SOLVING QUADRATIC EQUATIONS BY FACTORING


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1 6.6 Solving Qudrti Equtions y Ftoring (6 31) 307 In this setion The Zero Ftor Property Applitions 6.6 SOLVING QUADRATIC EQUATIONS BY FACTORING The tehniques of ftoring n e used to solve equtions involving polynomils. These equtions nnot e solved y the other methods tht you hve lerned. After you lern to solve equtions y ftoring, you will use this tehnique to solve some new types of prolems. The Zero Ftor Property In this hpter you lerned to ftor polynomils suh s x x 6. The eqution x x 6 0 is lled qudrti eqution. helpful hint The prefix qud mens four. So why is polynomil with three terms lled qudrti? Perhps it is euse qudrti polynomil n e ftored into produt of two inomils, whih ontin totl of four terms. Qudrti Eqution If,, nd re rel numers with 0, then x x 0 is lled qudrti eqution. The min ide used to solve qudrti equtions is the zero ftor property, whih sys tht if produt is zero, then one or the other of the ftors is zero. The Zero Ftor Property The eqution 0 is equivlent to 0 or 0. In our first exmple, we use the zero ftor property nd ftoring to redue qudrti eqution into two liner equtions. Solving the liner equtions gives us the solutions to the qudrti eqution. E X A M P L E 1 Using the zero ftor property Solve x x 6 0. First ftor the polynomil on the left side: x x 6 0 (x 3)(x ) 0 Ftor the left side. x 3 0 or x 0 Zero ftor property x 3 or x Solve eh eqution. We now hek tht 3 nd stisfy the originl eqution.
2 308 (6 3) Chpter 6 Ftoring For x 3: For x : x x 6 ( 3) ( 3) 6 x x 6 () () The solutions to x x 6 0 re 3 nd. A sentene suh s x 3 or x, whih is mde up of two or more equtions onneted with the word or is lled ompound eqution. In the next exmple we gin solve qudrti eqution y using the zero ftor property to write ompound eqution. E X A M P L E Using the zero ftor property Solve the eqution 3x 3x. First rewrite the eqution with 0 on the righthnd side: 3x 3x 3x 3x 0 Add 3x to eh side. 3x(x 1) 0 3x 0 or x 1 0 Zero ftor property x 0 or x 1 Solve eh eqution. Ftor the lefthnd side. Chek 0 nd 1 in the originl eqution 3x 3x. For x 0: For x 1: 3(0) 3(0) 3( 1) 3( 1) There re two solutions to the originl eqution, 0 nd 1. CAUTION If in Exmple you divide eh side of 3x 3x y 3x, you would get x 1 ut not the solution x 0. For this reson we usully do not divide eh side of n eqution y vrile. The si strtegy for solving n eqution y ftoring follows. helpful hint We hve seen qudrti polynomils tht nnot e ftored. So not ll qudrti equtions n e solved y ftoring. Methods for solving ll qudrti equtions re presented in Chpter 10. Strtegy for Solving n Eqution y Ftoring 1. Rewrite the eqution with 0 on the righthnd side.. Ftor the lefthnd side ompletely. 3. Use the zero ftor property to get simple liner equtions. 4. Solve the liner equtions. 5. Chek the nswers in the originl eqution. 6. Stte the solution(s) to the originl eqution.
3 6.6 Solving Qudrti Equtions y Ftoring (6 33) 309 E X A M P L E 3 Using the zero ftor property Solve (x 1)(x 1) 14. study tip To write the eqution with 0 on the righthnd side, multiply the inomils on the left nd then sutrt 14 from eh side: Set shortterm gols nd rewrd yourself for omplishing them. When you hve solved 10 prolems, tke short rek nd listen to your fvorite musi. (x 1)(x 1) 14 Originl eqution x x 1 14 Multiply the inomils. x x 15 0 Sutrt 14 from eh side. (x 5)(x 3) 0 Ftor. x 5 0 or x 3 0 Zero ftor property x 5 or x 3 Chek 5 x 5 or x 3 nd 3 in the originl eqution: ( 5 1) 5 ( 4) 7 14 ( 3 1)(3 1) (7)() 14 So the solutions re 5 nd 3. CAUTION In Exmple 3 we strted with produt of two ftors equl to 14. Beuse there re mny pirs of ftors tht hve produt of 14, we nnot mke ny onlusion out the ftors. If the produt of two ftors is 0, then we n onlude tht one or the other ftor is 0. If perfet squre trinomil ours in qudrti eqution, then there re two identil ftors of the trinomil. In this se it is not neessry to set oth ftors equl to zero. The solution n e found from one ftor. E X A M P L E 4 An eqution with repeted ftor Solve 5x 30x Notie tht the trinomil on the lefthnd side hs ommon ftor: 5x 30x (x 6x 9) 0 Ftor out the GCF. 5(x 3) 0 Ftor the perfet squre trinomil. (x 3) 0 Divide eh side y 5. x 3 0 Zero ftor property x 3
4 310 (6 34) Chpter 6 Ftoring You should hek tht 3 stisfies the originl eqution. Even though x 3 ours twie s ftor, it is not neessry to write x 3 0 or x 3 0. The only solution to the eqution is 3. CAUTION To simplify 5(x 3) 0 in Exmple 4, we divided eh side y 5. If we hd used the zero ftor property, we would hve gotten 5 0 or (x 3) 0. Sine 5 0 hs no solution, we n ignore it nd ontinue to solve (x 3) 0. If the lefthnd side of the eqution hs more thn two ftors, we n write n equivlent eqution y setting eh ftor equl to zero. E X A M P L E 5 helpful hint Compre the numer of solutions in Exmples 1 through 5 to the degree of the polynomil. The numer of rel solutions to ny polynomil eqution is less thn or equl to the degree of the polynomil. This ft is known s the fundmentl theorem of lger. An eqution with three solutions Solve x 3 x 8x 4 0. We n ftor the fourterm polynomil y grouping: x 3 x 8x 4 0 x (x 1) 4(x 1) 0 Ftor out the ommon ftors. (x 4)(x 1) 0 Ftor out x 1. (x )(x )(x 1) 0 Differene of two squres x 0 or x 0 or x 1 0 Zero ftor property x or x or x 1 Solve eh eqution. You should hek tht ll three numers stisfy the originl eqution. The solutions to this eqution re, 1, nd. Applitions There re mny prolems tht n e solved y equtions like those we hve just disussed. E X A M P L E 6 Are of grden Merid s grden hs retngulr shpe with length tht is 1 foot longer thn twie the width. If the re of the grden is 55 squre feet, then wht re the dimensions of the grden? If x represents the width of the grden, then x 1 represents the length. See Fig Beuse the re of retngle is the length times the width, we n write the eqution x(x 1) 55. x 1 ft FIGURE 6.1 x ft
5 6.6 Solving Qudrti Equtions y Ftoring (6 35) 311 helpful hint To prove the Pythgoren theorem, drw two squres with sides of length, nd prtition them s shown. We must hve zero on the righthnd side of the eqution to use the zero ftor property. So we rewrite the eqution nd then ftor: x x 55 0 (x 11)(x 5) 0 Ftor. x 11 0 or x 5 0 Zero ftor property 11 x or x 5 The width is ertinly not 1 1. So we use x 5 to get the length: x 1 (5) 1 11 We hek y multiplying 11 feet nd 5 feet to get the re of 55 squre feet. So the width is 5 ft, nd the length is 11 ft. The next pplition involves theorem from geometry lled the Pythgoren theorem. This theorem sys tht in ny right tringle the sum of the squres of the lengths of the legs is equl to the squre of the length of the hypotenuse. Ersing the four identil tringles from eh piture will sutrt the sme mount of re from eh originl squre. Sine we strted with equl res, we will hve equl res fter ersing the tringles: The Pythgoren Theorem The tringle shown in Fig. 6. is right tringle if nd only if. Hypotenuse FIGURE 6. Legs E X A M P L E 7 Using the Pythgoren theorem The length of retngle is 1 meter longer thn the width, nd the digonl mesures 5 meters. Wht re the length nd width? 5 x + 1 FIGURE 6.3 x If x represents the width of the retngle, then x 1 represents the length. Beuse the two sides re the legs of right tringle, we n use the Pythgoren theorem to get reltionship etween the length, width, nd digonl. See Fig x (x 1) 5 Pythgoren theorem x x x 1 5 Simplify. x x 4 0 x x 1 0 Divide eh side y. (x 3)(x 4) 0 x 3 0 or x 4 0 Zero ftor property x 3 or x 4 The length nnot e negtive. x 1 4
6 31 (6 36) Chpter 6 Ftoring To hek this nswer, we ompute 3 4 5, or So the retngle is 3 meters y 4 meters. CAUTION The hypotenuse is the longest side of right tringle. So if the lengths of the sides of right tringle re 5 meters, 1 meters, nd 13 meters, then the length of the hypotenuse is 13 meters, nd M A T H A T W O R K Cn you suessfully invest money nd t the sme time e soilly responsile? Geet Bhide, president nd founder of Wlden Cpitl Mngement, nswers with n emphti yes. Ms. Bhide helps lients integrte their soil vlues with portfolio of stoks, onds, sh, nd sh equivlents. With the lient s onsent, Ms. Bhide might invest in onds ked y the Deprtment of Housing nd Urn Development for housing in speifi innerity res. Other hoies might e environmentlly onsious ompnies or ompnies tht hve proven reord of equl employment prties. In mny ses, tegorizing prtiulr ompny s soilly onsious is judgment ll. For exmple, mny oil ompnies provide good return on investment, ut they might not hve n unlemished reord on oil spills. In this instne, piking the est of the worst might e the orret hoie. Beuse suh trdeoffs re neessry, lients re enourged to define oth their investment gols nd the soil idels to whih they susrie. As ny investment dvisor would, Ms. Bhide tries to minimize risk nd mximize rewrd, or return on investment. In Exerises 81 nd 8 of this setion you will see how solving qudrti eqution y ftoring n e used to find the verge nnul return on n investment. INVESTMENT ADVISOR WARMUPS True or flse? Explin your nswer. 1. The eqution x(x ) 3 is equivlent to x 3 or x 3.. Equtions solved y ftoring lwys hve two different solutions. 3. The eqution d 0 is equivlent to 0 or d If x is the width in feet of retngulr room nd the length is 5 feet longer thn the width, then the re is x 5x squre feet. 5. Both 1 nd 4 re solutions to the eqution (x 1)(x 4) If,, nd re the sides of ny tringle, then. 7. If the perimeter of retngulr room is 50 feet, then the sum of the length nd width is 5 feet. 8. Equtions solved y ftoring my hve more thn two solutions. 9. Both 0 nd re solutions to the eqution x(x ) The solutions to 3(x )(x 5) 0 re 3,, nd 5.
7 6.6 Solving Qudrti Equtions y Ftoring (6 37) EXERCISES Reding nd Writing After reding this setion, write out the nswers to these questions. Use omplete sentenes. 1. Wht is qudrti eqution?. Wht is ompound eqution? 3. Wht is the zero ftor property? 4. Wht method is used to solve qudrti equtions in this setion? 5. Why don t we usully divide eh side of n eqution y vrile? 6. Wht is the Pythgoren theorem? Solve eh eqution. See Exmple (x 5)(x 4) 0 8. ( 6)( 5) 0 9. (x 5)(3x 4) (3k 8)(4k 3) w 9w t 6t y y q 3q m m h h 3 0 Solve eh eqution. See Exmples nd m 7m 18. h 5h p p 4 1. x 5x 3. 3x 10x 7 3. (x )(x 6) 1 4. (x )(x 6) 0 5. ( 3)( 1) ( 3)(3 4) (4 5h) 3h 8. w(4w 1) 1 Solve eh eqution. See Exmples 4 nd x 50 0x 30. 3x 48 4x 31. 4m 1m y 0y x 3 9x x x w 3 4w 4w n 3 3n 3 n 38. w 3 w 5w y 3 9y 0y m 3 m 3m 0 Solve eh eqution. 41. x x x x x 3 4x 47. 3x 15x x x z 1 1 z m 8 3 m (t 3)(t 5) x(x 1) (x ) x (x 3) (x ) x 1 8 x 1
8 314 (6 38) Chpter 6 Ftoring h 1 h m 4 m 3 100m 100m Solve eh prolem. See Exmples 6 nd Dimensions of retngle. The perimeter of retngle is 34 feet, nd the digonl is 13 feet long. Wht re the length nd width of the retngle? 60. Address ook. The perimeter of the over of n ddress ook is 14 inhes, nd the digonl mesures 5 inhes. Wht re the length nd width of the over? ADDRESS BOOK 5 in. 66. Missing ges. Molly s ge is twie Anit s. If the sum of the squres of their ges is 80, then wht re their ges? 67. Skydiving. If there were no ir resistne, then the height (in feet) ove the erth for sky diver t seonds fter jumping from n irplne t 10,000 feet would e given y h(t) 16t 10,000. ) Find the time tht it would tke to fll to erth with no ir resistne, tht is, find t for whih h(t) 0. A sky diver tully gets out twie s muh free fll time due to ir resistne. ) Use the ompnying grph to determine whether the sky diver (with no ir resistne) flls frther in the first 5 seonds or the lst 5 seonds of the fll. ) Is the sky diver s veloity inresing or deresing s she flls? FIGURE FOR EXERCISE Violl s throom. The length of Violl s throom is feet longer thn twie the width. If the digonl mesures 13 feet, then wht re the length nd width? 6. Retngulr stge. One side of retngulr stge is meters longer thn the other. If the digonl is 10 meters, then wht re the lengths of the sides? x m 10 m x m FIGURE FOR EXERCISE Conseutive integers. The sum of the squres of two onseutive integers is 13. Find the integers. 64. Conseutive integers. The sum of the squres of two onseutive even integers is 5. Find the integers. 65. Two numers. The sum of two numers is 11, nd their produt is 30. Find the numers. Height (thousnds of feet) Time (seonds) FIGURE FOR EXERCISE Skydiving. If sky diver jumps from n irplne t height of 856 feet, then for the first 5 seonds, her height ove the erth is pproximted y the formul h 16t 856. How mny seonds does it tke her to reh 8000 feet. 69. Throwing sndg. If lloonist throws sndg downwrd t 4 feet per seond from n ltitude of 70 feet, then its height (in feet) ove the ground fter t seonds is given y S 16t 4t 70. How long does it tke for the sndg to reh the erth? (On the ground, S 0.) 70. Throwing sndg. If the lloonist of the previous exerise throws his sndg downwrd from n ltitude of 18 feet with n initil veloity of 3 feet per seond, then its ltitude fter t seonds is given y the formul S 16t 3t 18. How long does it tke for the sndg to reh the erth? 71. Glss prism. One end of glss prism is in the shpe of tringle with height tht is 1 inh longer thn twie the se. If the re of the tringle is 39 squre inhes, then how long re the se nd height?
9 6.6 Solving Qudrti Equtions y Ftoring (6 39) 315 x + 1 in. x in. FIGURE FOR EXERCISE Ares of two irles. The rdius of irle is 1 meter longer thn the rdius of nother irle. If their res differ y 5 squre meters, then wht is the rdius of eh? 73. Chnging re. Lst yer Otto s grden ws squre. This yer he plns to mke it smller y shortening one side 5 feet nd the other 8 feet. If the re of the smller grden will e 180 squre feet, then wht ws the size of Otto s grden lst yer? 74. Dimensions of ox. Rosit s Christms present from Crlos is in ox tht hs width tht is 3 inhes shorter thn the height. The length of the se is 5 inhes longer thn the height. If the re of the se is 84 squre inhes, then wht is the height of the pkge? 77. Crpeting two rooms. Virgini is uying rpet for two squre rooms. One room is 3 yrds wider thn the other. If she needs 45 squre yrds of rpet, then wht re the dimensions of eh room? 78. Winter whet. While finding the mount of seed needed to plnt his three squre whet fields, Hnk oserved tht the side of one field ws 1 kilometer longer thn the side of the smllest field nd tht the side of the lrgest field ws 3 kilometers longer thn the side of the smllest field. If the totl re of the three fields is 38 squre kilometers, then wht is the re of eh field? 79. Siling to Mimi. At point A the ptin of ship determined tht the distne to Mimi ws 13 miles. If she siled north to point B nd then west to Mimi, the distne would e 17 miles. If the distne from point A to point B is greter thn the distne from point B to Mimi, then how fr is it from point A to point B? MIAMI 13 mi B x in. N A x + 5 in. x 3 in. FIGURE FOR EXERCISE Flying kite. Imeld nd Gordon hve designed new kite. While Imeld is flying the kite, Gordon is stnding diretly elow it. The kite is designed so tht its ltitude is lwys 0 feet lrger thn the distne etween Imeld nd Gordon. Wht is the ltitude of the kite when it is 100 feet from Imeld? 76. Avoiding ollision. A r is trveling on rod tht is perpendiulr to rilrod trk. When the r is 30 meters from the rossing, the r s new ollision detetor wrns the driver tht there is trin 50 meters from the r nd heding towrd the sme rossing. How fr is the trin from the rossing? FIGURE FOR EXERCISE Buried tresure. Ahmed hs hlf of tresure mp, whih indites tht the tresure is uried in the desert x 6 pes from Cstle Rok. Vness hs the other hlf of the mp. Her hlf indites tht to find the tresure, one must get to Cstle Rok, wlk x pes to the north, nd then wlk x 4 pes to the est. If they shre their informtion, then they n find x nd sve lot of digging. Wht is x? 81. Emerging mrkets. Ctrin s investment of $16,000 in n emerging mrket fund grew to $5,000 in yers. Find the verge nnul rte of return y solving the eqution 16,000(1 r) 5, Venture pitl. Henry invested $1,000 in new resturnt. When the resturnt ws sold yers lter, he reeived $7,000. Find his verge nnul return y solving the eqution 1,000(1 r) 7,000.
SOLVING EQUATIONS BY FACTORING
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