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# Class Note for MATH 1310 at UH

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Date Created: 02/06/15

Secl39nn 2 sum Quzdn c Equz nnshszcmrEng Page I nf23 Solving by Factoring A quadratic equz nn ls an equatronthat ean be wntten tn the form ax bx e 0 where a b anol are real numbers wrth a 0 To solve a quaolratre equatron by factonng rewnte the equatron lf neeessary so that one slde ls equal to 0 anol use the Zernrl rnduct Prnperty ab0lfandonlylfa0orb0 Example 1 Solve the equatlon 2x2 7n 3 0 by faetonng Snlutinn to Example 1 The equauon ls tn the eorreet form srnee the nghtrhand slde ls 0 Faetor the le rhand slde anol use the zerorproduct property 2x2 7n 3 0 2x1x30 Faetor 2x1 or x 0 ZerorProduct Property 1orx3 r i one 2 Thesoluuonsarex andx 3 Example 2 Solve the equatron x2 5x100by faetonng Snlutinn to Example 2 Rewnte the equataon so that one slde ls equal to 0 Next factor anol use the zerorproduct property Secl39nn 2 sum Quzdn c Equz nnsh y raemmg x25x100 5x500 a 5x7 500 0 x725x200 Factor x7 25 0 or n20 0 ZerorProductPropeny 20 The soluuons are x7 5 max x25 or x 20 Example 3 Solve the equatlon x2 Ex gt12 by factonng Snlntinn m Example 3 x28x712 x28x1271212 x28x120 x6x2 0 Ayyly me Zemer ml pmpmy x60 or x20 Solve both equauom Elm equauon Second equauon MS 0 x2 0 x6 o5 x27 02 x 2 76 andx The soluuons are x Example 4 Solve the nquatlon 5x2 14x3 by laczonng Snlntinn m Example 4 Page 2 am Secl39nn 2 Sulving Quzdn c Equz nnsh y Picturing 5x244 5x214x714x314x 5x214x3 5f14xe3 7 5 14xe3 splxxn 0 mm Zemerm PmyenY 5x710 or x30 Solve both equations First equauon Second equation 5x710 x30 5x711 01 x373073 5 3 x The soluuons are F and x Example 5 Solve the equauon 4x2 81 by factoring Snlutinn m Example 5 2x92xe9 mum Zemer deyeny 2x9 0 or 2x79 0 Pig 3 nf23 Secl39nn 2 Sulving Quzdn c Equatinnshyraetnring Pigeonm Solve both equations Fust equatton Seeond equation 2i9 0 2i 9 0 2i9e 2x79909 The solutions are Example 5 Solve the equation 6 e i 4x by faetonng Snllltinn to Example 5 6x2 7 i 4x 6x2 7x74i4x74x 6x2 7 5x 0 none 5 0 innuun z nepnuntpnn M or one From the rst equation we see that one solution 15 i 0 Solve the seeond equation Seeond equation oneso oi75505 ops g Secl39nn 2 sum Quzdn c Equz nnsh y raemrhrg The soluuohs are x 0 andx 5 Example 7 The length ofarectangle ls 3 feet longer than lts wldth Kthe area ofthe rectangle ls 40 square feet whatls lts wldth7 Snlntinn m Example 7 Let x wldLh of the rectangle m feet Then x3 the length ofthe rectangle Wnte an equatlon wldLhlenth area xx3 x2 3x x2 3x740 x2 3x740 ramps 0 meme Zaraeeraaarerarmy x8 0 or x750 Solve both equauorrs Flrst equatlon Second equation x80 x750 x878 078 x75505 x 8 x5 We mustrule out 78 for the wldth smee the wldth cannot be hegauve Page 5 am Page a nf23 Secl39nn 2 Sulving Quzdn c Equz nnsh y raeuriug Sulving Quzdn c Equz nnshy Cumpkling the Square Check the results x5 x3 53 8 Wldth 5a Length 8a Area 5 mm rt 40 sq 1quot Answer the problem m a eomplete sehtehee The wrolth othe reetarrgle ls 5 feet Secl39nn 2 Sullving Quzdn c Equz nnsh y Camp leliug the Square Solving by Completing the Square To solve a quaolratre equatrorr x2 bxc by the methool ofcnmple ngthe square rewrrte the equatrorr as x2 bx m and make the le rhand slde a perfeet square 2 To make x2bx aperfectsquare mag the square ofhalfthe eoemereht ofx to 2 2 getx2bx 5 r 2 2 To solve ar bxc 0 where a as 0 anda as 1 by the methool ofcnmpledngthe square rewrrte the equatrorr as ar bx m and therr dlvlde both sldes by a to get 2 x2 x 7 Next make 5 aperfeet square by addmg i the square of a a a 211 halfthe coef clent ofx Example 1 Solve the equatror x2 10x 3 0 by eompletmg the square Snlutinn tn Example 1 Secl39nn 2 Sulving Quzdn c Equz nnsh y Camp lethrg the Square Page 7 m x 710x30 x2710 Subtract 3 from both states 2 x2710x257325 Completethesquareby addlng 25 x75 22 r4145 x5m Faetor the perfeet square on the LHS The solutrorrs are x 5JZ and r5745 Example 2 Solve the equatrorr 2r2 8x1 0 by eompletmg the square Snlutinn tn Example 2 2x2 8x1 0 2r28r71 Subtraet 1 from both sldes x24x 7 Dlvlde both sldes by 2 2 r 4r4 e m Complete the square by addlng Slmpllfy on the RHS Faetor the perfeet square on the LHS m T Secl39nn 2 Sulving Quzdn c Equz nnsh y Camp leling the Square Page s m Example 3 Solve the equatlon x2 4x7 5 0 by eompleung be square Snlntinn m Example 3 x274x750 x274x75505 x274x5 l cemplee be same m x2 e My leange 4 2 x723 or x7 Solve eael equauon by addlng 2 to both sldes Flrst equatlon Second equatlon x7 2 3 x7 2 e 7 x72232 x72 x 5 The soluuons are x 5 andx Example 4 Solve the equatlon x2 10xl3 0 by eompleung the square Snlntinn m Example 4 Secl39nn 2 Snlving Quzdn c Equz nnsh y Camp Eling the Square Pig 9 nf23 x210x1 0 224402443713 0713 X210X713 2 221022571325 Campletethesquarefmxzl x byaddnngl m 2 221022512 x5212 45 12 2375 7523 The solutions are 27572 md2752J Example 5 2 Solve the equauonx E2 by eompleung the square Snlntinn m Example 5 2 1 1 2 2 1 2 2 exeele campxeumeemerm eexbymng e e 9 9 2 2 2 1 E Secl39nn 2 Snlving Quzdn c Equz nnsh y Camp Eling the Square Pig In an gm Example 5 Solve the equauon 2x2 8x3 0 by completing the square Snlntinn m Example 5 2x28x30 2x28x37 073 2 2 Secl39nn 2 Snlving Quzdn c Equz nnsh y Camp Eling the Square Page 11 an Sulving Quzdn c Equz nnsh y the Quzdn c Furnmlz x24x4 x22 E 2 E 2 The solutions ofthe equauon ax bxc 0 where a a 0 ean be found by usmgthe quadratic fnrmulz Example 1 Solve the equauon 5x2 2x 2 0 by the quadratic formula Snlntinn m Example 1 Subsmuce a 5 b 72 and 772 mo the quadratic formula Secl39nn 2 sum Quzdn c Equz nnsh y the Quadnlic Funnle Page l2 an 2 72274572 21W721m7212J 1 J17 25 10 10 10 5 ml lee andx The soluuohs are x 5 5 Example 2 Solve the equatlon 4x2 ex 3 by the quadratlc formula Snllltinn m Example 2 Rewnte the equatlon so thatthe nghtrhand slde ls 0 4x27 4 X30 Substltute a 4 b 71 and J 12 4 24 73 mm the quadratic formula 7 8 The soluuohs are x andx Example 3 Solve the equahon x2 7 4x 5 by the quadratic formula Secl39nn 2 sum Quzdn c Equz nnsh y the Quadnlic Funnle Page IS an Snlntinn m Example 3 To solve the equatlon uslng the quadratic formula rewnte me equauon so that the RHS ls equal to 0 274x5 750 Substltute a 1 and 75 mm the quadratic formula 75 ill 74a 2a 4gt1lt gt2 41 5 7 21 4t1620 2 7443 x The soluuons arex5 andxrl Example 4 Solve the equatlon x2 10xl3 0 by the quadratic formula Snlntinn m Example 4 Substltute a 1 b 10 and 13lntothe quadraue formula Secl39nn 2 Sulving Quzdn c Equz nnshy the Quadnlic Furnmlz 75 xbz 7 4a 2a 710L011 74mm 21 10 ere100 752 2 The solutions are e esez and e 752J Example 5 Solve the equauon 2x2 8x3 0 by me quadrath formula Snlutinn m Example 5 Subsmutea 2 b 8 and 3 mo the quadraue formula 75 tel 7411 2421 78L 82 7 423 22 781647 24 4 Pig u an Secl39nn 2 Sulving Quzdn c Equz nnsh y the Quadnlic Furnmlz Fig 15 um The soluuohs are x imam aim Example 6 The sum 6 the squares 6 two posmve consecuuve even mtegers 15 1460 Whatxs the smallermtege Snlutinn tn Example 6 Let x the smaller hteger Then rd 2 the larger hteger whte an equauon square of the smaller hteger square 6f the larger hteger 1460 x2x221460 x2x24x41460 2x24x 460 2r24r4e146 46071460 2x24xe14560 Subsmute a 75 t 452 7 4a 2a 7 445 42 74271456 22 4 L0 6 11648 4 4t ll 1664 4 4108 4 1456 mto the quadratn formula Deterrruhe the s61uu6hs t6 the quadraue equauon 2x2 4xe1456 0 Seeinn 2 Slhing Quzdn c Equz nnsh y die Quadnlic Furnmlz Page 16 m The Discriminanl 74741087112 728 or X 74108 26 4 4 We rnustruie out the solution x e28 einee the problem eans for positive integers Check the results x 26 rd 2 e 28 2o2 676 28 7784 575 784 1450 39 Answer the problem in a eornpiete sentenee The srnaiierinteger i526 Seeinn 2 The Discriminanl The Discriminant The discriminant ofthe equation ax bxc 0 a e 0 is given by 13524 1m gt 0 thentne equation ax bxc Uhas two distinet real solutions 1m 0 thentne equation ax bxc has exaetiy one real solution 1m lt 0 thentne equation ax bxc Uhas no real solution Example 1 re s e 0 determine how many reai solutions without solving the equation x2 the equation has Secl39nn 2 The Diserlminam Page I7 om Solution to Example Flnd the dlscnmlnant D 71 waxes e 24 e 5 gt 0 The equatloh has two dlstlnctreal solutlohs The gure below shows why the equatloh x2 7x76 has two real solutlohs 6 paths The two solutaohs to the equatloh x2 xeo 0 are x 2 andx3 whleh are the xahtereepts of the graph ofthe equatlony x2 e x7 6 Example 2 wlthout solvlng the equataoh 4 12x9 detel mlne how manyreal solutlohs the equatloh has Solution to Example Flnd the dlscnmlnant 13022 74mm 1447144 0 The equatloh has exaetly one real solutloh The gure below shows why the equatloh 4x2 12x9 Uhas one real solutloh Secl39nn 2 The Distriminanl Fig m an x 2 y4x 12x9 M r 35325215Vn5 The solution to the equahon 4x2 12x9 0 he x 715 whxchxs the xehteheepc ofthe graph ofthe equanony 4 12x 9 Example 3 Wxthout solmhg the equamh x5 0 determan how manyreal solutions the equamh has Snllltinn m Example Fmd the ducnmmant 7415172 719 lt0 The equauoh has no real soluuoh The gure below shows whythe equamh x2 n5 0 has no real solumh Seernn 2 The Diserlminam Page 19 an The equatron x2 x5 0 has no real solutron The graph of the equatron y x5 does not cross the xraxls The graph has no xantereepts Example Use the dlscnmmant to determlne the number of real soluuons to the equanon 4x2 573x Do not solve the equahon Snllltinn tn Example Rewnte the equatron so that the RHS ls equal to 0 4x2 573x 4x23x573x3x 4x2 3x5 4x23x75575 4x23x750 Substltute a 4 A 3 and 5 lnto the dlscnmmant Dbz 74a 32 James 980 89 The equatlon has two dlstlnct real solutlons slnee D 89 gt 0 Example 5 Use the dlscnmmant to determlne the number of real soluuons to the equanon x26x 13 Do not solve the equanon Snllltinn tn Example 5 Secl39nn 2 The Discriminanl Pagem an Rewnte the equatmh so that the RHS is equal to 0 Substxtute a 1 A 6 and 13 mto the ducnmmant Dbz 74a 6274113 79 716 The equatmh has no real soluuons smee D 716 lt 0 Example 5 Use the ducnmmant to detemuhe the number of real solutions to the equation 25x2 60x 736 Do not solve the equahoh Snllltinn tn Example 5 Rewnte the equatmh so that the RHS is equal to 0 25x2 60x736 25x2 60x36 736 36 25x260x360 Substxtute a 25 b 60 and 36 mto the ducnmmant 13427415 602 742536 36007 3600 0 The equatmh has exactly one real solutmh smee D 0 Secl39nn 23 The Distriminanl Pig 21 um Example 7 Use the ducnmmant to nd all values ofk so thatthe equation 21a 8xk 0 has exactly one real solution Snllltinn m Example 7 Subsmutea 2k 5 8 and e k mto the ducnmmant D 52 7 4a 82 742kk 647 812 The equauoh has exactly one soluuoh when D 0 SeLD 0 and solve the resulting equamh for k 06478k2 Exercise Set 23 Quadratic Equations Solve the following equations by factoring 1 10 11 12 13 14 x2 710x210 x2 13x400 xx7235 xx820 x2 l4x72 x276011x 2x277x7150 3x277x40 6x2 l7x712 10x277x6 x2 7250 x2 7490 4x2790 36x2 725 0 Solve the following equations by factoring To simplify the process remember to first factor out the Greatest Common Factor GCF 15 16 17 18 19 20 21 22 23 24 x2 78x0 x2 10x0 73x2 21x0 5x2 730x0 3x27120 77x270 75x2715x900 74x220x240 80x2 230x7300 12x2775x180 Find all real solutions of the following equations by completing the square 25 x28x120 26 x276x7400 27 xx710718 28 x24x78 29 x214x600 30 xx12728 31 x25x750 32 x277x20 33 3x2712x750 34 2x212x73 35 74x28x76 36 75x2740x7780 37 3x275x72 3s 8x276x750 Find all real solutions of the following equations by using the quadratic formula 39 x25x20 40 x277x30 41 x276x80 42 x272x15 43 x25x70 44 x278x716 45 x210x250 46 4x276x50 47 3x275x71 4s 2x23x760 49 72x276x730 50 75x28x710 Exercise Set 23 Quadratic Equations Find all real solutions of the following equations by using a method of your choice 51 x2710x160 52 x276x780 53 2x274x5 54 3x212x36 Use the discriminant to determine the number of real solutions of each equation Do not solve the equation 55 x275x20 56 2x23x70 57 x24x74 58 3x276x73 59 x24x75 60 3x27x720 61 xzicxid0wheredgt0 62 3x277xt0wheretlt0 Use the discriminant to find all values ofk so that each of the following equations has exactly one solution 63 kx26x3k0 64 4x2kx490 For each of the following problems a Model the situation by writing appropriate equations b Solve the equations and then answer the question posed in the problem 65 The length of a rectangular frame is 5 cm longer than its width If the area of the frame is 36 cm2 find the length and width of the frame 66 The height of a right triangle is 4 inches longer than its base If its diagonal measures 20 inches find the base and the height of the triangle 67 68 69 70 71 72 The height of a triangle is 3 cm shorter than its base If the area of the triangle is 90 cm2 find the base and height of the triangle Find x if the area of the figure below is 26cm2 Note that Ihe gure may natbe drawn to scale 8cm XCIH Find two consecutive odd integers that have a product of 255 Find two numbers that have a sum of 39 and a product is 350 John can paint the fence in 3 hours less time than Chris If it takes them of an hour when working together how long does it take each of them to paint the fence individually Marie takes 1 hour more time to clean the house than Tina If it takes them 1 hours when working together how long would it take each of them to clean the house individually

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