Solutions for College Algebra Enhanced With Graphing Utilities 8th Us Edition by Sullivan

Page 1


Chapter 1 Graphs, Equations, and Inequalities

Section 1.1

0

3. x-coordinate; y-coordinate;quadrants

4. False;pointsthatlieinQuadrantIVwillhavea positive x-coordinateandanegative y-coordinate. Thepoint 1,4liesinQuadrantII.

5. d 6. c

7. (a)QuadrantII (b) x-axis

(c)QuadrantIII (d)QuadrantI (e) y-axis (f)QuadrantIV

8. (a)QuadrantI (b)QuadrantIII (c)QuadrantII (d)QuadrantI (e) y-axis (f) x-axis

9. Thepointswillbeonaverticallinethatistwo unitstotherightofthey-axis.

10. Thepointswillbeonahorizontallinethatis threeunitsabovethex-axis.

11. 1,4;QuadrantII

12. (3,4);QuadrantI

13. (3,1);QuadrantI

14. 6,4;QuadrantIII

15. min11 max5 scl1 min3 max6 scl1 X X X Y Y Y

16. min3 max7 scl1 min4 max9 scl1 X X X Y Y Y

17. min30 max50 scl10 min90 max50 scl10 X X X Y Y Y

18. min90 max30 scl10 min50 max70 scl10 X X X Y Y Y

19. min10 max110 scl10 min10 max160 scl10 X X X Y Y Y

20. min20 max110 scl10 min10 max60 scl10 X X X Y Y Y

21. min6 max6 scl2 min4 max4 scl2 X X X Y Y Y

22. min3 max3 scl1 min2 max2 scl1 X X X Y Y Y

23. min6 max6 scl2 min1 max3 scl1 X X X Y Y Y

24. min9 max9 scl3 min12 max4 scl4 X X X Y Y Y

25. min3 max9 scl1 min2 max10 scl2 X X X Y Y Y

26. min22 max10 scl2 min4 max8 scl1 X X X Y Y Y

Chapter 1: Graphs, Equations, and Inequalities

27. 4 y xx

(0,0)isonthegraphoftheequation.

28. 32 y

(0,0)and(1,–1)areonthegraphofthe equation.

29. 229yx 22 309

30. 31

(0,1)and(–1,0)areonthegraphofthe equation.

31. 224xy 22 024 44

(0,2)and2,2areonthegraphofthe equation.

32. 2244xy 22 0414 44   22 2404 44    22 1 2 244 54   (0,1)and(2,0)areonthegraphoftheequation.

33. (–1,0),(1,0)

34. (0,1)

35. ,0,,0,(0,1)22 

36. (–2,0),(2,0),(0,–3)

37. 1,0, 0,2,  0,2

38. (2,0),(0,2),(–2,0),(0,–2)

39. (–4,0),(–1,0),(4,0),(0,–3)

40. (–2,0),(2,0),(0,3)

41. 2 yx 42. 6 yx

44. 39yx
45. 21 yx
46. 29 yx
47. 24 yx
48. 21 yx
49. 236 xy
50. 5210 xy

Chapter 1: Graphs, Equations, and Inequalities

2 9436 xy

2 44 xy

53. 213yx

Thex-interceptis6.5 x  andthey-interceptis 13 y

54. 314yx

Thex-interceptis4.67 x  andthey-interceptis 14 y  .

55. 2 215yx

Thex-interceptsare2.74 x  and2.74 x  They-interceptis15 y

56. 2 319yx

Section 1.1: Graphing Utilities; Introduction to Graphing Equations

59. 2 25375337or 3 x xyy

Thex-interceptsare2.52 x  and2.52 x  They-interceptis19 y  .

57. 3243343 or 2 x xyy 

Thex-interceptis14.33 x  andthey-intercept is21.5 y 

58. 4582482 or 5 x xyy 

Thex-interceptis20.5 x  andthey-interceptis 16.4 y 

Thex-interceptsare2.72 x  and2.72 x  They-interceptis12.33 y  .

60. 2 22352335or 3 x xyy

Thex-interceptsare4.18 x  and4.18 x  They-interceptis11.67 y  .

61. If   2,5isshifted3unitsrightthenthex coordinatewouldbe23  .Ifitisshifted2units downthenthey-coordinatewouldbe5(2) 

Chapter 1: Graphs, Equations, and Inequalities

Thus the new point would be

( ) 23,5(2)(5,3) ++ - =

( ) 1,61(2)- +64 +

62. If is shifted 2 units left then the x coordinate would be . If it is shifted 4 units up then the y-coordinate would be . Thus the new point would be .

( ) 1(2),64(3,10)- + - += -

63. a. 25 feet

b. 23.2 ft; 34.1 ft

c. (0,6), The shot is released at a height of 6 feet; (48.7, 0), the shot hits the ground after traveling a horizontal distance of 48.7 feet.

64. a. 20 meters

b. 12 seconds; 36 meters

c. (0, 2), The discus is released at a height of 2 meters; (18, 0), the discus hits the ground after 18 seconds.

65. a. $19.95; $19.95

b. $182.45

c. (0, 19.95), The membership plan costs $19.95 per month

66. a. 1.5 miles

b. 1 mile

c. (0, 2), Caleb’s friend lives 2 miles from his house; (28, 0), it takes Caleb 28 minutes to ride home.

67.

68.

(1,0),(1,0),(0,1),(0,1) (2,0),(3,0),(0,2),(0,0),(0,2)

69. Answers will vary.

b. Since for all , the graphs of are the same.

c. For , the domain of the variable is ; for , the domain of the variable is all real numbers. Thus,

2 xx = x 2 and yxyx == ( ) 2 yx = x 0 x ³ yx = x ( ) 2 only for 0. xxx = ³

d. For , the range of the variable is ; for , the range of the variable is all real numbers. Also,

2 yx = y 0 y ³ yx = y 2 only if 0. xxx = ³

71. Answers will vary.

70. a.

Section 1.2: Solving Equations Using a Graphing Utility; Linear and Rational Equations

72. Answerswillvary.Acompletegraphpresents enoughofthegraphtotheviewersotheycan “see”therestofthegraphasanobvious continuationofwhatisshown.

73. Answerswillvary.

Section 1.2

1. Usingthedistributiveproperty:

2.  23512(3)35[(3)1] 635(4)6320 31017 xx

3. No,4 x  isnotinthedomainof3 4 x because itmakesthedenominatorequal0.

4. LCM=  2 1323 x xxx 

5. b

6. identity

7. linear;first-degree

8. d

9. True;equationsthatarecontradictionshaveno solution.

10. False;whenusingagraphingutility,weoften obtaindecimalapproximationstosolutions.

11. Dividebothsidesby7.Thesolutionsetis  3

12. Dividebothsidesby6.Thesolutionsetis  4

13. Subtract35frombothsides,thendivideboth sidesby5.Thesolutionsetis  7

14. Subtract18frombothsides,thendivideboth sidesby6.Thesolutionsetis 3.

15. Add3tobothsides,thendividebothsidesby2. Thesolutionsetis  3 2

16. Subtract4frombothsides,thendivideboth sidesby3.Thesolutionsetis  4 3

17. Multiplybothsidesby4.Thesolutionsetis7 5

18. Multiplybothsidesby3 2 .Thesolutionsetis27 4

19. 3420xx ;UseZERO(orROOT)onthe graphof3142 yxx

Thesolutionsetis  2.21,0.54,1.68

20. 3810xx ;UseZERO(orROOT)onthe graphof3181y xx  .

Thesolutionsetis  2.89,0.13,2.76

21. 4 2532 xx  ;UseINTERSECTonthe graphsof4125yx and232 yx .

.

Chapter 1: Graphs, Equations, and Inequalities

Thesolutionsetis  1.55,1.15

22. 42123xx  ;UseINTERSECTonthe graphsof411yx and2223yx

Thesolutionsetis  1.11,1.11.

23. 43 25840 xxx ;UseZERO(orROOT) onthegraphof4312584 yxxx

Thesolutionsetis  1.23,0.60.

24. 432 3450 xxx ;UseZERO(orROOT) onthegraphof4321345 yxxx

Thesolutionsetis 0.85,0.94.

25. 325720 32 xxx  ; UseZERO(orROOT)onthegraphof 32 15/37/22 yxxx 

Thesolutionsetis  2.69,0.49,1.51

26. 432715 320 32 xxxx  ;UseZERO(or ROOT)onthegraphof

432 137/315/22 yxxxx 

Thesolutionsetis  1.64,0.31,1.31,3.02

27. 432 2521 2 3232 xxxx 

UseINTERSECTonthegraphsof  43 12/325/2 yxxx  and 2 2(2/3)1/2 yx

Section 1.2: Solving Equations Using a Graphing Utility; Linear and Rational Equations

Thesolutionsetis  2.86,1.34,0.20,1.00

28. 321154 45 xxx

UseINTERSECTonthegraphsof 3 1/45y xx and22/54yx anda standardviewingwindow(10to10forboth x and y).

Thesolutionsetis  4.47,0.80,4.47

29. 4252110xxx

UseZERO(orROOT)onthegraphof 42 15211 yxxx

Therearenorealsolutions.

30. 42 38290 xxx  UseZERO(orROOT)onthegraphof 42 13829 yxxx 

Therearenorealsolutions.

31. 34 3444 34 34 24 24 22 2 xx xx xx x xxx x x x  

 Thesolutionsetis  2

32. 295 29959 259 25595 39 39 33 3 xx xx xx x xxx x x x    

 Thesolutionsetis  3

33. 263tt  26636 29 29 39 39 33 3 tt tt tttt t t t  

Thesolutionsetis  3

34. 5618 y y  566186 524 524 624 624 66 4 yy yy yyyy y y y 

Thesolutionsetis 4.

Chapter 1: Graphs, Equations, and Inequalities

35. 629 66296 23 2232 33 33 33 1 xx xx xx x xxx x x x

Thesolutionsetis  1

36.

322 32323 21 21 1 111 1 xx xx xx x xxx x x x

Thesolutionsetis 1.

37. 3(53)8(1)  xx 15988 159158815 9823 988238 23

xx xx xx x xxx x

Thesolutionsetis 23.

38. 3(2)21 x x  6321 636216 327 32272 57 57 55 7 5 xx xx xx x xxx x x x

Thesolutionsetis  7 5

39.  832310 832310 52310 5223102 538 53383 28 xxx xxx xx xx xx x xxx x

28 22 4 x x  

Thesolutionsetis 4.

40.  72110 72110 2810 288108 22 22 22 1 x x x x x x x

Thesolutionsetis  1

41.  2356334556 610618121556 12282018 82818 810 5 4 xxxx xxxx xx x x x

Thesolutionsetis5 4

.

42.  52231412 5106248 1247 6124 616 8 3 x xxx x xxx xx x x x

Thesolutionsetis8 3

43.

45. 211 323 pp

Thesolutionsetis 2.

46. 114 233 p  11466 233 328 32383 25 25 22 5 2 p p p p p p

Thesolutionsetis  5 2

47. 0.20.90.5  mm 0.20.50.90.50.5 0.30.9 0.30.9 0.30.3 3 

mmmm m m m

Thesolutionsetis  3

48. 0.91tt  0.91 0.11 0.11 0.10.1 10 tttt t t t   

Thesolutionsetis  10

49.  122 37 12 2121212 37 773642 xx xx xx

101342 1013134213 1029 1029 1010 29 10 x x x x x 

Thesolutionsetis29 10

Chapter 1: Graphs, Equations, and Inequalities

53. 123 24 12344 24 283 28232 8or8 x xx x xx xxxx xx

Thesolutionsetis 8. 54. 311 36 31166 36 182 18222 183 183 33 6or6 x xx x xx xxxx x x xx

  

Thesolutionsetis  6

55. 7(1)(1)2xxx  22 22 2222 7721 6721 6721 6721 672212 471 47717 48 482 44 xxxxx xxxx x xxxxx xx xxxx x x x x x















 

Thesolutionsetis  2

Thesolutionsetis

56.

Section 1.2: Solving Equations Using a Graphing Utility; Linear and Rational Equations

59. 23 33 3333 312 312 312 312 312 33 4

     ppp ppp ppppp p p p

Thesolutionsetis  4

60. 23 33 48 48 48 2 www www w w 

Thesolutionsetis

57. (23)(21)(4) xxxx 22 2222 23274 2322742 374 37747 44 441 44 xxxx x xxxxx xx xxxx x x x

Thesolutionsetis 1.

58. (12)(21)(2) xxxx  22 2222 2252 222522 52 5525 62 621 663 xxxx x xxxxx xx xxxx x x x

Thesolutionsetis   1 3 .

61.

Thesolutionsetis   2.

 32 22 2322 22 322 362 462 48 2 x xx x xx xx xx xx x x x 

 

Thissolutionisnotinthedomainofthevariable soitmustbediscarded.Theequationhasno solution.Thesolutionsetis  or  .

62.

262 33 26 332 33 2623 2626 2212 412 3 x xx x xx xx xx xx xx x x 

Thissolutionisnotinthedomainofthevariable soitmustbediscarded.Theequationhasno solution.Thesolutionsetis  or  .

Chapter 1: Graphs, Equations, and Inequalities

63.

22 22 22 243 442 24344 442 2432 2436 2103 510 2 x x xx x xx x xx xx xx xx x x

Thissolutionisnotinthedomainofthevariable soitmustbediscarded.Theequationhasno solution.Thesolutionsetis  or  .

64.

Thissolutionisnotinthedomainofthevariable soitmustbediscarded.Theequationhasno solution.Thesolutionsetis   or 

65.  3 22 232 236 6 6 x x xx xx x x  

Thesolutionsetis  6

66.  32 1 321 322 2 x x xx xx x 

Thesolutionsetis  2

67.

 72 3103 732310 721620 7641 41  

  xx xx xx xx x Thesolutionsetis   41.

68.  43 46 4634 424312 4312 12 12 xx xx xx xx x x  

Thesolutionsetis  12

69.  6738 4124 67243841 tt tt tttt    22 22 12142428123238 12102812298 1028298 102920 3920 20 39 20 39 tttttt tttt tt tt t t t 

Thesolutionsetis  20 39 .

70.

22 22 8543 10757 855743107 4025563540302821 408135405821 ww ww wwww wwwwww wwww

81355821 815814 13914 14 139 ww ww w w

Thesolutionsetis  14 139

Section 1.2: Solving Equations Using a Graphing Utility; Linear and Rational Equations

411 231231 xx

Chapter 1: Graphs, Equations, and Inequalities

2 22 22 43103 7121030 7121030

Thesolutionsetis  14 76. 222 143 232 xx xxxxxx

Thesolutionsetis

77. 54 254 25 2 5 yx a a a 

78. 2 2 3 23(2)4610 yxx b

79. 236 xy 236 326 22 3 ab ba ba

80. 0(2) 5(0)5 ymxb mb mbb    Therefore, 025 25 5 2 m m m   

81. ,0axbca axbbcb axcb axcb aa bc x a   

 

82. 1,0 axba  111 1 1 11 axb axb ax b aa bb x aa 

Section 1.2: Solving Equations Using a Graphing Utility; Linear and Rational Equations

83. ,0,0, xx cabab ab  () () xx ababc ab bxaxabc x ababc xab abc abab abc x ab

84. ,0 ab cc xx  ab x xc xx abcx abcx cc ab x c

suchthat ab

85. 112 1 x axax   MultiplybothsidesbytheLCD,  1 xaxax ,toget:

 

bxcxabacbxcxabac bxcxabacbxbxcxabacbx

 2 2 22 22 22 cxabaccxabac cxabaccxcxabaccx cxabacabac cxabacabacabacabac cxab cxab cc ab x c 

 

   

where0,0 ac ,and bc 

87. 2166 x aaxa when4 x  :  421646 421646 42162 412 3 aaa aaa aa a a     

88. 242 x bxbx when2 x  :



222422 22244 2224 42 42 2 bb bb bb b bb

89. SolvingforR: 12 1212 12 1221 1221 1221 2121 12 12 111 111 () () RRR RRRRRR RRR RRRRRR RRRRR RRRRR RRRR RR R RR

Chapter 1: Graphs, Equations, and Inequalities

90. Solvingfor r : (1) A Prt A PPrt APPrt Prt

91. SolvingforR: 2

92. SolvingforT:

93. Solvingfor r:

94. Solvingfor t: 0 000 0 0 00 or vgtv vvgtvv vvgt vv gt gg vvvv tt g g

95. Let x =amountinvestedinbonds. InvestedinCDs:3000 x 3,00020,000 23,00020,000 223,000 11,500 xx x x x

$11,500willbeinvestedinbondsand$8,500 willbeinvestedinCD's.

96. Let x =amountthatSeanreceives. AmountforGeorge:3000 x 3,00010,000 23,00010,000 213,000 6,500 xx x x x 

 Seanwillreceive$6,500andGeorgewillreceive $3,500.

97. Let x Sandra’sregularhourlywage.  401.58910 4012910 52910 91017.50 52     xx xx x x Sandra’sregularhourlywageis$17.50.

98. Let x =Leigh’sregularhourlywage.  4061.5421083 40981083 571083 108319 57

   xxx xxx x x Leigh’sregularhourlywageis$19.00.

99. Letx=finalexamscore. Computethefinalaverageandsetequalto80.  1808371619580 7 xx

Nowsolvefor x:

Brookeneedstoscorean85onthefinalexamto getanaverageof80inthecourse.

100. Let x =finalexamscore.

Mikemustscore78onthefinaltoearnaB. Followingthesameapproachasabove,weget

34081290 34081080 8740

Mikemustscore93onthefinaltoearnanA.

101. Let x representtheoriginalpriceofthephone. Then0.15 x representsthereductionintheprice ofthephone. originalprice–reduction=newprice 0.15799 0.85799 940

 xx x x Theoriginalpriceofthephonewas$940. Theamountofthesavingsis 0.15($940)=$141.

102. Let x representtheoriginalpriceofthecar. Then0.12 x representsthereductionintheprice ofthecar. originalprice–reduction=newprice 0.1228,160 0.8828,160 32,000 xx x x    Theoriginalpriceofthecarwas$32,000.

Theamountofthesavingsis $32,000–$28,160=$3,840.

103. Let x representthepricethetheaterpaysforthe candy. Then2.75 x representsthemarkuponthecandy. Thesellingpriceofthecandyis$3.00. suppierpricemarkupsellingprice 2.754.50 3.754.50

Thetheaterpaid$1.20forthecandy.

104. Let x representthedealer'scostofthenewcar. 0.85($34,000)$28,900 x  Ifthedealeraccepts$100overcost,thenyou willpay $28,900+$100=$29,000.

105. Let a =numberofadultpatrons. Childpatrons:5200 a  8.506.00520032,200 8.5031,2006.0032,200 2.5031,20032,200 2.501,000 2.501,000 2.502.50 400 aa aa a a a a

Thetheaterhad400adultpatrons.

106. Let x =originalpriceoftheboots. Discount:0.3 x originalpricediscountsaleprice 0.3399 0.7399 0.7399 0.70.7 570 xx x x x

Theoriginalpriceofthebootswas$570.

107. Let w =width. Length:8 lw Perimeter:22Plw  2260 2(8)260 216260 41660 444 11 lw ww ww w w w   

Chapter 1: Graphs, Equations, and Inequalities

Therectanglehasawidthof11feetandalength of19feet. 108.Let w =width.

Length:2lw 

Perimeter:22Plw

Therectanglehasawidthof7metersanda lengthof14meters.

109. Let x =thetotalnumberofpeoplewhoowneda smartphonein2019.Then0.534 x represents thenumberofpeoplewhoownedasmartphone thatrantheGoogleAndroidOS. 0.534124.4 124.4 233 0.534

x x

Therewereabout233millionpeoplewhoowned asmartphonein2019.

110. LetTbethepercentagethatusedTwitter.Then: 6729

Therefore,38%ofU.S.adultsaged18-29used Twitter.

111. Let x =amountthatScottgets.

AmountforAlice:3 4 x

AmountforTricia:1 2 x

Scottreceives$400,000;Alicereceives $300,000;Triciareceives$200,000.

112. Let x =amountthatJudypays. AmountTompays:2 3 x

Judypays$10.80andTompays$7.20.

113. Anisoscelestrianglehasthreeequalsides. Therefore:410240318  xxx .Solve eachsetseparately: 410240 230 15 410318 8 240318 22 22

x x xx x xx x x Since22isthelargestofthenumbersthenthe largestperimeteris:  422102224032218266 

114. 311114 316 4524205 313134 1 41058025

xxx xxx MultiplybothsidesbytheLCD80toclear fractions. 608488012064 1087258 10716 16 107

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

Thusr=2inchesistheradiusofthecirclewhere theperimeterofthehexagonis10inchesmore thantheradius.

116. Inordertosolve58 3 33 x xx    ,wemultiply eachtermbytheexpression“3 x  ”toget

58 333 33 x xx xx

Now,provided3 x  ,wecancancelthe denominatorstoget

5338 5398263 xx xxxx

However,wealreadystatedthat3 x  .Sowe haveacontradiction.

117. Answerswillvary.Oneexampleis 3136 xx

118. Answerswillvary.Ingeneral,‘solve’meansto findvaluesforavariablethatmaketheequation true,‘evaluate’meanstopluginaspecificvalue foravariableinanexpressionandsimplify,and ‘simplify’meanstocollectliketermsandwrite anexpressioninasimplerform.

Ex:Solve  328 xx .

Ex:Evaluate 32

119. Inobtainingstep(7)wedividedbyx–2.Since x=2fromstep(1),weactuallydividedby0.

Section 1.3: Quadratic Equations

Section 1.3

1.  25661 xxxx 

2.  2 23231 xxxx 

3.  5,3 3

4. True

5. 2 2 2 2 1552525 5;;5 22244 255 5 42 xx xxx

6. discriminant;negative

7. False;aquadraticequationmayhavenoreal solutions.

8. False;If2  x p 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}.

Chapter 1: Graphs, Equations, and Inequalities

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

   xx xx xx 20or50 2or5   xx xx

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 x x xx x x x     

Thesolutionsetis  3 2

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



  

Thesolutionsetis  4 5 .

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

320or230 23 or 32 p p p p

Thesolutionsetis23 , 32

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 x xx x xx xx xx 

320or230 23 or 32 x x x x

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis23 , 32

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

29.

Section 1.3: Quadratic Equations

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis{3,4}.

4233 33 423333 33 x xx xx x xxxx xx xx

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis

Chapter 1: Graphs, Equations, and Inequalities

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

Thesolutionsetis  0,4 3

37.  2 2 2 421 44214 225 225 25 25 3or7 xx xx x x x x xx

Thesolutionsetis7,3.

38.  2 2 2 613 69139 322 322 322 xx xx x x x 

Thesolutionsetis322,322. 

39. 2 2 2 2 130 216 13 216 1131 2161616 11 44 xx xx xx x   

  111 442 11 42 31 or 44 x x xx    The13 solutionsetis,. 44

40. 2210 33 xx

Section 1.3: Quadratic Equations

The317317 solutionsetis,. 44

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

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. 

Chapter 1: Graphs, Equations, and Inequalities

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 x x xx 

abc x 

  

  

 2 2,2,1 224(2)(1)248 2(2)4 21222313 442



  

Thesolutionsetis1313,.22

53. 2 2 49 490 (49)0 x x xx xx 

0or490 09 or 4 xx xx

Thesolutionsetis  9 0,. 4

54. 2 2 54 045 0(45) xx x x 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

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

Section 1.3: Quadratic Equations

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

The3solutionsetis,3. 2

Chapter 1: Graphs, Equations, and Inequalities

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

Thesolutionsetis329329 , 1010

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 abc x

Thesolutionsetis323323,.33

63.  2 22 2 2 11 40 11 40 410 x x xx x x x x 

4,1,1abc

2 11441 24 1116117 88 x

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis 117117 , 88

64. 2 83 20

x x

 xx x x xx abc x

22 2 2 2 83 20 2830 2,8,3 88423 22 86424840 44 8210410 42

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis 410410 , 44 

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

(13)(13)4(2)(3) 2(2) 131692413145 44 x

Neitherofthesevaluescausesadenominatorto equalzero,sothesolutionsetis 1314513145 , 44

67. 24.12.20xx

Thesolutionsetis  0.63,3.47

Section 1.3: Quadratic Equations

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 x x 

Thesolutionsetis  3.37,0.53.

69. 2330xx

  2 1,3,3 33413 21 3312315 22 1.07or2.80 abc x xx



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 x x     

 

 

 Thesolutionsetis  0.85,1.17.

Chapter 1: Graphs, Equations, and Inequalities

72. 220xx 

2 2 ,,2 42 2 8 2 0.44or1.44 abc x x x

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

85.

86.

87.

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

2 2 2 2 121 2 1210 2 1 22120 2 2220 x x xx xx xx

Section 1.3: Quadratic Equations

Thesolutionsetis22,22. 

89. 2 2 4 40 xx xx 

2 (1)(1)414 2(1) 1116117 22

The117117 solutionsetis,. 22

90. 2 2 1 10 xx xx   1,1,1abc

2 (1)(1)411 2(1) 11415 22 x  

 The1515 solutionsetis,. 22

Chapter 1: Graphs, Equations, and Inequalities

(2)(1)(2)(1)21(2)(1) (1)2(2)71 2471 3471 450 (1)(5)0

Thevalue1 x  causesadenominatortoequalzero,sowedisregardit.Thus,thesolutionsetis{5}. 92. 2 3147 212 x x

310or20 1or2 3 xx xx 

Thevalue2 x  causesadenominatortoequalzero,sowedisregardit.Thus,thesolutionsetis 1 . 3

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

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

13 w  or11 w 

Discardthenegativesolutionsincewidthcannot benegative.Thewidthoftherectangular windowis11feetandthelengthis13feet.

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

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

18 w  or17 w 

Discardthenegativesolutionsincewidthcannot benegative.Thewidthoftherectangular windowis17centimetersandthelengthis18 centimeters.

97. Let l representthelengthoftherectangle. Let w representthewidthoftherectangle. Theperimeteris26metersandtheareais40

squaremeters. 2226 13so13 lw lwwl

2 2 40 (13)40 1340 13400 (8)(5)0 lw ll ll ll ll

8or5 58 ll ww

Thedimensionsare5metersby8meters.

98. Let r representtheradiusofthecircle. Sincethefieldisasquarewitharea1250square feet,thelengthofasideofthesquareis 1250252  feet.Thelengthofthediagonal is2r . UsethePythagoreanTheoremtosolvefor r :

222 2 2 2 (2)252252 412501250 42500 625 25 r r r r r

Theshortestradiussettingforthesprinkleris25 feet.

99. Let x =lengthofsideoforiginalsheetinfeet. Lengthofbox:2 x feet Widthofbox:2 x feet Heightofbox:1foot

0or4xx Discard0 x  sincethatisnotafeasiblelength fortheoriginalsheet.Therefore,theoriginal sheetshouldmeasure4feetoneachside.

Chapter 1: Graphs, Equations, and Inequalities

100. Let x =widthoforiginalsheetinfeet.

Lengthofsheet:2 x

Lengthofbox:22 x feet

Widthofbox:2 x feet

Heightofbox:1foot

Discard0 x  sincethatisnotafeasiblelength fortheoriginalsheet.Therefore,theoriginal sheetis3feetwideand6feetlong.

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

2 2 2 9680160 1680960 560 610 tt tt tt

Discardthenegativesolutionsincethetime offlightmustbepositive.Theballwill strikethegroundafter6seconds.

b. Whentheballpassesthetopofthebuilding, itwillbe96feetfromtheground.Therefore, wesolve

2 2 2 96801696 16800 50 50 tt tt tt tt

Theballisatthetopofthebuildingattime 0

whenitisthrown.Itwillpassthetop ofthebuildingonthewaydownafter5 seconds.

102. a. Tofindwhentheobjectwillbe15meters abovetheground,wesolve 2 2 4.92015 4.920150 tt tt   4.9,20,15abc

0.99or3.09tt  Theobjectwillbe15metersabovethe groundafterabout0.99seconds(ontheway up)andabout3.09seconds(ontheway down).

b. Theobjectwillstrikethegroundwhenthe distancefromthegroundis0.Therefore,we solve  2 4.9200 4.9200 tt tt   0 t  or4.9200 4.920 4.08 t t t   

Theobjectwillstrikethegroundafterabout 4.08seconds.

c. 2 2 4.920100 4.9201000 tt tt   4.9,20,100abc

 2 202044.9100 24.9 201560 9.8 t    

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    

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. 2 2 2 (12)(7)(3)0.8(12)(7)(3) 357252201.6 35750.40 1916.80 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 (5)2 T Ax

Theareaofthepoolis2(5)25 AP 

Section 1.3: Quadratic Equations

106. Let x representthewidthoftheborder measuredinfeet.Theradiusofthepoolis5 feet.Then5 x  representstheradiusofthe circle,includingboththepoolandtheborder. Thetotalareaofthepoolandborderis (5)2 T Ax Theareaofthepoolis2(5)25 AP  . Theareaoftheborderis 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) AT xx

BTP AAAx

Theareaoftheborderis 2 (5)25

Sincetheconcreteis3inchesor0.25feetthick, thevolumeoftheconcreteintheborderis

 2 0.250.25(5)25 B Ax

Solvingthevolumeequation:

 2 2 10(10)4()(108) 2() 31.42100432 6.28 2.71or12.71 x xx

Discardthenegativesolution.Thewidthofthe borderisroughly2.71feet.

Theareaofthegardenis61060 AG  . 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





  

2.56or10.56 x xx

  2 884(1)(27)8172 2(1)2

 Discardthenegativesolution.Thewidthofthe borderisapproximately2.56feet.

Chapter 1: Graphs, Equations, and Inequalities

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

Vlwhxx x xx Thedimensionsofthepatioare18feetby36 feet.

109. Let x =thelengthofa12.9-inchiPadProina 16:94:3format.

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

Sincethelengthcannotbenegative,thelengthof theSurfaceProis10.23inchesandthewidthis 21361.61 6.82 313  inches.Thus,theareaofthe 12.3-inch3:2formatSurfaceProis 1361.611361.61 2 13313 69.83squareinches.

TheiPadProformathasthelargerscreensince itsareaislarger.

110. Let x =thelengthofa7.9-inchiPadMiniina 4:3format.

Then3 4 x =thewidthoftheiPad.Thediagonal ofthe7.9-inchiPadis7.9inches,sobythe Pythagoreantheoremwehave:

Sincethelengthcannotbenegative,thelengthof theiPadis42600.96337inchesandthewidthis

42600.96 337 9 6.32 16

inches.Thus,theareaofthe iPadis42600.96942600.963371633771.11

square inches.

Let y =thelengthofa12-inch3:2format MicrosoftSurfacePro.Then 2 3 y =thewidthof theSurfacePro.Thediagonalofa12-inch SurfaceProis12inches,sobythePythagorean theoremwehave:

Sincethelengthcannotbenegative,thelengthof theiPadis6.32inchesandthewidthis  3 6.324.74 4  inches.Thus,theareaofthe iPadis(6.32)(4.74)29.9568  squareinches. Let y =thelengthofa8-inch16:10format AmazonFireHD8™.Then 10 16 y =thewidthof theFire.Thediagonalofa8-inchFireis8 inches,sobythePythagoreantheoremwehave:

Section 1.3: Quadratic Equations

113. Wewillsetg=2.97andsolveforhusingthe QuadraticFormula.

0.00060.0153.04

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

TheiPadMini™4:3formathasthelargerscreen sinceitsareaislarger.

111. Let h be1.1.Then

1.10.000250.04



00.000250.041.1



0.04(0.04)4(0.00025)(1.1) 2(0.00025)

35.3ftor124.7ft

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

112. Sincedisexpressedin1000’swewillsetd=15 andsolveforxusingtheQuadraticFormula. 2 2 2 2 0.828(0.828)4(0.012)(9.25) 2(0.012) 0.828.241584 0.024 0.0120.82815.750 250.0120.82815.750 00.0120.8289.25

54.98or14.02

Sothenearestyearwhenthedifferencewas $25,000occurredabout14yearsafter1980or 1994.Thevalue55hasnomeaningsinceitisin thefuture.

2.970.00060.0153.04

00.00060.0150.07 29or4.02 gxx

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

Sincexisthenumberofmembers,itmustbe positivesothenumberofmembersis49.

Chapter 1: Graphs, Equations, and Inequalities

115. Let a betheagetheindividualisabletostart savingmoney.Thenweneedtofindwherethe modelsareequal.Solvingthesetwoequations together:

2 2240(2240)4(25)(38540) 2(25)

22401163600 50 252400307001607840 252240385400

22401078.7 50

22401078.722401078.7 or 5050

66.4or23.2

Since x istheagetostartsaving,itmakessense thattheanswerisapproximateatage23.

116.Wewillsettheequationequalto10andsolve:

0.0030.0348.08610

0.0030.0341.9140

2(0.003)

0.034.024124 0.006

0.034.15532 0.006

0.0340.155320.0340.15532 or 0.0060.006 20.22

x or31.55  x Thepercentagewillreach10%approximately32 yearsafter1960whichis1992.

Sincethenumberofconsecutiveintegerscannot benegative,wediscardthenegativevalue.We mustadd37consecutiveintegers,beginningat1, inordertogetasumof703. 118.

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

 2 33411603646 212 n 

Neithersolutionisaninteger,sothereisno polygonthathas80diagonals.

119. Therootsofaquadraticequationare

120. Therootsofaquadraticequationare

121. Inordertohaveonerepeatedsolution,weneed thediscriminanttobe0.

122. Inordertohaveonerepeatedsolution,weneed thediscriminanttobe0.

Section 1.3: Quadratic Equations

bbac x cc bbacbbac c bbac bbac ac cbbaccbbac a bbac x 

123. For20 axbxc : 24 2

For20 axbxc :

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

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

2 2 12  y y y Equatingtheratiosgives

Chapter 1: Graphs, Equations, and Inequalities

3.

126. a. 29 x  and3 x  arenotequivalent becausetheydonothavethesamesolution set.Inthefirstequationwecanalsohave 3 x 

b. 9 x  and3 x  areequivalentbecause 93 

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

Thefirstequationhasthesolutionset  1 whilethesecondequationhasnosolutions.

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

128. Answerswillvary.Knowingthediscriminant allowsustoknowhowmanyrealsolutionsthe equationwillhave.

129. Answerswillvary.Onepossibility:

Twodistinct:23180 xx

Onerepeated:214490 xx Noreal:240 xx

130. Answerswillvary.

Section 1.4

1. Integers:  3,0

Rationals:  3,0,6 5

2. True;thesetofrealnumbersconsistsofall rationalandirrationalnumbers.

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

Section 1.4: Complex Numbers; Quadratic Equations in the Complex Number System

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

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

28. 2 2 222 222 2(1)121 2(1)22 iiiii iii i i i i

22. (53)(2)105632 103(1) 13

iiiii i i

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

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

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

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

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

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

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

33. 22 (1)1212(1)2 iiiii  34. 22 (1)1212(1)2 iiiii

11 2322122211 (1) iiiiiiii

36.

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

Chapter 1: Graphs, Equations, and Inequalities

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

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 x i    

Thesolutionsetis 2,2. ii

60. 240 x 

 (2)(2)0 2or2 Thesolutionsetis2,2. xx xx  

61. 2160 x    440 4or4

Thesolutionsetis4,4. xx xx  

62. 2250 x   225 255 Thesolutionsetis5,5. x xi ii  

63. 26130xx

Thesolutionsetis32,32.

64. 2480xx

.

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

Thesolutionsetis  3,3ii

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 .

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

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



Thesolutionsetis  3131 , 10101010 ii

69. 2 2 512 5210 x x 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 x x xx   22 13,6,1 4(6)4(13)(1)365216 (6)166432 2(13)261313 abc bac i xi  

  

Thesolutionsetis  3232 , 13131313 ii

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

bac i x i 

 

Thesolutionsetis1313 , 2222 ii    

72. 210xx 1,1,1abc 224(1)4(1)(1)143 (1)31313 2(1)222 bac i x i



 

Thesolutionsetis1313 , 2222 ii     .

Chapter 1: Graphs, Equations, and Inequalities

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







abc bac i x i

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 x i

Thesolutionsetis3333333,,. 2222 ii   

75. 4 4 16 160 x x  

22 2 440 (2)(2)40 xx xxx

2 2 20or20or40 2or2or4 2or2or42 xxx xxx x xxi

Thesolutionsetis  2,2,2,2. ii

76. 4 4 1 10 x x  

22 2 110 (1)(1)10 xx xxx





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

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

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

 2 2 10or10or40 1or1or4 1or1or42 xxx xxx x xxi  

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

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



 Theequationhastwocomplexsolutionsthatare conjugatesofeachother.

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 x x xx 



 22 1,2,6 4(2)4(1)(6)42420 abc bac

 Theequationhastwocomplexsolutionsthatare conjugatesofeachother.



2 2 10or10or10 1or1or1 1or1or1 xxx xxx x xxi

Thesolutionsetis  1,1,,. ii

Section 1.4: Complex Numbers; Quadratic Equations in the Complex Number System

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 Z ZZiiii iii ii iii

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













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

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













97.    22 2222 222222 2222 2()2() 22 22 

 abiabi aabibiaabibi aabibiaabibi aabibaabib

Chapter 1: Graphs, Equations, and Inequalities

2222 22 22 () 2()0 



 abab ab abba

Anycomplexnumberoftheform  aai or aai willwork.

98. Let32  u in320  x sothat330 xu

     uuu x uuuu ii

Then,22()()0  xuxuxu .Fromthefirst factorwefind32 xu .Fromthesecond factor,usethequadraticformulatoget 22 233 ()()41 21 33223 22222

Thesolutionsetis: 33 3223 2,22

99. 2 (5)(5)();  x yxy let5 ux (so 5 xu and5 vy so5 yv

Substitutinggives2 () uvuv or 220 uuvv whichisquadraticinu.Using thequadraticformulagives

223 41 212     vv vvv 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

1.5 1. True

33 x x 

322 2 4936494 49 433 xxxxxx xx xxx 

5. Thedistancefromtheoriginto4onareal numberlineis4units.

6. False;youcanalsousetheQuadraticFormulaor completingthesquare.

7. quadraticinform

8. True

9. a 10. c 11. 64 y  2264 61610   y yy Check:223255  Thesolutionsetis  22

Section

Section 1.5: Radical Equations; Equations Quadratic in Form; Absolute Value Equations; Factorable Equations

12. 37 t 

2237 34952 t tt 

Check:523497

Thesolutionsetis52.

13. 211 t

22211 211221 t

Check:2(1)111

Thesolutionsetis1.

14. 342 t 

22342 344300 t

Check:3(0)442

Thesolutionsetis{0}.

15. 346 t 

Sincetheprincipalsquarerootisnevernegative, theequationhasnorealsolution.

16. 532 t 

Sincetheprincipalsquarerootisnevernegative, theequationhasnorealsolution.

17. 31230 x 

3 333 123 123 122722613 x x xxx

Check:3312(13)32730

Thesolutionsetis{13}.

18. 31210 x 

3 333 121 121 121200 x x xxx

Check:3312(0)1110

Thesolutionsetis{0}.

19.

20.

4 444 542 542 5416 520 4 x x x x x 

 

Check:  44544162  T

Thesolutionsetis   4.

5 555 231 231 231 22 1 x x x x x

Check:  5521311  T

Thesolutionsetis   1.

21.

22.

52 5 55 2 2 2 2 21 21 21 210 10 10 1 xx xx xx xx x x x

Check:  52551211211  T

Thesolutionsetis  1

 42 444 2 2 2 165 165 1625 9 3 x x x x x  

Chapter 1: Graphs, Equations, and Inequalities

Thesolutionsetis

Thesolutionsetis

Check–5:152(5)25 55

Check3:152(3)933

Thesolutionsetis{3}.

26. 12 x x 

22 22 12 12120 (4)(3)04or3 xx xxxx xxxx

Check–4:12(4)16 44

Check3:123933

Thesolutionsetis{3}.

27.

28.

    

 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

Check7:3(710)49417

Check2:3(210)436422

 

Discard7 x  asextraneous.

Thesolutionsetis{2}.



 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. 242 xxx

   222 22 42 444 858 5 xxx xxxx x x 

Check:





Section 1.5: Radical Equations; Equations Quadratic in Form; Absolute Value Equations; Factorable Equations

2 88842 555 6482 4 2555 42 255 22 55

Thesolutionsetis  8 5

30. 285 xxx

2 22 22 85 81025 1133 3 xxx xxxx x x

Check:

2 33835 93835 42 22

 T

Thesolutionsetis 3.

31. 331 x x 

22 2 2 313 31(3) 3169 098 0(1)(8) 1or8 xx xx xxx xx xx xx 



32. 2122 x x 





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











Check2:2+122(2)21662

Check4:2122(4)2444



Discard2 x  asextraneous.

Thesolutionsetis{4}

33. 3572 xx

 

 22 22 2 2 2 2 3527 3527 354477 21647 (216)47 46425616(7) 46425616112 4801440 20360 (2)(18)02or18 xx xx xxx xx xx xxx xxx xx xx xxxx

 









 



Check2:3(2)527 191322

Check18:3(18)5187

 

49257522



Discard2 x  asextraneous.

Check1:33(1)13451

Check8:33(8)132588

 

Discard1 x  asextraneous.

Thesolutionsetis 8.

Thesolutionsetis  18

Chapter 1: Graphs, Equations, and Inequalities

34. 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.

35. 3112 xx

22 22 2 2 2 3121 3121 314411 2241 (22)41 48416(1) 2144 650 (1)(5)01or5 xx xx xxx xx xx xxx xxx xx xxxx

Check1:3(1)111 402022

Check5:3(5)151 1644222

Thesolutionsetis  1,5

36. 2311 xx



 22 22 2 2 2 2311 2311 231211 121 (1)21 214(1) 2144 230 (1)(3)01or3 xx xx xxx xx xx xxx xxx xx xxxx

Check–1:2(1)311101 Check3:2(3)331321

Thesolutionsetis  1,3

37.    22 32 32 32 23 xx x x xx xx  

  22 2 2 23 469 0109 091 xx xxx xx xx     1or9xx Check: 3211 321 11 11     T 3299 3233 33   

Discard9 x  asextraneous.Thesolutionsetis {1}

Section 1.5: Radical Equations; Equations Quadratic in Form; Absolute Value Equations; Factorable Equations

41. 1/3 522 x     3 1/33522 5285102 x xxx  

Check:  1/3 1/3 5222 82 22   

Thesolutionsetis{2}.

42. 1/3 211 x     3 1/33211 211221 x xxx  

Check:

Discard4 x  asextraneous.Thesolutionsetis

39. 1/2 314 x 

1/2 1/2 3514 164 44

  T

Thesolutionsetis{5}

40. 1/2 352 x     2 1/22352 354393 x xxx 

Check:

1/2 1/2 3352 42 22    T

Thesolutionsetis{3}

1/3 1/3 2111 11 11  

Thesolutionsetis 1.

43. 21/295 x 

 21/222 22 95 92516 4or4 x xx xx

Check:

21/2 1/2 495 255 55   

Thesolutionsetis 4,4.

21/2 1/2 495 255 55   

Chapter 1: Graphs, Equations, and Inequalities

Thesolutionsetis

240hasnorealsolution t  ,soweonlyneed toconsider 20 2 t t

Thesolutionsetis  2,2

46. 440 y 

22 2 22 220 2220 202,whichisimpossible or2202 yy yyy yy yyy

Thesolutionsetis  2,2.

47. 42540xx

22 22 410 40or10 2or1 xx xx xx



Thesolutionsetis  2,1,1,2.

48. 4210240xx

 22640xx 2 2 60 6 6 x x x    or 2 2 40 4 2 x x x   

Thesolutionsetis  6,2,2,6.

49. 42 6510 xx



 22 22 2 6110 610or10

 

61,whichisimpossible or1

 xx xx x x Thesolutionsetis 1,1.

50. 42 25120 xx

 22 22 2 2340 230or40



23,whichisimpossible or2 xx xx x x 

 

Thesolutionsetis   2,2.

51. 63780xx

 33 33 3 3 810 80or10 82 or11 xx xx xx x x 







Thesolutionsetis 2,1.

52. 63780xx

 33 33 3 3 810 80or10 82 or11 xx xx xx x x 







Thesolutionsetis   1,2.

Section 1.5: Radical Equations; Equations Quadratic in Form; Absolute Value Equations; Factorable Equations

53.  2 272120xx

 22 2 22 7120 340 30or40 3235 or4246 letpxpx pp pp pp pxx pxx 

Thesolutionsetis   6,5.

54.  2 252560 xx



22 2 2525 60 320 3020 32531 22527 2 letpxpx pp pp porp pxx orpxx



Thesolutionsetis  7 2,1.

55.  2 21513 ss



 22 2 2 11 253 2530 2130 210or30 113 1 222 or3132 letpsps pp pp pp pp pss pss













Thesolutionsetis 3,2. 2

56.  2 315120 yy    22 2 11 3520 3210 320or10 225 1 333 or1112 letpypy pp pp pp pyy pyy













Thesolutionsetis 5,2. 3

57. 40xx

   22 22 4 4 16016 0160 or16016 xx xx x xxx xxx x x    



Check:  0:0400 00 16:164160 16160 00 x x     

Thesolutionsetis  0,16 58. 80xx

   22 22 8 8 64064 0640or64 xx xx xxxx xxxx  

 

Check: 0:0800 00

64:648640 64640 x x   



Thesolutionsetis  0,64

Chapter 1: Graphs, Equations, and Inequalities

59. 20 xx  2

62. 1/21/4210tt

2 2 20 200 540 50or40 5=5non-real or4=4=16

letpxpx pp pp pp pp px pxx   

Check: 16:161620 16420 x

Thesolutionsetis 16.

60. 6 xx

1/421/2 2 1/4 210 110 10 111 letptpt pp pp p ptt 

   

Check:  1:1/21/412110 121000 t

Thesolutionsetis  1

2

letpxpx pp pp pp pp px pxx

2 2 6 60 320 30or20 3=3non-real or2=24

Check:

4:446 426 x 



Thesolutionsetis  4

61. 1/21/4440zt

 1/421/2 2 1/4 440 220 20 2216 letpzpz pp pp p pzz 

Check:

 16:1/21/41641640 484000 z  

Thesolutionsetis 16.

63. 1/21/4 4940 xx 1/421/2 2 4 1/4 let 4940 98164917 88 917917 88 pxpx pp p xx  

Check:

1/21/444 2 2 2 917917 494 648 64064 917917 4940 88 917917 4940 88 917917 4940 648

Section 1.5: Radical Equations; Equations Quadratic in Form; Absolute Value Equations; Factorable Equations

Check:

 2 121 22 2:112

64. 1/21/4320xx

Check:

1/21/4 1/21/4 16:1631620 462000 1:13120 132000 x x

Thesolutionsetis   1,16.

65.  2 112 11 x x   

2 2 22 let11 11 220 1201or2 1 11112 1 or 1122122 12 pp xx pppp pppp pxx x pxx x

2/3 2/33/23/2 3/2 22 2 2whichisnotarealnumber px x x 

Check:

4/32/333 99 11 93 3520 3520 1520 33 00

4/32/333 99 11 93 3520 3520 1520 33 00

  Thesolutionsetis33 , 99 

.

71. 235 x  235or235 22or28 1or4 xx xx xx 





Thesolutionsetis{–4,1}.

72. 312 x  312or312 33or31 11 or 3 xx xx xx 





Thesolutionsetis  1,1 3

Chapter 1: Graphs, Equations, and Inequalities

73. 14813145 tt  145or145 44or46 13 or 2 tt tt tt

Thesolutionsetis  1,3 2

74. 1269123 zz  123or123 22or24 1or2 zz zz zz

Thesolutionsetis  1,2

75. 721 x 721or721 3or3

xx xx

Thesolutionsetis{–3,3}.

76. 1 x  1or1xx  Thesolutionsetis{–1,1}.

77. 42321 xx  21or21 11 or 22 xx xx 



Thesolutionsetis  11 , 22 .

78. 11532 22 x x  112or2 22 4or4 xx xx  

Thesolutionsetis  4,4

79. 83 7  x 212121 or 888 xxx

Thesolutionsetis  2727 , 22 .

80. 39 4 x  1212or12xxx Thesolutionsetis{–12,12}.

81. 22 35 x  222or2 3535 5630or5630 524or536 2436 or 55 xx xx xx xx



Thesolutionsetis  3624 , 55 .

82. 11 23 x  111or1 2323 326or326 38or34 84 or 33 xx xx xx xx 

  

Thesolutionsetis  48 , 33 .

83. 21 2 u  impossible,sinceabsolutevaluealwaysyieldsa non-negativenumber.

84. 21 v  impossible,sinceabsolutevaluealwaysyieldsa non-negativenumber.

85. 290 x  2 2 90 9 3 x x x   

Thesolutionsetis 3,3.

Section 1.5: Radical Equations; Equations Quadratic in Form; Absolute Value Equations; Factorable Equations

86. 2160 x  2 2 160 16 4 x x x  

87. 223xx

88. 212xx

22 22 12or12 120or120 3401148 or 2 147norealsol. 2 3or4 xxxx xxxx xxx xx

Thesolutionsetis 4,3.

89.

3 2 90 90 330 xx xx xxx 

0 x

Thesolutionsetis

90.

42 22 2 810 810 990 xx xx xxx  

20 0 x x   or90 9 x x   or90 9 x x  

Thesolutionsetis  9,0,9

91.

92.

32 2 200 200 540 xxx xxx xxx    0 x  or50 5 x x   or40 4 x x  

Thesolutionsetis  5,0,4

32 2 670 670 710 xxx xxx xxx    0 x  or70 7 x x   or10 1 x x  

Thesolutionsetis  7,0,1

93.

 32 2 2 10 1110 110 1110 xxx xxx xx xxx     10 1 x x   or10 1 x x   Thesolutionsetis  1,1

94.     32 2 2 440 4140 410 4110 xxx xxx xx xxx

   40 4 x x   or10 1 x x   or10 1 x x   Thesolutionsetis   4,1,1.

95.

   32 2 2 316480 31630 3160 3440     xxx xxx xx xxx 30 3 x x   or40 4   x x or40 4   x x Thesolutionsetis  4,4,3

Chapter 1: Graphs, Equations, and Inequalities

100. 2/31/3420xx ;UseZERO(orROOT)on thegraphof2/31/3142 yxx

Thesolutionsetis  39.80,0.20

101. 42330xx ;UseZERO(orROOT)on thegraphof42133 yxx

Thesolutionsetis  1.03,1.03

102. 42220xx ;UseZERO(orROOT)on thegraphof42122 yxx .

Thesolutionsetis  0.93,0.93

103.  2 37200 3540 xx xx   350 35 5 3 x x x    or40 4 x x  

Thesolutionsetis5 4, 3 

.

108. 13 23 42 236 x

or236 23 3 2 x x x

Thesolutionsetis39 , 22

109.  433 4123 315 5 ww ww w w 

Thesolutionsetis 5.

110.  63212 618212 46 3 2 kk kk k k

Thesolutionsetis3 2

.

Chapter 1: Graphs, Equations, and Inequalities

7 u

or1 u

Gobackintermsof y andsolve: 7 1 77 67 7 6 y y yy y y

115.

Thesolutionsetis15 , 53

22 2 2 251 251 251 2521 40 220 xx xx xx x xx x xx

2 x  or2 x 

Check:

 22521 121 31 

Thesolutionsetis   2.

116.

22521 921 11

 T

22 2 2 3126 3126 3126 3142436 427350 4750 xx xx xx xxx xx xx

Thesolutionsetis

Check: 77 3126 44 57 6 22 16

Thesolutionsetis  5

351256 4106 66

T

Section 1.5: Radical Equations; Equations Quadratic in Form; Absolute Value Equations; Factorable Equations

117. 2 2 361 3610

118.

Check:

Thesolutionsetis

119.

2 2 322 340 410 4or1 xx xx xx xx

Sinceanevenrootcannotbenegative,wecan exclude1asasolution. Check:

Thesolutionsetis

Chapter 1: Graphs, Equations, and Inequalities

4 22 160 440 t tt 

Thesolutionsetis

2 2 390 1,3,9 33419 21 327333 22 zz abc z i

2 2 10 1,1,1 11411 21 1313 22 zz abc z i

Thesolutionsetis33313,,2222 ii

33313 ,,3,1 2222 ii

127. 2212120kkkk

430kk 4 34 3 3412 315 5 k x x xx x x 

  or3 33 3 339 3 2 k x x xx x 



Andsinceneitherofthese x valuescausesa denominatortoequalzero,thesolutionsetis

 3,5 2

128. 223283280kkkk

Andsinceneitherofthesexvaluescausesa denominatortoequalzero,thesolutionsetis

129. 8327

Thesolutionsetis  3,1

130. 53124 x x  53124 54123 99 1 x x xx x x

or53(124) 53124 54123 15 x x x x xx x

Thesolutionsetis   1,15.

131. Graphtheequations1/4/1100 yxx  and 24y  ;thenuseINTERSECTtofindthe x-coordinateofthepointsofintersection: 300 5 0 0

Thedistancetothewater'ssurfaceis approximately229.94feet.

132. 2 4 25 LH T 

Let4 T  and10 H  ,andsolvefor L

Thecrushingloadis64tons.

133. 2 32 l T  

Let16.5 T  andsolvefor l 22 2 2 16.52 32 16.5 232 16.5

Thelengthwasapproximately221feet. 134.

Chapter 1: Graphs, Equations, and Inequalities

Sincex>2,thenegativesolutionisextraneous.

136.

Thesolutionsetis

xxx xxx xxx xx

Tosolve1 2 121330 xx ,let1 2  ux 11 22 Then2121330 (43)(31)0 31 or 43 91 169

uu uu uxx xx

Thesolutionsetis  910,,169





137. 63 33 22 28270 (27)(1)0 (3)(39)(1)(1)0

 zz zz zzzzzz 30or10 3or1   zz zz   2 2 1,3,9 33419 2 327333 22 , 1,1,1 11411 2 1313 22 

   abc x i Also abc x i Thesolutionsetis133333,1,,. 22  

138. Answerswillvary,oneexampleis11 x 

139. Answerswillvary,oneexampleis 20 xx

140. Answerswillvary.

141. x representsthedistancefrom x totheorigin. Sincedistanceisnevernegative,2 x  hasno realsolution.

142. Janedidnotcheckhersolutionsandincludedthe extraneoussolution,1 x 

  2 2 2 2 230 23 23 23 230 310 3or1 xx xx x x xx xx xx x x

Check:

Section 1.6: Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications

Thesolutionsetis 3.

Section 1.6

1. mathematicalmodeling

2. interest

3. uniformmotion

4. False;theamountchargedfortheuseof principalistheinterest.

5. True;thisistheuniformmotionformula.

6. a

7. b

8. c

9. Let A representtheareaofthecircleand r the radius.Theareaofacircleistheproductof π timesthesquareoftheradius:2 A r 

10. Let C representthecircumferenceofacircle and r theradius.Thecircumferenceofacircle istheproductof π timestwicetheradius: 2 Cr 

11. Let A representtheareaofthesquareand s the lengthofaside.Theareaofthesquareisthe squareofthelengthofaside:2 A s 

12. Let P representtheperimeterofasquareand s thelengthofaside.Theperimeterofasquareis fourtimesthelengthofaside:4Ps 

13. Let F representtheforce, m themass,and a theacceleration.Forceequalstheproductofthe masstimestheacceleration: F ma 

14. Let P representthepressure, F theforce,and A thearea.Pressureistheforceperunitarea: F P A 

15. Let W representthework, F theforce,and d thedistance.Workequalsforcetimesdistance: WFd 

16. Let K representthekineticenergy, m themass, and v thevelocity.Kineticenergyisone-half theproductofthemassandthesquareofthe velocity:2 1 2 K mv 

17. C  totalvariablecostindollars, x  number ofdishwashersmanufactured:150Cx 

18. R  totalrevenueindollars, x  numberof dishwasherssold:250Rx 

19. Let x representtheamountofmoneyinvestedin bonds.Then50,000 x representstheamount ofmoneyinvestedinCD's.Sincethetotal interestistobe$6,000,wehave:



0.150.07(50,000)6,000 1000.150.07(50,000)6,000100 157(50,000)600,000 15350,0007600,000 8350,000600,000 8250,000 31,250 xx xx xx xx x x x 

$31,250shouldbeinvestedinbondsat15%and $18,750shouldbeinvestedinCD'sat7%.

20. Let x representtheamountofmoneyinvestedin bonds.Then50,000 x representstheamount ofmoneyinvestedinCD's.Sincethetotal interestistobe$7,000,wehave: 

0.150.07(50,000)7,000 1000.150.07(50,000)7,000100 157(50,000)700,000 15350,0007700,000 8350,000700,000 8350,000 43,750 xx xx xx xx x x x

$43,750shouldbeinvestedinbondsat15%and $6,250shouldbeinvestedinCD'sat7%.

Chapter 1: Graphs, Equations, and Inequalities

21. Let x representtheamountofmoneyloanedat 8%.Then12,000 x representstheamountof moneyloanedat18%.Sincethetotalinterestis tobe$1,000,wehave:

0.080.18(12,000)1,000

1000.080.18(12,000)1,000100

818(12,000)100,000

8216,00018100,000

10216,000100,000 10116,000 11,600

$11,600isloanedat8%and$400isat18%.

22. Let x representtheamountofmoneyloanedat 16%.Then1,000,000 x representstheamount ofmoneyloanedat19%.Sincethetotalinterest istobe$1,000,000(0.18),wehave:

0.16190,0000.19180,000

0.03190,000180,000 0.0310,000

0.160.19(1,000,000)1,000,000(0.18) 10,000 0.03 $333,333.33

Wendycanlend$333,333.33at16%.

23. Let x representthenumberofpoundsofEarl Greytea.Then100 x representsthenumberof poundsoforangepekoetea. 64(100)5.50(100) 64004550 2400550 2150 75

xx xx x x x 75poundsofEarlGreyteamustbeblendedwith 25poundsoforangepekoe.

24. Let x representthenumberofpoundsofthe firstkindofcoffee.Then100 x representsthe numberofpoundsofthesecondkindofcoffee. 2.755(100)4.10(100) 2.755005410 2.25500410 2.2590 40

  xx xx x x x 40poundsofthefirstkindofcoffeemustbe blendedwith60poundsofthesecondkindof coffee.

25. Let x representthenumberofpoundsof cashews.Then60 x  representsthenumberof poundsinthemixture. 94.50(60)7.75(60) 92707.75465

1.25195 156  

xx xx x x 156poundsofcashewsmustbeaddedtothe60 poundsofalmonds.

26. Let x representthenumberofcaramelsinthe box.Then30 x representsthenumberof cremesinthebox.

RevenueCostProfit

12.500.250.45(30)3.00

12.500.2513.50.453.00

12.5013.50.203.00

12.5013.500.203.00 1.000.203.00 0.204.00 20 xx xx x x x x x

Theboxshouldcontain20caramelsand10 cremes.

27. Let r representthespeedofthecurrent.

RateTimeDistance

Upstream16

Downstream16 r r r r   

2016 1 6033 1516 1 6044

Sincethedistanceisthesameineachdirection: 1616 34

4(16)3(16) 644483 167 162.286 7 rr rr rr r

Thespeedofthecurrentisapproximately2.286 milesperhour.

28. Let r representthespeedofthemotorboat.

RateTimeDistance

Upstream3553

Downstream32.52.53 rr rr 

Thedistanceisthesameineachdirection:

5(3)2.5(3) 5152.57.5 2.522.5

Thespeedofthemotorboatis9milesperhour.

29. Let r representthespeedofthecurrent.

(25)(25)4(12)(75)

Sincethetotaltimeis1.5hours,wehave:

10(15)10(15)1.5(15)(15) 15010150101.5(225)

3001.5(225) 200225 250 (5)(5)0 5or5

Speedmustbepositive,sodisregard5

Thespeedofthecurrentis5milesperhour.

30. Let r representtherateoftheslowercar.Then 10 r  representstherateofthefastercar.

RateTimeDistance

Slowercar3.53.5 Fastercar103310

3.53(10) 3.5330 0.530 60 rr

Theslowercartravelsatarateof60milesper hour.Thefastercartravelsatarateof70miles perhour.Thedistanceis(70)(3)=210miles.

31. Let r representKaren’snormalwalkingspeed.

RateTimeDistance

With50walkway2.550 2.5

Against50walkway2.550 2.5

Sincethetotaltimeis48seconds:

Karen’normalwalkingspeedisapproximately 3.75feetpersecond.

32. Let r representthespeedoftheairportwalkway.

RateTimeDistance

Walking280with1.5280 1.5

Walkingwiththewalkwaytakes60secondsless timethanstandingstillonthewalkway:

Speedmustbepositive,sodisregard7 2 r

Thespeedoftheairportwalkwayis2metersper second.

33. Let w representthewidthofaregulationdoubles tenniscourt.Then26 w  representsthelength. Theareais2808squarefeet:

Chapter 1: Graphs, Equations, and Inequalities

Thewidthmustbepositive,sodisregard39 w 

Thewidthofaregulationdoublestenniscourtis36 feetandthelengthis2(36)+6=78feet.

34. Let t representthetimeittakestheBrotherHLL8350CDWtocompletetheprintjobalone. Then9  t representsthetimeittakestheXerox VersaLinkC500tocompletetheprintjobalone.

todojobinoneminute

Timemustbepositive,sodisregard5  t . TheBrotherHL-L8350CDWtakes36minutesto completethejobalone,printing144040 36  pagesperminute.XeroxVersaLinkC500takes 36+9=45minutestocompletethejobalone, printing144032 45  pagesperminute.

35. Let t representthetimeittakestodothejob together.

Lois20

   Workingtogether,thejobcanbedonein12 minutes.

36. Let t representthetimeittakesApriltodothe jobworkingalone. 1 10 1 1 6

todojobinonehour

Aprilwouldtake15hourstopainttherooms.

37. l  lengthofthegarden w  widthofthegarden

a. Thelengthofthegardenistobetwiceits width.Thus,2lw  . Thedimensionsofthefenceare4 l  and 4 w  . Theperimeteris46feet,so: 2(4)2(4)46 2(24)2(4)46 482846 61646 630 5 lw ww ww w w w

     Thedimensionsofthegardenare5feetby 10feet.

b. Area51050 lw  squarefeet

c. Ifthedimensionsofthegardenarethesame, thenthelengthandwidthofthefenceare alsothesame(4) l  .Theperimeteris46 feet,so:

Section 1.6: Problem Solving: Interest, Mixture, Uniform Motion, Constant Rate Job Applications

2(4)2(4)46 282846 41646 430 7.5 ll ll l l l

Thedimensionsofthegardenare7.5feetby 7.5feet.

d. Area7.5(7.5)56.25 lw  squarefeet.

38. l  lengthofthepond w  widthofthepond

a. Thepondistobeasquare.Thus, lw  . Thedimensionsofthefencedareaare6 w  oneachside.Theperimeteris100feet,so: 4(6)100 424100 476 19 w w w

Thedimensionsofthepondare19feetby 19feet.

b. Thelengthofthepondistobethreetimes thewidth.Thus,3lw  .Thedimensionsof thefencedareaare6and6 wl .The perimeteris100feet,so: 2(6)2(6)100 2(6)2(36)100 212612100 824100 876 9.5 3(9.5)28.5 wl ww

Thedimensionsofthepondare9.5feetby 28.5feet.

c. Ifthepondiscircular,thediameteris d and thediameterofthecirclewiththepondand thedeckis6 d  d 3 3

Theperimeteris100feet,so:

Thediameterofthepondis25.83feet. d. 2 square Area19(19)361ft lw

. 2 rectangle Area28.5(9.5)270.75ft lw

. 2 22 circle 25.83 Area=524ft 2 r

Thecircularpondhasthelargestarea.

39. Let t representthetimeittakesforthedefensive backtocatchthetightend.

Def. Back TimeRate

1002525 1233

Tight12sec End 10sec1010 tt tt

100 10 TimetorunDistance 100yards

Sincethedefensivebackhastorun5yards farther,wehave:

Thedefensivebackwillcatchthetightendatthe 45yardline(15+30=45).

40. Let x representthenumberofhighwaymiles traveled.Then30,000 x representsthenumber ofcitymilestraveled.

30,000900 4025

30,000 200200900 4025 5240,0008180,000 3240,000180,000 360,000 20,000

x x x

Thereseisallowedtoclaim20,000milesasa businessexpense.

41. Let x representthenumberofgallonsofpure water.Then1 x  representsthenumberof gallonsinthe60%solution.

Chapter 1: Graphs, Equations, and Inequalities

2 3gallonofpurewatershouldbeadded.

42. Let x representthenumberofliterstobe drainedandreplacedwithpureantifreeze.

%liters%liters%liters

5litersshouldbedrainedandreplacedwithpure antifreeze.

43. Let x representthenumberofouncesofwater tobeevaporated;theamountofsaltremainsthe same.Therefore,weget 2 3 0.04(32)0.06(32) 1.281.920.06

0.060.64

0.64643210 0.0663 x x x x

2 3 1010.67  ouncesofwaterneedtobe evaporated.

44. Let x representthenumberofgallonsofwater tobeevaporated;theamountofsaltremainsthe same.

0.03(240)0.05(240) 7.2120.05 0.054.8 4.896 0.05 x x x x 

96gallonsofwaterneedtobeevaporated.

45. Let x representthenumberofgramsofpure gold.Then60 x representsthenumberof gramsof12karatgoldtobeused.

12(60)(60)23

300.540 0.510 20 xx xx x x

  20gramsofpuregoldshouldbemixedwith40 gramsof12karatgold.

46. Let x representthenumberofatomsofoxygen. 2 x representsthenumberofatomsofhydrogen. 1 x  representsthenumberofatomsofcarbon. 2145 444 11 xxx x x

Thereare11atomsofoxygenand22atomsof hydrogeninthesugarmolecule.

47. Let t representthetimeittakesforMiketo catchupwithDan.Sincethedistancesarethe same,wehave: 11(1) 69 322 2 tt

MikewillpassDanafter2minutes,whichisa distanceof1 3 mile.

48. Let t representthetimeofflightwiththewind. Thedistanceisthesameineachdirection: 330270(5) 3301350270 6001350 2.25 tt tt

Thedistancetheplanecanflyandstillreturn safelyis330(2.25)=742.5miles.

49. Let t representthetimetheauxiliarypump needstorun.Sincethetwopumpsareemptying onetanker,wehave: 31 49 27436 49 92.25 4 t t t

Theauxiliarypumpmustrunfor2.25hours.It mustbestartedat9:45a.m.

50. Let x representthenumberofpoundsofpure cement.Then20 x  representsthenumberof poundsinthe40%mixture.

54. Let r representthespeedoftheeastbound cyclist.Then5 r  representsthespeedofthe westboundcyclist.

RateTimeDistance Eastbound66 Westbound566(5) rr rr 

5poundsofpurecementshouldbeadded.

51. Let t representthetimeforthetubtofillwith thefaucetsonandthestopperremoved.Since onetubisbeingfilled,wehave: 1 1520 4360 60

60minutesisrequiredtofillthetub.

52. Let t bethetimethe5horsepowerpumpneeds toruntofinishemptyingthepool.Sincethetwo pumpsareemptyingonepool,wehave:

The5horsepowerpumpmustrunforan additional1.75hoursor1hourand45minutesto emptythepool.

53. Let t representthetimespentrunning.Then 5 t representsthetimespentbiking.

RateTimeDistance Run66 Bike25525(5) tt tt

Thetotaldistanceis87miles: 625(5)87 61252587 1912587 1938 2 tt tt

Thetimespentrunningis2hours,sothe distanceoftherunis6(2)12  miles.The distanceofthebicycleraceis25(52)75  miles.

Thetotaldistanceis246miles: 66(5)246 6630246 1230246 12216 18 rr rr r r r     

Thespeedoftheeastboundcyclistis18miles perhour,andthespeedofthewestboundcyclist is18523  milesperhour.

55. Burke'srateis10012meters/sec.In9.81seconds, Burkewillrun100(9.81)81.75 12  meters.Bolt wouldwinby100-81.75=18.25meters.

56. 2 22 A rrh .Since58.9  A square inchesand6.4  h inches, 2 2 2 22(6.4)58.9 212.858.90 212.858.90      rr rr rr 2 12.8(12.8)4(2)(58.9) 2(2) 12.8635.04 4 3.1or9.5      r rr . Theradiusofthecoffeecanis3.1inches.

57.Lettheindividualtimestocompletetheproject beEforElaine,BforBrian,andDforeither daughter.Usingtherespectiveratesgives 1111111 , 22  EBEDD (or121 2  ED ), and111 4  BD .Fromthefirsttwoequations, 12  BD .Substitutingintothethirdequation gives211 4  DD or 3112hours. 4  D D Then1213hours 122  E E and

Chapter 1: Graphs, Equations, and Inequalities

1116hours. 124  B B Thecombinedrate ofElaine,Brian,andoneoftheirdaughtersis 1117 361212

 projectperhour,soitwilltake them127hourstocompletetheproject.

58. If x =litersoforiginalsolution,thentherewere originally0.2x litersofsaltand0.8litersofpure water.Overtime,thesolutionloses

0.25(0.8)0.2  x x litersofpurewater.Sheadds 20litersofsaltsothetotalamountofsaltis 0.220  x liters.Shealsoadds10litersofpure water,sothetotalamountofpurewateris 0.80.2100.610  xxx liters.Theresulting concentrationis331/3%whichmeans 0.22010.2201 or 0.2200.61030.8303

xx xxx or 0.6600.830150  xxx .Therewere initially150litersofsolutioninthevat.

59. Thespeedofthetrainrelativetothemanis30–4=26milesperhour.Thetimeis 5551 secminhh. 603600720 

Thefreighttrainisabout190.67feetlong.

60. Answerswillvary.

61. Let x betheoriginalsellingpriceoftheshirt. ProfitRevenueCost 40.4020240.6040 xxxx

Theoriginalpriceshouldbe$40toensurea profitof$4afterthesale.

Ifthesaleis50%off,theprofitis: 400.50(40)204020200

At50%offtherewillbenoprofit.

62. Let12 and tt representthetimesforthetwo segmentsofthetrip.SinceAtlantaishalfway betweenChicagoandMiami,thedistancesare equal.

TheaveragespeedforthetripfromChicagoto Miamiis49.5milesperhour.

63. Thetimetraveledwiththetailwindwas: 9191.67091hours 550 t

Sincetheywere20minutes  1 3hourearly,the timeinstillairwouldhavebeen:

1.67091hrs20min1.670910.33333hrs 2.00424hrs

Thus,withnowind,thegroundspeedis 919458.53 2.00424  .Therefore,thetailwindis 550458.5391.47knots 

64. Itisimpossibletomixtwosolutionswitha lowerconcentrationandendupwithanew solutionwithahigherconcentration.

AlgebraicSolution: Let x =thenumberoflitersof25%solution.

Section 1.7

1. 2 x 

2. True;theabsolutevalueofanumberisthe distancethenumberliesfrom0onarealnumber line.Negativenumberslietotheleftof0ona realnumberline,butstillhaveapositive distancefrom0.

3.  ,,,,,,, A Babcdeiou

4. negative

5. closedinterval

6. aua

7. (,] a 

8. True;thisfollowsfromtheadditionpropertyfor inequalities.

9. True;thisfollowsfromthemultiplication propertyforinequalities.

10. False;sincebothsidesoftheinequalityarebeing dividedbyanegativenumber,thesense,or direction,oftheinequalitymustbereversed. Thatis, ab cc 

11. a

12. c

Section 1.7: Solving Inequalities

13. Interval:   0,2 Inequality:02 x 

14. Interval:  1,2

Inequality:12 x 

15. Interval:   2,  Inequality:2 x 

16. Interval:   ,0 Inequality:0 x 

17. Interval:   0,3 Inequality:03 x 

18. Interval:  1,1 Inequality:11 x 

19. a. 35 3353 68   

b. 35 3555 20   

c.  35 3335 915   

d.  35 2325 610   

20. a. 21 2313 54   

b. 21 2515 34   

c.  21 3231 63   

Chapter 1: Equations and Inequalities

21 2221 42

21. a. 212 21323 245 x x x 

b. 212 21525 243 x x x

c.

d.

212 32132 636 x x x

212 22122 424 x x x

22. a. 125 12353 428 x x x

b. 125 12555 240 x x x

c.  125 31235 3615 x x x

d.  125 21225 2410 x x x

23. [0,4]

24. (–1,5)

25. [4,6)

41. If4,then40. xx

42. If6,then60. xx

43. If4,then312. xx

44. If3,then26. xx

45. If6,then212. xx

46. If2,then48. xx 

47. If840,then5. xx

48. If312,then4. xx

49. If13,then6. 2 xx 

50. If11,then4. 4 xx 

51. 372 39 3 x x x    3or(3,) xx   

52. 251 24 2 x x x    2or(2,) xx 

53. 3518 515 3    x x x 3or[3,)  xx

54. 235 33 1 x x x    1or[1,) xx 

55. 313 24 2 x x x x   

Section 1.7: Solving Inequalities

2or[2,) xx   

56. 223 5 x x x  

5or[5,) xx 

57. 2(3)8 268 214 7 x x x x    

7or(7,) xx 

58. 3(1)12 3312 315 5 x x x x     5or(,5) xx   

59. 43(1)3 4333 313 32 2 3 x x x x x      22 or, 33 xx

2 3  

Chapter 1: Equations and Inequalities

60. 84(2)2 8842 42 60 0 x x x x x x x x

65. 5432 932 32 3 x x x

223or,333 xx

61. 1(4)8 2

62. 1 34(2) 3 3412 33 9122 814 7 4 xx xx xx

33or3,3xx

67. 21 30 4 12210 1121 111 22 x x x x

111111 or, 2222 xx

1 2  11 2

68. 32 04 2 0328 236 22 3 x x x x 

63. 1375 6312 24

x x x

xx 64. 42210 228 14 x x x

44or2,4

222or,233 xx

2 3 

69. 1 114 2 1 03 2

06or60 x x xx

70. 1 011 3 1 10 3 30or03 x x xx 

03or0,3xx

71. 22 (2)(3)(1)(1) 61 61 5 5 xxxx xxx x x x 

5or,5 xx 

72. 22 (1)(1)(3)(4) 112 112 11 11 xxxx xxx x x x 

 11or,11 xx 

73. 28 x  828 44 x x    44or4,4xx

74. 315 x  15315 55 x x 

55or5,5xx

75. 742  x 742or742 6or6   xx xx

Section 1.7: Solving Inequalities

 6or6or,66, xxx

76. 26 x  26or26 3or3 xx xx    3or3or,33,xxx  

77. 324 t  4324 236 22 3 t t t    222or,233 tt 

2 3  

78. 257 u  7257 1222 61 u u u      61or6,1uu   

79. 223 21 x x   121 13 x x    13or1,3xx  

80. 435 42 x x   242 62 x x    62or6,2xx   

Chapter 1: Equations and Inequalities

81. 32 x  32or32 1or5 xx xx

82. 42 x  42or42 6or2 xx xx

83. 527 527

 33 122 or1, xx

 3 2

86. 1241 123 x x   3123 422 42 22 21or12 x x x xx

 12or1,2xx

87. 952  x Nosolutionsinceabsolutevalueisalwaysnonnegative. 

1or6or,16, xxx 84. 231 231 x x

231or231 33or31 11 or 3 xx xx xx



11or1or,1,33 xxx

88. 340 x  Allrealnumberssinceabsolutevalueisalways non-negative.

 or, xx 

89. 3411 48 2 x x x   

|2xx  or  2,   

90. 137 36 2 x x x     |2xx  or  2,     

335 244 335 444 244 635 38 x x x x

 2 22 4321 43441 1 1 xxx x xxx x x

 2 22 9531 95961 561 1 xxx x xxx xx

Chapter 1: Equations and Inequalities

99.  1 327 2 1 39 2 x x   11 39 22 1719 3 22 1719 66 x x x

 1719 |66xx

or 1719 , 66

17 6 19 6 100.  1 4111 4 1 412 4 x x 

11 412 44 4749 4 44 4749 1616 x x x

 4749 |1616xx

or 4749 , 1616

49 16 47 16 101. 35211 826 43 34 x x x x

|34xx or  3,4

102.  23218 23228 2128 127 17 22 71 22 x x x x x x

71 |22xx

7 2 1 2 

103. 714 13 13 313 24 x x x x x

  |24xx or  2,4   

104. 935 34 34 434 71 x x x x x 

|71xx

or7,1 

105. 352 x x  35 8 8 x x x  

and52 5 5 x x x x    Weneedboth8 x  and5 x  .Therefore, 5 x  issufficienttosatisfybothinequalities.  |5xx  or   5,   

106. 232 x x  23 5 5 x x x    and32 3 3 x x x x   

Weneedboth5 x  and3 x  .Therefore, 5 x  issufficienttosatisfybothinequalities.

|5xx  or  5,   

107. 2215 x xx 221 21 3 3 x x x x x   

and215 310 31 1 3 x x x x x 

Weneedboth1 3 x  and3 x  .Therefore, 3 x  issufficienttosatisfybothinequalities.

|3xx  or   3,   

108. 213557 xxx  2135 15 6 6 x x x x x     and3557 257 212 6 xx x x x 

Weneedboth6 x  and6 x  .Therefore, 6 x  issufficienttosatisfybothinequalities.

 |6xx  or  6,   

109. 20.5 x  0.520.5 0.520.52 1.52.5 x x x   

 Solutionset:1.52.5 xx

110. 11 x   111 111 1111 20 x x x x 

 Solutionset:20 xx

111. 32 x  32or32 32or32 5or1 xx xx xx   

 Solutionset:5or1 xxx

112. 23 x  23or23 1or5 xx xx 



 Solutionset:1or5 xxx

113. 21<youngadult'sage<30

114. 40 ≤ middle-aged<60

115. Atemperature x thatdiffersfrom98.6 Fbyat least1.5F  .

98.61.5

98.61.5or98.61.5

97.1or100.1 x xx xx

Thetemperaturesthatareconsideredunhealthy arethosethatarelessthan97.1˚Forgreaterthan 100.1˚F,inclusive.

116. ThelengthLmustbewithin0.0025of5.375 inches..

5.3750.0025

0.00255.3750.0025

5.37255.3775

Thelengthsmustbebetween5.3725and5.3775 inches,inclusive.

117. Thepercentagemustbewithin3.9percentage pointsof64percent.Theinequalitythat representsthiswouldbe:

643.9

3.9643.9 60.167.9    x x x

Theactualpercentageislikelytofallbetween 60.1%and67.9%inclusive.

118. Thespeed x variesfrom707mphbyupto55 mph.

a. 70755 x 

Chapter 1: Equations and Inequalities

b. 5570755 5570755 652762 x x x 

Thespeedofsoundisbetween652and762 milesperhour,dependingonconditions.

119. a. Let x =ageatdeath.

122. Let C representthecommission. Calculatethecommissionrange: 250.4(200)250.4(3000) 1051225 C C  

Thecommissionsareatleast$105andatmost $1225.

3051.7

81.7   x x

Therefore,theaveragelifeexpectancyfora 30-year-oldmalein2019willbegreater thanorequalto81.7years.

b. Let x =ageatdeath.

123. Let W =weeklywagesand T =taxwithheld. Calculatingthewithholdingtaxrange,giventhe rangeofweeklywages:

3055.5

85.5   x x

Therefore,theaveragelifeexpectancyfora 30-year-oldfemalein2019willbegreater thanorequalto85.5years.

c. Bythegiveninformation,afemalecan expecttolive85.581.73.8  yearslonger.

120. 20 VT 

80º120º

80º120º 20 16002400 T V V

Thevolumerangesfrom1600to2400cubic centimeters,inclusive.

121. Let P representthesellingpriceand C representthecommission. Calculatingthecommission:

45,0000.25(900,000)

45,0000.25225,000

0.25180,000

P P

Calculatethecommissionrange,giventheprice range:

225,000180,0000.25180,000275,000180,000 900,0001,100,000

0.25(900,000)0.250.25(1,100,000) 225,0000.25275,000

45,00095,000 P P P P C

Theagent'scommissionrangesfrom$45,000to $95,000,inclusive.

45,000 900,000 0.055% to95,0001,100,0000.0868.6%,  inclusive.

Asapercentofsellingprice,thecommission rangesfrom5%to8.6%,inclusive.

14.9687.340.2283287.3458.9687.34 9001100 9008328321100832 68832268

0.22(68)0.228320.22(268) 14.960.2283258.96 102.3146.3

W W W W W W T

Theamountwithheldvariesfrom$102.30to $146.30,inclusive.

124. Let x representthelengthoftimeSueshould exerciseontheseventhday. 20040450502535300 200195300 5105 x x x 

SuewillstaywithintheACSMguidelinesby exercisingfrom5to105minutes.

125. Let K representthemonthlyusageinkilowatthoursandlet C representthemonthlycustomer bill.

Calculatingthebill:0.093630 CK

Calculatingtherangeofkilowatt-hours,given therangeofbills: 137.64221.88

137.64300.0936221.88

107.640.0936191.88 11502050

C W K K

Theusagevariesfrom1150.00kilowatt-hoursto 2050.00kilowatt-hours,inclusive.

126. Let W representtheamountofsewer/waterused (inthousandsofgallons).Let C representthe customercharge(indollars).

Calculatingthecharge: 23.550.40 CW

Calculatingtherangeofwaterusage,giventhe rangeofcharges:

30.3536.75

30.3523.550.4036.75 6.80.4013.2 1733

Therangeofwaterusagerangedfrom17,000to 33,000gallons.

127. Youhavealreadyconsumed21gramsoffat. LetCrepresentthenumberofcookies.Thenwe havethefollowingequation: 21546 525 5

Youmayeatupto5cookiesandkeepthetotal fatcontentofyourmealnotmorethan46g.

128. Youhavealreadyconsumed150gramsof sodium.Let H representthenumberof hamburgers.Thenwehavethefollowing equation: 1503801290 3801140 3

Youmayeatupto3hamburgersandkeepthe totalsodiumcontentofyourmealnotmorethan 1290g.

129. a. Let T representthescoreonthelasttestand G representthecoursegrade.

Calculatingthecoursegradeandsolvingfor thelasttest: 68828789 5 326 5 5326 5326 T G T G GT TG

Calculatingtherangeofscoresonthelast test,giventhegraderange: 8090 4005450 745326124 74124 G G G T

TogetagradeofB,youneedatleasta74 onthefifthtest.

b. Let T representthescoreonthelasttestand G representthecoursegrade. Calculatingthecoursegradeandsolvingfor thelasttest:

688287892 6 3262 6 163 3 3163 T G T G T G TG

Calculatingtherangeofscoresonthelast test,giventhegraderange: 8090 2403270 773163107 77107 G G G T    

TogetagradeofB,youneedatleasta77 onthefifthtest.

130. Let T representthetestscoresofthepeoplein thetop2.5%. 1.96(12)100123.52 T  Peopleinthetop2.5%willhavetestscores greaterthan123.52.Thatis,123.52 T  or (123.52,). 

131. Since ab  , and 2222 and 22222222 and 22 abab aaababbb abab ab    

So,2ab ab   .

132. Fromproblem131,2ab ab   ,so 2 ,2222 daaababababa     and 2 ,2222 dbbababbabba     Therefore,2ab  isequidistantfrom a and b.

133. If0,then ab

  22 22 22 0and0 and and ababab ababab ababab

   Therefore, aabb 

Chapter 1: Equations and Inequalities

134. Showthat 2 ab ab  

2 12 22 10,since. 2 ab abaabb abab

135. For1111 0,2 ab hab

11 2 11 2 2 ba hh hab ba h ab ab h ab 

22() 2 ()0 abaab ab haa abab abaababa abab aba ab

22

Therefore, ha

22 2()2 2 ()0 babab ab bhb abab abbabbab abab bba ab

Therefore, hb

,andwehave ahb

136. Showthat

2 (geometric2 mean) arithmeticmean1() 2 ab

2 1111 2 211 2 2 1() 2 hab ba habab hab ab ab ab h ab ab

137. 4235 3 312695 31269and695 33514 114 4

Thisisequivalentto14 1 5  x .Thesolution set,inintervalnotation,is14 1,5 

138. Thelargestvalueof223 x occursatthelargest valuefor x 259 34 43 34or43

x x x xx

Thelargestvaluefor223 x is 2 2(4)332329  .

139. Answerswillvary.Onepossibility: Nosolution:  46252 x xx Onesolution:  35231321 xxx 

140. Answerswillvary.

141. Since20 x  ,wehave 2 2 101 11 x x   Therefore,theexpression21 x  canneverbe lessthan5

142. Answerswillvary.

Chapter 1 Review Exercises

Chapter 1 Review Exercises

1. 28 3 624 18 x x x   

Thesolutionsetis  18

2. 2(53)845 106845 6245 6 x x x x x x x

Thesolutionsetis  6

3. 1 5 31 4312 941 51 xx xx x x 

Thesolutionsetis  1 5

4. 6 15 566 6 x x xx x 

Since x =6doesnotcauseadenominatorto equalzero,thesolutionsetis{6}.

5. 2 2 (1)6 6 06 xx xx xx

2241416 12423 bac

Therefore,therearenorealsolutions.

6.   1361 432 1361 1212 432 3(13)4(6)6 394246 1327 27 13 xx xx xx xx x x

Thesolutionsetis  27 13 7. 2 2 (1)(23)3 233 260 (23)(2)03or2 2 xx xx xx xxxx

Thesolutionsetis  2,3 2 8. 2 2 234 0423 2448252 88 2213113 84 xx xx x 

Thesolutionsetis113113 , 44

9. 

32 333 2 2 2 12 12 18 93 x x x xx 

Check:3 x  Check:3 x 

32 3 3 312 912 82 22

Thesolutionsetis  3,3

10. 2 (1)20 20 11817 22 xx xx x     Norealsolutions.

11. 2 310 1112111 66 xx x    Norealsolutions.

Chapter 1: Equations and Inequalities

12.

42 22 22 540 410 40or10 2or1 xx xx xx xx

Thesolutionsetis  2,1,1,2

13. 2 2 233 233 2396 8120(2)(6)0 2or6 xx xx xxx xxxx xx

Check2: 2(2)32123

Check6: 2(6)369693 x x

Thesolutionsetis  2 14.

4 444 232 232 2316 213 13 2 x x x x x 

444

Check13: 2 13 23133162 2 x

Thesolutionsetis

22 +1121 +1121 12+11121 22+1121 2+111 xxx xxx xxxxx xxxx xx

Check5: 2 555 +1121 222 1.798907439951.79890743995

Check5: 2 555 +1121, 222 x x

Thesecondsolutionisnotpossiblebecauseit makestheradicandnegative.

Thesolutionsetis5

Thesolutionsetis

19.

Thesolutionsetis

222 222 102360 5180 5920 axabxb axabxb axbaxb 

590 59 9 5 axb axb b x a    or20 2 2 axb axb b x a   

Check1:

20.

Thesolutionsetis  92 5,,0. bb a aa 

  22 22 22 22 222 2 373920 37392 37392 37394394 66439 xxxx xxxx xxxx xxxxxx xxx  

22 2 22 22 61439 36211639 3672361648144 xxx xxxx xxxx

2 2 20241080 56270 59309or3 5 xx xx xxxx

Chapter 1: Equations and Inequalities

Check9:

Thesolutionsetis{–5,2}.

22. 2329 237 x x 

237 35 5 3 x x x

23.

or237 39 3 x x x

Thesolutionsetis  5,3 3

32 32 2 2 23 230 230 00 2303 2 xx xx xx xx xx

Thesolutionsetis   0,3 2

24.

 32 2 2 2 258200 254250 2540 250or40 5or2 2 xxx xxx xx xx xx

Thesolutionsetis  5,2,2 2

25.

2 1311 122 23111 1212 23111 23311 4111 412 3 xx xx xx xxxx xx xx x x x



Since3doesnotmakeanydenominatorequalto 0,thesolutionsetis  3

26.  2 29 29 23 23 x x x x     5 x  or1 x 

Thesolutionsetis   1,5.

27. 3530xx UsetheZerooptionfromtheCALCmenu.

Thesolutionsetis   2.49,0.66,1.83.

28. 4321 x x  UsetheIntersectoptionontheCALCmenu.

Thesolutionsetis  1.14,1.64

29. 23 2 52

2(23)10(2)5 46205 14 14 x x x x x x x x 

30. 23 97 4 362328 33231 3331 22 3133 22 x x x x x  

 31333133 or, 2222 xx

31. 33 26 12 243372 21369 723 x x x x 

 237or23,7xx

32. 1 34 2 11 34 22 97 3 22 37 26 x x x x

3737 or, 2626

Chapter 1: Equations and Inequalities

38. 2 3339393 33310 933 93 1010 iii iii iii i

39.

40.

12 504824212111 iiiii

32 2 2 232323 412923 51223 10152436 469 iii iii ii iii i

41. 210xx

22 1,1,1, 41411143 131313 21222 abc bac i x i

Thesolutionsetis 1313 , 2222 ii 

42. 2 220 xx

22 2,1,2, 4142211617 117117 224 abc bac x 

Thesolutionsetis 117117 , 44

43. 23 x x 

2 22 30 1,1,3, 4141311211 111111111 21222 xx abc bac i x i

Thesolutionsetis 111111 , 2222 ii

44. (1)2xx 220 (2)(1)02or1 xx xxxx 

Thesolutionsetis

45. 215 yx

46.  4,0,0,0,2,0 ,  0,2,0,0,0,2 x-intercepts:4,0,2 y-intercepts:2,0,2

47. 236 326 22 3 xy yx yx

48. 29 yx

49. 2 2 2 216 216 18 2 xy yx yx

50. 22 plw  51.

90000.071$630

52. Let x representtheamountofmoneyinvestedin bonds.Then70,000 x representstheamount ofmoneyinvestedinCD's. Sincethetotalinterestistobe$5000,wehave:

Chapter 1: Equations and Inequalities

0.080.05(70,000)5000

1000.080.05(70,000)5000100

$50,000shouldbeinvestedinbondsat8%and $20,000shouldbeinvestedinCD'sat5%.

53. Using s vt  ,wehave3and1100 tv .

Findingthedistance s infeet: 1100(3)3300 s  Thestormis3300feetaway.

54. 16003600 I 

RaftHelicopter Rate590 Time Dist.590 tt tt

Sincethetotaldistanceis150miles,wehave: 590150 95150 1.58hours1hourand35minutes tt t t

Thehelicopterwillreachtheraftinabout1hour and35minutes.

57. Let d representthedistanceflownbythebee travelingat3meterspersecond.

(Timesneededtomeetareequal.) 150 35 54503 8450

56.25 56.25meters18.75seconds 3 dd dd d dt

Thebeesmeetforthefirsttimeafter18.75 seconds.

Therangeofdistancesisfrom0.5metersto0.75 meters,inclusive.

55. Let s representthedistancetheplanecantravel. 22 WithwindAgainstwind

Rate2503028025030220 Time(/2)(/2) 280220

Dist. ss ss 

Sincethetotaltimeisatmost5hours,wehave:

/2/2 5 280220 5 560440 11145(6160) 2530,800 1232 ss ss ss s s

Theplanecantravelatmost1232milesor616 milesonewayandreturn616miles.

56. Let t representthetimeittakesthehelicopterto reachtheraft.

Thebeeswillmeetasecondtimeonthesecond lap.Thefirstbeewillhavetraveled 150+ x metersandthesecondbeewillhave traveled150+(150– x)meters. Solvingfortime,wehave: 150150(150) 35 150300 35 75059003 8150 18.75metersintothesecondlap 168.7556.25seconds 3 x x xx xx x x t

Thebeesmeetthesecondtimeafter56.25 seconds(37.5secondslater).

58. Giventhat2 12803216 s tt , a. Theobjecthitsthegroundwhen0 s 

2 2 012803216 2800 108010,8 tt tt tttt 

Theobjecthitsthegroundafter8seconds. b. After4seconds,theobject’sheightis  2 1280324164896 s  feet.

59. Let t representthetimeittakesClarissato completethejobbyherself.

ClarissaShawna Timetodo5 jobalone Partofjob11 done5 in1day Timeonjob66 (days) Partofjob66 donebyeach5 person

Sincethetwopeoplepaintonehouse,wehave:

x

62. Letthelengthofleg1= x. Thenthelengthofleg2=17– x BythePythagoreanTheoremwehave

222 22 2 2 1713 34289169 2341200 17600 125012or5 xx xxx xx xx xxxx

thelegsare5cmand12cmlong.

63. a. Considerthefollowingdiagram:

IttakesClarissa10daystopaintthehousewhen workingbyherself.

60. Let x representtheamountofthe$8-per-pound coffee.

4650

Thepaintingis6.5inchesby6.5inches. 612.5 s  ,sotheframeis12.5inchesby 12.5inches.

b. Considerthefollowingdiagram: w

Add263poundsof$8/lbcoffeetoget2 263 poundsof$5/lbcoffee.

61. Let x representtheamountofwaterevaporated.

12 33 41221250 626 26428 6 ww w wlw

Thepaintingis283inchesby143inches. Theframeis2143inchesby1103inches.

Chapter 1: Equations and Inequalities

64. Let t representthetimeittakesthesmaller pumptofinishfillingthetank. 3hpPump8hpPump Timetodo128 jobalone Partofjob11 done128 in1hr Timeonjob44 (hrs) Partofjob44 donebyeach128 pump

Sincethetwopumpsfillonetank,wehave: 441

Ittakesthesmallpumpatotalof2more hourstofillthetank.

65. Let4 w  .Solveforthelength:

Thelengthoftheplasterboardshouldbecut toalengthofapproximately6.47feet.

66. Let x representthenumberofpassengersover 20.Then20 x  representsthetotalnumberof passengers,and150.1x representsthefarefor eachpassenger.Solvingtheequationfortotal cost,$482.40,wehave:

the114.Therefore,201636  peoplewenton thetrip;eachpersonpaid  150.116$13.40  .

67. Let rS representScott'srateandlet rT represent Todd'srate.ThetimeforScotttorun95meters isthesameasforToddtorun100meters.

95100

0.95 0.950.95 ST ST SsTT rr rr dtrtrd

IfToddstartsfrom5metersbehindthestart: 105

0.950.95(105)99.75 T ST d dd

a. Theracedoesnotendinatie.

b. Toddwinstherace.

c. Toddwinsby0.25meters.

d. Toendinatie: 1000.95(100) 100950.95

50.95 5.263meters x x x x 

e. 95=0.95(100)Therefore,theraceendsina tie.

f,g,h. Raceisatie.

68. Ineachproblem,weneedtousetheLeast CommonMultipleoftheexpressions 22and4xx ,namely24. x

a. Weuse24 x astheLCDinorderto combinethegivenexpressions.

b. Wemultiplyeachsideoftheequationby 24 x inordertoclearoutthe denominatorsbeforesolvingtheequation.

c. Weuse24 x astheLCDinorderto combinethetermsonthelefthandside beforesolvingtheinequality.

Sincethecapacityofthebusis44,wediscard

Chapter 1 Test

1.

2.

Sinceneithersolutionisarestrictedvalue,the solutionsetis

4.  58416 58446 58410 918 2 xx xx xx x x 

Thesolutionsetis 2. 5. 53278 53215 323 b b b

  323 20 0 b b b    or323 26 3 b b b

  

Thesolutionsetis 0,3.

6. 422 42 38 280 xxx xx 

 Let2 ux  .Then  2 224 uxx  ,andwehave

 2280 420 uu uu 

 4or2uu

Sincewearesolvingfor x,weget 24 2 x x   or 22 2 norealsolution x x  

Thesolutionsetis 2,2.

7. 2420 1,4,2 xx abc     2 2 4 2 44412 21 4168 2 48 2 422 2 22 bbac x a           

Thesolutionsetis  22,22 

Chapter 1: Equations and Inequalities

8.  2 22 2 2172 2172 630 xxxx xxxx xx

Thesolutionsetis

9. 2721 7221 23 7 xy yx yx

2 7 2 7 2 7 3, 77357,5 00330,3 2 77317,1 7 x yxxy y y y

Theinterceptsare

2 2 2 2 2 2 5, 33543,4 11541,4 00550,5 11541,4 33543,4 x yxxy y y y y y

Theinterceptsare

11. 32 2210 xxx Sincethisequationhas0ononeside,wewill usetheZerooptionfromtheCALCmenu.Itwill beimportanttocarefullyselectthewindow settingssoasnottomissanysolutions.

Thesolutionstotheequationare1,0.5,and1.

12. 42580xx

Sincethisequationhas0ononeside,wewill usetheZerooptionfromtheCALCmenu.

Thesolutions,roundedtotwodecimalplaces, are2.50and2.50.

13. 327233xxxx 

Sincetherearenonzeroexpressionsonboth sidesoftheequation,wewillusetheIntersect optionfromtheCALCmenu.Entertheleftside oftheequationinY1andtherightsideinY2.

15. 2343 234434 237 x x x  

237 210 5 x x x    or237 24 2 x x x   

Solutionset:  |5or2xxx

Interval:     ,52, 

16. 7358 7335383 1055 1055 555 21 12 x x x x x x

Solutionset:  |12xx

Interval: 

1,2

Thesolutions,roundedtotwodecimalplaces, are2.46,0.24,and1.70

14.

17. 348 8348 1234 44 3 x x x x 

44 3 xx 

or 4,4 3

4 3 

18.  237411614411 641411 225 iiii ii i 

19.   2 310838108 2438010 2483101 1483 iiiii iii i i 

22. Wecanuseatabletosummarizethegiven information. minutesper#ofcust.in

Since8565204.63 27  ,itwilltakeJamieand Scottabout204.63minutes(3.41hours)tocheck out65customersworkingtogether.

23. Let x =poundsofbananachips.Thenthetotal poundsofthemixwillbe40 x  Sincethereistobenolossinrevenue,weget

newcher. newmixbananacherrymix ban. pricelbspricelbspricelbs 4100 25 revenuerevenuerevenue 6.75402.759.2540 6.752702.75370

Theretailerneedstomix25poundsofbanana chipswiththe40poundsoftheoriginalmix.

24. Let x =saleprice.Weknowthatthediscountis  0.42275.00115.50  .Therefore, salepriceoriginalpricediscount 275115.50 159.50 x x 

 Thesalepriceis$159.50.

25. Hereweneedthesimpleinterestformula, I Prt .Inthiscasewehave0.04 r  , 3months1 12yr months/yr4 t  ,and10,000 P    1 10,0000.044 100

IPrt    Glennwillearn$100.00ininterestafter3 months.

Chapter 1 Projects

Project I

Internet-based Project

Project II

1.  ,3,5,1,0.2 331 0.23510.660.22 n TnLMC CnpLM T ppp

2. Allofthetimesgiveninproblem1werein seconds,so T =0.1boardpersecondneedsto usedasthevaluefor T intheequationfoundin problem1.

3. T =0.15boardpersecond

Thus,only23partsperboardwillwork.

Forproblems4–6, C isrequested,sosolvefor C first:

n T CnpLM CnpLMTn CnpTLTMTn CnpTnLTMT

4. T =0.06, n =3, p =100, M =1, L =5 

350.0610.06 0.147sec 31000.06 C 

5. T =0.06, n =3, p =150, M =1, L =5



350.0610.06 0.098sec 31500.06 C 

6. T =0.06, n =3, p =200, M =1, L =5

  350.0610.06 0.073sec 32000.06 C 

7. Asthenumberofpartsperboardincreases,the tacttimedecreases,ifalltheotherfactorsremain constant.

Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.