2(5)25 PA  .

Theareaoftheborderis

108. Let x =thewidthand2x =thelengthofthe patio.Theheightis13footandtheconcrete availableis  827216  cubicfeet.. 2 2 1 (2)216 3 2216 3 32418 

2 (5)25

BTP AAAx .

Sincetheconcreteis4inches=13footthick,the volumeoftheconcreteintheborderis

112 33(5)25 B Ax

Solvingthevolumeequation:



2 2 2 1(5)2527 3 10252581 10810 x xx xx

  

 2 2 10(10)4()(81) 2()

2.13or12.13 x xx

31.42100324

6.28

  

Discardthenegativesolution.Thewidthofthe borderisapproximately2.13feet.

107. Let x representthewidthoftheborder measuredinfeet.

Thetotalareais(62)(102) T Axx 

Theareaofthegardenis61060 GA  .

feet.

109. Let x =thelengthofa12.9-inchiPadProina 16:9format.

Then 9 16 x =thewidthoftheiPad.Thediagonal ofthe12.9-inchiPadis9.7inches,sobythe Pythagoreantheoremwehave:

Theareaoftheborderis (62)(102)60

BTG AAAxx 

Sincetheconcreteis3inchesor0.25feetthick, thevolumeoftheconcreteintheborderis

 0.250.25(62)(102)60 B Axx

Solvingthevolumeequation:  2 2 2 0.25(62)(102)6027 6032460108 4321080 8270 xx xx xx xx     2 884(1)(27)8172 2(1)2 2.56or10.56 x xx

Discardthenegativesolution.Thewidthofthe borderisapproximately2.56feet.

Sincethelengthcannotbenegative,thelengthof theiPadis42600.96337inchesandthewidthis 42600.96 337 9 6.32 16  inches.Thus,theareaofthe iPadis42600.96942600.963371633771.11  square inches.

Let y =thelengthofa14.4-inch3:2format MicrosoftSurfacePro.Then 2 3 y =thewidthof theSurfacePro.Thediagonalofa14.4-inch SurfaceProis14.4inches,sobythe Pythagoreantheoremwehave:

Sincethelengthcannotbenegative,thelengthof theSurfaceProis11.98inchesandthewidthis 21866.24 7.99 313  inches.Thus,theareaofthe 14.4-inch3:2formatSurfaceProis 1866.241866.24 2 13313 95.7squareinches.  

TheSurfaceProformathasthelargerscreen sinceitsareaislarger.

110. Let x =thelengthofa8.3-inchiPadMiniina 4:3format.

Then3 4 x =thewidthoftheiPad.Thediagonal ofthe8.3-inchiPadis8.3inches,sobythe Pythagoreantheoremwehave:

Sincethelengthcannotbenegative,thelengthof theiPadis6.64inchesandthewidthis

3

4  inches.Thus,theareaofthe iPadis(6.64)(4.98)33.1  squareinches. Let y =thelengthofa8-inch16:10format AmazonFireHD8™.Then 10 16 y =thewidthof theFire.Thediagonalofa8-inchFireis8 inches,sobythePythagoreantheoremwehave:

16384

Sincethelengthcannotbenegative,thelengthof theFireis 16384 6.78399 356  inchesandthe widthis 1016384 4.240 16356  inches.Thus,thearea oftheAmazonFireis

6.783994.24028.8squareinches.  TheiPadMini™4:3formathasthelargerscreen sinceitsareaislarger.

111. Let h be1.1.Then 2 2 2 0.04(0.04)4(0.00025)(1.1) 2(0.00025) 1.10.000250.04



00.000250.041.1 35.3ftor124.7ft



   xx xx x 124.7ftdoesnotmakesenseinthecontextof theproblem,sotheansweris35.3ft.

112. Sincedisexpressedin1000’swewillsetd=12 andsolveforxusingtheQuadraticFormula. 2 2 2 2 0.26(0.26)4(0.33)(9.71) 2(0.33) 0.2613.0772 0.66 0.330.262.29 120.330.262.29 00.330.269.71

5.085or5.873 dxx xx xx x xx  

      Sothenearestyearwhentheearningswere12 occurredabout5yearsafter2018or2023.The negativevalue-5.873hasnomeaning.

Chapter 1: Equations and Inequalities

113. Wewillsetg=2.97andsolveforhusingthe QuadraticFormula. 2

0.00060.0153.04

2.970.00060.0153.04

0.015(0.015)4(0.0006)(0.07) 2(0.0006)

0.0150.000393 0.0012

Sotheestimatednumbersofhoursworkedbya studentwithaGPAof2.97is29hours.The value-4.02hasnomeaningsinceitisnegative.

114. Letxbethenumbersofmembersinthe fraternityandsbethesharepaidbyeach member.Then1470 s x  . Ifthereare7 memberswhocannotcontributethentheshare goesupby$5.Sowehavethefollowing equation: 51470 7 s x

or

571470sx

Solvingthesetwoequationstogether:

2 5714701470 and 1470571470 10290 14705351470 10290 5350 535102900 sxs x x x x x x x xx

2 535102900 720580 (42)(49)0 42or49 xx xx xx xx

Sincexisthenumberofmembers,itmustbe positivesothenumberofmembersis49.

115. Let a betheagetheindividualisabletostart savingmoney.Thenweneedtofindwherethe modelsareequal.Solvingthesetwoequations together: 2 2 2 2240(2240)4(25)(38540) 2(25)

22401163600 50 252400307001607840 252240385400

22401078.7 50

aa a a a aa aa

22401078.722401078.7 or 5050 66.4or23.2 

Since x istheagetostartsaving,itmakessense thattheanswerisapproximateatage23.

116.  

 2 1365 2 3130 31300 13100 nn nn nn nn     13or10nn

Sincethenumberofsidescannotbenegative,we discardthenegativevalue.Apolygonwith65 diagonalswillhave13sides.   2 1 380 2 3160 31600 nn nn nn    1,3,160abc   2 33411603646 212 n   

Neithersolutionisaninteger,sothereisno polygonthathas80diagonals.

117. Therootsofaquadraticequationare

118. Therootsofaquadraticequationare

119. Inordertohaveonerepeatedsolution,weneed thediscriminanttobe0.

121. For20 axbxc : 24 2   bbac x a For20 axbxc :

120. Inordertohaveonerepeatedsolution,weneed thediscriminanttobe0.

122. For20 axbxc : 2 1 4 2 bbac x a  and 2 2 4 2 bbac x a

For20 cxbxa :

123. If x =originalwidthand y =originallength,then 11 or xyx y .Theratioofsidelengthsis

2 1  x yy .Foldingalongthelongestsideresults insidesoflength1and2  y x y whoseratiois

Thefirstequationhasthesolutionset  1 whilethesecondequationhasnosolutions.

125. Answerswillvary.Methodsmayincludethe quadraticformula,completingthesquare, graphing,etc.

126. Answerswillvary.Knowingthediscriminant allowsustoknowhowmanyrealsolutionsthe equationwillhave.

127. Answerswillvary.Onepossibility: Twodistinct:23180 xx Onerepeated:214490 xx Noreal:240 xx

128. Answerswillvary.

124. a. 29 x  and3 x  arenotequivalent becausetheydonothavethesamesolution set.Inthefirstequationwecanalsohave 3 x  . b. 9 x  and3 x  areequivalentbecause 93  .

c.  1212xxx  and21 xx are notequivalentbecausetheydonothavethe samesolutionset.

Section 1.3

1. Integers:  3,0

Rationals:  3,0,6 5

2. True;thesetofrealnumbersconsistsofall rationalandirrationalnumbers.

3.

22 3323 232323 323 23 323 43 323

4. real;imaginary;imaginaryunit

5. False;theconjugateof25i  is25i .

6. True;thesetofrealnumbersisasubsetoftheset ofcomplexnumbers.

7. False;if23i isasolutionofaquadratic equationwithrealcoefficients,thenits conjugate,23i  ,isalsoasolution.

8. b

9. a 10. c

11. (23)(68)(26)(38)85 iiii 

12. (45)(82)(4(8))(52) 47 iii i 

13. (32)(44)(34)(2(4)) 76 iii i 

14. (34)(34)(3(3))(4(4)) 606 iii i 

15. (25)(86)(28)(56) 611 iii i 

16. (84)(22)(82)(4(2)) 106 iii i 

17. 3(26)618 ii

18. 4(28)832 ii

19. 2 3(76)21182118(1) 1821   iiiii i

20. 2 3(34)912912(1)129 iiiiii 

21. (34)(2)63842 654(1) 105 iiiii i i  

25. 2 1010343040 3434349121216 30403040 916(1)25 3040 2525 68 55 ii iiiiii ii i i

26. 2 1313512 512512512 65156 256060144 6515665156 25144(1)169 65156 169169 512 1313 i iii i iii ii i i

27. 2 2 222 2(1)1212 (1)1 iiiii iiii ii i

28. 2 2 222 222 2(1)121 2(1)22 iiiii iiii ii i

22. (53)(2)105632 103(1) 13  

iiiii i i

iiiii

23. (5)(5)25552 25(1) 26

24. (3)(3)9332 9(1) 10 iiiii

29. 2 2 66166 1111 67(1)5757 1(1)222 iiiiii iiiiii iii

30. 2 2 232312233 1111 253(1)1515 1(1)222 iiiiii iiiiii iii

31. 2 1311332 2 224224 13313 (1) 42422 iii ii

Chapter 1: Equations and Inequalities

32. 2 3133112 2 224224 33113 (1) 42422 iii ii

33. 22 (1)1212(1)2 iiiii

34. 22 (1)1212(1)2 iiiii

35.

11 2322122211 (1) iiiiiiii

7 1427(1)1 ii

37. 20 2020210 10 111 () 111 (1)1

i iii

38. 23 2322122211 112 1111 () 111 (1)(1) i iiiiii iii i iii ii

39. 3 62355(1)5156ii

40. 32 444(1)4 iiiii 

41. 3532 2 64(64) (64(1))1(10)10 iiii iiii

42. 3222 421421 4(1)2(1)1 421 34 iiiii i i i    

43. 32 2 (1)(1)(1)(1)(12)(1) (121)(1)2(1) 2222(1) 22 iiiiiii iiii iii i    

44. 44 (3)181181(1)182 ii

45. 7277 (1)(1(1))(0)0 iiii

46. 42 2(1)2(1)(1(1))2(0)0 ii

47.

 432 86422222 432 (1)(1)(1)1 1111 0

    iiiiiiii 48.  32 753222 32 (1)(1)(1) 0 iiiiiiiiiii iiii iiii 

  

49. 42i 

50. 93i 

51. 255 i

52. 648 i 

53. 124323 ii

54. 189232 ii

55. 2001002102 ii

56. 459535 ii

57. 2 (34)(43)1291612 916(1) 25

5 iiiii i 

58. 2 (43)(34)1216912 169(1) 25

5 iiiii i    

59. 2 2 40 4 4 2 x x x xi  

Thesolutionsetis 2,2. ii





66. 2250xx 22 1,2,5 4(2)4(1)(5)42016 (2)162412 2(1)2 abc bac i xi

Thesolutionsetis  12,12 ii .

60. 240 x   (2)(2)0 2or2

Thesolutionsetis2,2.

61. 2160 x    440 4or4

67. 2 251020 xx 22 25,10,2 4(10)4(25)(2)100200100 (10)100101011 505055 

abc bac i xi

Thesolutionsetis  1111 , 5555 ii

Thesolutionsetis4,4.

62. 2250 x   225 255

Thesolutionsetis5,5. x xi ii 

63. 26130xx  22 1,6,13, 4(6)4(1)(13)365216 (6)166432 2(1)2

Thesolutionsetis32,32. abc bac i xi ii

64. 2480xx 22 1,4,8 444(1)(8)163216 41644 22 2(1)2 abc bac i xi 

Thesolutionsetis

65. 26100xx 22 1,6,10 4(6)4(1)(10)36404 (6)4623 2(1)2 abc bac i xi





68. 2 10610 xx 22 10,6,1 464(10)(1)36404 646231 2(10)201010

abc bac i xi  

Thesolutionsetis  3131 , 10101010 ii

69. 2 2 512 5210 xx xx    22 5,2,1 424(5)(1)42016 (2)162412 2(5)1055 abc bac i xi

  

Thesolutionsetis  1212 , 5555 ii .

70. 2 2 1316 13610 xx xx   22 13,6,1 4(6)4(13)(1)365216 (6)166432 2(13)261313

abc bac i xi  

  

Thesolutionsetis  3232 , 13131313 ii .

Chapter 1: Equations and Inequalities

71. 210xx 1,1,1,abc 22414(1)(1)143 131313 2(1)222

bac i xi

Thesolutionsetis1313 , 2222 ii

72. 210xx 1,1,1abc

224(1)4(1)(1)143 (1)31313 2(1)222 bac i xi

Thesolutionsetis1313 , 2222 ii

73. 3640  x  2 2 (4)4160 404 or4160 

75. 4 4 16 160 x x     22 2 440 (2)(2)40 xx xxx

2 2 20or20or40 2or2or4 2or2or42 xxx xxx xxxi    Thesolutionsetis  2,2,2,2. ii

 

76. 4 4 1 10 x x     22 2 110 (1)(1)10 xx xxx   2 2 10or10or10 1or1or1 1or1or1 xxx xxx xxxi

 Thesolutionsetis  1,1,,. ii

abc bac i xi

xxx xx xx 22 1,4,16 444(1)(16)166448 448443 223 2(1)2

Thesolutionsetis  4,223,223. ii

74. 3270 x   2 2 (3)390 303 or390 xxx xx xx

22 1,3,9 4(3)4(1)(9)93627 (3)27333333 2(1)222 abc bac i xi

Thesolutionsetis3333333,,. 2222 ii 

77. 4213360xx  22 22 22 940 90or40 9or4 9or4 3or2 xx xx xx xx xixi







  Thesolutionsetis  3,3,2,2. iiii

78. 42340xx   22 2 140 (1)(1)40 xx xxx 

 2 2 10or10or40 1or1or4 1or1or42 xxx xxx xxxi

 

 Thesolutionsetis  1,1,2,2. ii

79. 2 3340 xx 22 3,3,4 4(3)4(3)(4)94839 abc bac



 Theequationhastwocomplexsolutionsthatare conjugatesofeachother.

Section 1.3: Complex Numbers; Quadratic Equations in the Complex Number System

80. 2 2410 xx 22 2,4,1 4(4)4(2)(1)1688 abc bac  

Theequationhastwounequalrealnumber solutions.

81. 2 2 234 2340 xx xx   22 2,3,4 434(2)(4)93241 abc bac   Theequationhastwounequalrealsolutions.

82. 2 2 62 260 xx xx   22 1,2,6 4(2)4(1)(6)42420 abc bac

Theequationhastwocomplexsolutionsthatare conjugatesofeachother.

83. 2 91240 xx 22 9,12,4 4(12)4(9)(4)1441440 abc bac 

Theequationhasarepeatedrealsolution.

84. 2 41290 xx 22 4,12,9 4124(4)(9)1441440 abc bac 

Theequationhasarepeatedrealsolution.

85. Theothersolutionis2323. ii

86. Theothersolutionis44. ii

87. 343434346zziiii 

88. 8383 83(83) 8383 066 wwii ii ii ii    





89. 2 (34)(34) (34)(34) 9121216 916(1)25 zzii ii iii



90. 34(83) 3483 57 57 zwii ii i i 

91. 2 2 181834 343434 54723454754 9121216916 5075 23 25 Viii Z Iiii iiii iii i i 

Theimpedanceis23i  ohms.

92. 12 2 11111(43)(2) 243(2)(43) 626262 8643823112 ii ZZZiiii iii iiiii 

So, 2 2 11211262 626262 662212466104 3612124364 701071 4044 iii Z iii iiii iii i i

Thetotalimpedanceis71 44 i  ohms.

93. ()() 2 zzabiabi abiabi a    () () 2 zzabiabi abiabi abiabi bi 

  

94. zabiabiabiz 

95. ()() ()() ()() ()() zwabicdi acbdi acbdi abicdi abicdi zw 

    

Chapter 1: Equations and Inequalities





96. 2 ()() ()() ()() zwabicdi acadibcibdi acbdadbci acbdadbci



 2 ()() ()() zwabicdi abicdi acadibcibdi acbdadbci











97. 





 abiabi aabibiaabibi aabibiaabibi aabibaabib 2222 22 22 () 2()0  

22 2222 222222 2222 2()2() 22 22

 abab ab abba Anycomplexnumberoftheform  aai or aai willwork.

98. Let32  u in320  x sothat330 xu . Then,22()()0  xuxuxu .Fromthefirst factorwefind32 xu .Fromthesecond factor,usethequadraticformulatoget 22 233 ()()41 21 33223 22222

     uuu x uuuuii

Thesolutionsetis: 33 3223 2,22   

i

99. 2 (5)(5)();  xyxy let5 ux (so 5 xu and5 vy so5 yv . Substitutinggives2 () uvuv or 220 uuvv whichisquadraticinu.Using thequadraticformulagives 223 41 212    vvvvv x .Since x isarealnumber, u mustalsobearealnumber.

Thisisonlypossibleifv=0whichthenmakes u =0.Therefore,055  x and 055 y ,so5510 xy

100 – 102. Answerswillvary.

103. Answerswillvary.Acomplexnumberisthe sumordifferenceoftwonumbers(realand imaginarypartsofthecomplexnumber)justasa binomialisthesumordifferenceoftwo monomialterms.Wemultiplytwobinomialsby usingtheFOILmethod,anapproachwecanalso usetomultiplytwocomplexnumbers.

104. Althoughthesetofrealnumbersisasubsetof thesetofcomplexnumbers,notallrulesthat workintherealnumbersystemcanbeusedin thelargercomplexnumbersystem.Therulethat allowsustowritetheproductoftwosquare rootsasthesquarerootoftheproductonly worksintherealnumbersystem.Thatis, abab  onlywhen a and b arereal numbers.Inthecomplexnumbersystemwe mustfirstconverttheradicalstocomplexform. Inthiscasethismeansweneedtowrite9as 19913i  .Thenwecanmultiplyto get

2 99339919 iii

False;youcanalsousetheQuadraticFormulaor completingthesquare.

quadraticinform

c

Section 1.4

9. 211 t

Check:2(1)111

Thesolutionsetis{1}.

Check:3(0)442

Thesolutionsetis{0}.

11. 346 t 

Sincetheprincipalsquarerootisnevernegative, theequationhasnorealsolution.

12. 532 t 

Sincetheprincipalsquarerootisnevernegative, theequationhasnorealsolution.

13. 31230 x

Check:3312(13)32730

Thesolutionsetis{13}.

16.

52551211211  Thesolutionsetis{1}.

42 444 2 2 2 165 165 1625 9 3 x x x x x

Check3:316916255

Thesolutionsetis   3,3.

17.

  22 2 2 8 8 64 640 640 xx xx xx xx xx      0or64xx

Check0:080 00   Check64:64864 6464  

Thesolutionsetis  0,64

18.    22 2 2 3 3 9 90 90 xx xx xx xx xx      0or9xx

Chapter 1: Equations and Inequalities

Check0:030 00   Check9:939 99  

Thesolutionsetis  0,9 19. 152 xx

Check–5:152(5)2555

Check3:152(3)933

Disregard5asextraneous. Thesolutionsetis{3}. x

Check–4:12(4)1644

Check3:123933

Disregard4asextraneous. Thesolutionsetis{3}. x

Check:2221 22

Thesolutionsetis{2}.

xx

22 2 2 2 2 21

Check:22(2)1 22

Theequationhasnorealsolution.

23. 242 xxx

2 22 22 42 444 85 8 5 xxx xxxx x x

Check: 2 88842 555 6482 4 2555 42 255 22 55

Thesolutionsetis

24. 2 32 xxx

8 5

2 22 22 32 344 31 1 3 xxx xxxx x x

Check: 2 111 32 333 3115 393

Sincetheprincipalsquarerootisalwaysanonnegativenumber;1 3 x  doesnotcheck. Thereforethisequationhasnorealsolution.

Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations

25. 331xx  22 2 2 313 31(3) 3169 098

0(1)(8) 1or8 xx xx xxx xx xx xx 

Check1:33(1)13451

Check8:33(8)132588

Discard1 x  asextraneous. Thesolutionsetis{8}.

26. 2122 xx

22 2 2 1222 122(2) 12244 028 (2)(4)0 2or4 xx xx xxx xx xx xx

28.

  

 22 2 2 132 15 1(5) 11025 01124 0(3)(8) 3or8 xx xx xx xxx xx xx xx

   

Check3:1(3)33211

Check8:1(8)38206

  Discard8 x  asextraneous. Thesolutionsetis{-3}.

29. 3572 xx



Check2:2+122(2)21662

Check4:2122(4)2444

Discard2 x  asextraneous. Thesolutionsetis{4}.





27.  22 2 2 3(10)4 3(10)4 3(10)(4) 330816 0514 0(7)(2) 7or2 xx xx xx xxx xx xx xx

 

22 22 2 2 2 2 3527 3527 354477 21647 (216)47 46425616(7) 46425616112 4801440 20360 (2)(18)0 xx xx xxx xx xx xxx xxx xx xx xx  

       2or18xx



Check2:3(2)527 191322

Check18:3(18)5187

 

49257522

 Discard2 x  asextraneous. Thesolutionsetis{18}.

Check7:3(710)49417

Check2:3(210)436422

 

Discard7 x  asextraneous. Thesolutionsetis{2}.

Chapter 1: Equations and Inequalities

30. 3721 xx

22 22 2 2 3712 3712 371222 2422 22 (2)2 442 320 (1)(2)0 1or2 xx xx xxx xx xx xx xxx xx xx xx



Check–1:3(1)712 412131

Check2:3(2)722 101011

Discard1 x  asextraneous.

Thesolutionsetis{2}.

31. 3112 xx

22 22 2 2 2 3121 3121 314411 2241 (22)41 48416(1) 2144 650 (1)(5)0 xx xx xxx xx xx xxx xxx xx xx

Check1:3(1)111 402022

Check5:3(5)151 1644222

Thesolutionsetis

32. 2311 xx

 



  22 22 2 2 2 2311 2311 231211 121 (1)21 214(1) 2144 230 (1)(3)0 xx xx xxx xx xx xxx xxx xx xx

1or3xx

Check–1:2(1)311 101011

Check3:2(3)331 943211 

Thesolutionsetis  1,3

22 32 32 32 23 xx xx xx xx

  22 2 2 23 469 0109 019 xx xxx xx xx 

1or9xx Check1:Check9: 3211 321 11 11

Discard9 x  asextraneous. Thesolutionsetis{1}.

Discard4 x  asextraneous.

Thesolutionsetis{25}.

35. 1/2 314 x 

1/22

Check:  1/21/2 351164 

Thesolutionsetis{5}.

36. 1/2 352 x     2 1/22352 354 39 3 x x x x    

Check:  1/21/2 33542 

Thesolutionsetis{3}.

37. 1/3 522 x     3 1/33522 528 510 2 x x x x    

Check:  1/31/3 52282 

Thesolutionsetis{2}.

38. 1/3 211 x     3 1/33211 211 22 1 x x x x    

Check:    1/31/3 21111 

Thesolutionsetis{1}.

39. 21/295 x 

  21/222 2 2 95 925 16 164 x x x x      

Check4:  21/21/2 49255 

Check4:  21/21/2 49255 

Thesolutionsetis  4,4

40. 21/2169 x    21/222 2 2 169 1681 97 97 x x x x      

Check97:  21/21/2 9716819  

Check97:  21/21/2 9716819  

Thesolutionsetis 97,97.

Chapter 1: Equations and Inequalities

41.  3/21/2 1/2 30 30 xx xx   1/20or30 0or3 xx xx  

Check0:3/21/2030000 

Check3:3/21/233333330 

Thesolutionsetis  0,3

42.  3/41/4 1/41/2 90 90 xx xx   1/40 0 x x   or 1/29 81 x x  

Check0:3/41/4090000 

Check81:3/41/48198127270 

Thesolutionsetis  0,81

43.  42 22 22 540 410 40or10 2or1 xx xx xx xx

Thesolutionsetis  2,1,1,2.

44.

 42 22 2 10250 550 50 5 xx xx x x    

Thesolutionsetis  5,5.

45.  42 22 6510 6110   xx xx 22 22 610or10 61or1 Notrealor1 

 xx xx x Thesolutionsetis 1,1.

46.

 42 22 25120 2340 xx xx 

22 22 230or40 23or4 Notrealor2 xx xx x   

Thesolutionsetis 2,2.

47.

 63 33 780 810 xx xx   33 33 80or10 8or1 2or1 xx xx xx    Thesolutionsetis 2,1.

48.

 63 33 780 810 xx xx



 33 33 80or10 8or1 2or1 xx xx xx  

 Thesolutionsetis   1,2.

49.  2 272120xx

 Let22 2,sothat2.uxux 

 27120 340 uu uu 



 30or40 3or4 23or24 5or6 uu uu xx xx

 



Thesolutionsetis   6,5.

50.  2 252560 xx  Let22 25sothat25.uxux 

 260 320 uu uu   30or20 3or2 253or252 7 1or2 uu uu xx xx   

 

Thesolutionsetis  7,1. 2

51.  2 491049250 xx

 Let22 49sothat49.  uxux

Section 1.4: Radical Equations; Equations Quadratic in Form; Factorable Equations

uu u u u x x x

2 2 10250 50 50 5 495 414 7 2

Thesolutionsetis  7 . 2

52.  2 22200 xx 

Let22 2sothat2.uxux 

2200 540 uu uu

50or40 5or4 25or24 7or2 uu uu xx xx

Thesolutionsetis 2,7.

53.  2 21513 ss

Let22 1sothat1.usus   2 2 253 2530 2130 uu uu uu

210or30 1or3 2 11or13 2 3or2 2 uu uu ss ss

Thesolutionsetis 3,2. 2

54.

2 315120 yy

Let22 1sothat1.uyuy 

320or10 2or1 3 12or11 3 5or2 3 uu uu yy yy

Thesolutionsetis 5,2. 3

55.  40 140 xxx xx 

 1 4 122 4 1 16 0or140 14 xx x x x x     

Check:

 1111 16161616 111 16164 11 1616 0:04(0)00 00 :40 40 0 00 x x      

Thesolutionsetis  1 0,. 16

56. 80xx

  22 2 2 8 8 64 064 064 xx xx xx xx xx 



 0or64xx

Check:0:0800 00 64:648640 64640 x x    

Thesolutionsetis  0

Chapter 1: Equations and Inequalities

57. 20 xx

Let2sothat. uxux   2 2 20 200 540 uu uu uu    5040 or 54 or or 54 notorpossible16 uu uu xx x

Check:161620 16420 

Thesolutionsetis  16

58. 6 xx Let2sothat. uxux 

2 2 6 60 320 uu uu uu

3020 or 32 or or 32 notorpossible4 uu uu xx x

Check:446 426 

Thesolutionsetis   4.

59. 1/21/4210tt

Let1/421/2 sothat.utut 

2 2 1/4 210 10 10 1 1 1 uu u u u t t

Check:  1/21/4 12110 1210 00

Thesolutionsetis  1

60. 1/21/4440zt Let1/421/2 sothat.uzuz 

 2 2 1/4 440 20 20 2 2 16 uu u u u z z

    

Check:  1/21/4 1641640 4840 00   

Thesolutionsetis 16.

61. 1/21/4320xx Let1/421/2 sothat.uxux 

 2320 210 uu uu   1/41/4 2or1 2or1 16or1 uu xx xx   

Check:   1/21/4 1/21/4 16:1631620 4620 00 1:13120 1320 00 x x 

 

  

Thesolutionsetis 1,16.

62. 1/21/4 4940 xx Let1/421/2 sothat.uxux  2 4940 uu 2 (9)(9)4(4)(4)917 2(4)8 u    1/4 4 917 8 917 8 x x       

Chapter 1: Equations and Inequalities

Since5415410,isnotreal. 2 x 

Since x isafourthroot, 541 2 x   isalso notreal.Therefore,wehaveonlyonepossible solutiontocheck:541: 2 x

Check541:

22 2 2 3or2 33or32 Notpossibleor34 340 410 4or1 uu xxxx xx xx xx xx

Check4: x 

22 4344346 161216126 161246 66

Check1: x 

22 1311316 13136 446 66 

Thesolutionsetis 4,1. 66. 22 222 332 Let3sothat3. xxxx uxxuxx

2 2 2 20 120 uu uu uu

22 2 2 1or2 31or32 Notpossibleor34 340 410 4or1 uu xxxx xx xx xx xx

Check4: 43443416124 422

22 22

Check1: 131131134 422 x x 

Thesolutionsetis1,4.

34 The43 solutionsetis,. 69. 21 3760 xx

Let122 sothat.uxux

00

Thesolutionsetis341341 , 88

2/31/3 2530 xx Let1/322/3 sothat.uxux

2 2530 2130 uu uu

1/31/3 3 3331/31/3 1or3 2 1or3 2 1or3 2 1or27 8 uu xx xx xx

Check 2/31/3111 :2530 888 11 2530 42 1530 22 330 00

27:2/31/322752730 295330 181530 330 00 x

Thesolutionsetis

1,27 8