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# Introductory Algebra MTH 65

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This 36 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 65 at Portland Community College taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/224638/mth-65-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15

Mr Simonds MTH 65 class quadratic equation group work MTH 65 Group Work Quadratic Equations Please work each ofthese problems on your own paper Check your answers with your group mates and by expanding your result Problem set 1 Solve each equation using the quadratic formula a 2x2 4x 30 b x23x 20 c t3t 710 d 3x25x Problem set 2 Choosing an approach You39ve learned three different methods for solving quadratic equations Another skill you need to develop is the ability to make appropriate choices with regards to the solution method to apply to a given equation WfIauf cranana re equa ans in any way whafsaever decide what would be the best solution method for each of the following equations The options are as follows i use the zero principle ii use the square root property iii use the quadratic formula iv there39s no way of knowing which method to use without first doing some algebra The equations are a x25x 60 b x23x 80 c x23x 89 d x229 e x2 2x50 f xx 6 8 g x 6x 6 8 h 2x2 x40 i x249x0 j 2x 13x26x5x 4 Problem 3 Go ahead and solve each of the equations in problem set 2 Problem 4 Solve equations a b d f and i using a different method than you used in problem 3 Page 1 of1 x23x 20 2 0x2 3x2 a1 b 3 c2 t3t 710 t2 4t 2110 t2 4t 310 a1b 4c 31 Scrafch work divide 2 our of each Term MTH 65 Simonds class Quadratic formula group work solutions The solufions are 245 2 Scratch work divide 2 our of each ferm 3 1 31 x x20rx1 The solufions are 1 and 2 4 435 T 4r2 T 2ixE The solufions are 2 i V35 Page 1 of 6 MTH 65 Simonds class Quadratic formula group work solutions d 3x25x 3 3x2 x50 1i 12435 x a3b 1c5 23 1i J1 80 The equaTion has no real number 6 soluTions 1m 6 Problem 23 0 x2 5x 6 0 Since The IefT side of The equaTion easily facTors I39m going To go ahead and use The zero principle x2 5x 6 0 x6x 10 The soluTions To The equaTion are 1 and 6 x60 or x 10 x 6 or x 1 b x 23x 80 Since I have a facTored expression equaTed To zero I39d be nuTTy To noT jusT go ahead and use The zero principle x23x 80 x 2 0 or 3x 8 0 The soluTions To The equaTion are 2 and 8 x 2 or x A 0 x 23x 89 There39s no way To choose a meThod unTiI I FOIL ouT The IefT side and see whaT I39ve goT x23x 89 3x2 2x 169 3x2 2x 250 Hmm maybe The lefT side facTors and maybe iT doesn39T I Think I39ll jusT go ahead and use The quadraTic formula Page 2 of 6 MTH 65 Simonds class Quadratic formula group work solutions x b 25 2 i V304 c 6 The soluTions To The equaTion are d 2iJ1619 1219 24J19 6 1i2J19 3 1 2J1 9 3 x22 9 A perfecT square equals a posiTive number sounds like square rooT properTy maTeriaI To me x22 9 x23 orx23 The soluTions To The equaTIon are 1 and 5 xlor x 5 x2 2x50 The IefT side don39T facTor so iT39s Time for The quadraTic formula song all over Two aehhh a b 2 x 21 This equaTion has no frickin real number soluTions c 5 2 i 16 2 xx 6 8 I need To expand Things ouT before deciding on a soluTion meThod xx 6 8 x2 6 x 8 0 Since The IefT side of The equaTion facTored so nice and Tidyly I wenT ahead and used The zero principle Sorry for The lack of x2x40 heads up x 2 0 or X 4 0 The soluTions To The equaTion are 2 and 4 x2or x4 Page 3 of 6 MTH 65 Simonds class Quadratic formula group work solutions x 6x 6 8 You know whaT x 6x 6 is jusT The long way of wriTing x 62 So meThinks me besT be using The square rooT properTy Gee golly ThaT was shorT and sweeT I could square real x 6x 6 8 numbers from now unTil The cows come home and I ain39T x 62 8 never gonna geT 8 So I reckon This here liTTle equaTion jusT doesn39T have any real number soluTions 2x2 x 40 Since I39m noT wanTing To Think abouT wheTher 2x2 x 40 facTors or noT I Think I39ll mosey on over To The quadraTic formula 2 x2 x 40 2 x2 x 40 0 Z a 2 x W The soluTions To The equaTion are I 1 22 c 40 1ix321 1x321 andl xl321 4 4 4 x2 49x 0 A liTTle zero principle acTion is called for here x2 49x0 xx490 The soluTions To The equaTion are 0 and 49 x0 or x490 x0 or x 49 2x 13x 2 6x 5x 4 Some preliminary algebraic inTervenTion is definiTely on The menu TonighT 2x 13x26x5x 4 Aw man we were punked The equaTion wasn39T even 6x2 x 26x2 19x 20 quadpa cl x 2 19x 20 20x 2 18 The only soluTion To This lowly linear equaTion is 910 x0 Page 4 of 6 MTH 65 Simonds class Quadratic formula group work solutions Problem 4 0 x25x 60 a1 5r 52 41 6 x H 21 6 c M The solufions To The equa rion are 2 1 and 6 5 7 7 x OIquot x b x23x 80 3x2 2x 160 x a3 23 32 2 2iJ196 c 16 7 2 14 The solufions To The equa rion 2 8 6 are 2 andA d 2 4 4 41 5 x229 x 2 21 x 4x49 414 x2 4x 5 0 T The soluflons To The equa rlon are land 5 46 4 6 x or x c 5 xlor x 5 Page 5 of 6 xx7678 x276x80 74949 xT Ol39I MTH 65 Simonds class Quadratic formula group work solutions x 21 6 JZ The solutions To The equation are 2 2 and 4 6 2 6 7 2 x T or x x4or x2 7 749 i 492 7410 T 7 749 i 492 2 The solutions To The equa rion are 7 749 i 49 7 2 0 and 749 749749 x 2 x0 or39x749 Page 6 of 6 Mr Simonds MTH 65 Square roots and the square root property Key Concepts Simplifying square roots The square root property MartinGaye sections and practice problems 811 190dd 821 13odd 91 1 350dd Square roots If b is a positive real number the principal square root of b is the positive number a with the property that a2 b Positive real numbers in fact have two square roots one positive J3 and one negative xE I J6 0 u The square roots of a negative real number are not real numbers Find each square root R 41 KinC a m o g 43 q rl 15 quot l39 o rckl 39quotLtf a F 1 l 3 Page 1 of 4 Mr Simonds MTH 65 Square roots and the square root property The product rule for square roots If 6120 and b20 rhen abgE Simplify each expression J5 J5 3 m 1 1S 7 g L l JEE m 3quot WK Lm W70 ZrlaJxH a J39Jc W 294 1 C 39 4 r gumu r A cfr J MixE 4 mm Nth2 4 l 3 Mt L1 2103 L39 3f I Page 2 of 4 Mr Simonds MTH 65 Square roots and the square root property The square root property If 6120 and u2athenu or u x Use the square root property to find all solutions to each equation Lu Z X LL J 9 l e N W 04 i 1quot 3m Pg 1 l 1 6K WIU39IL 31 2 t 1 393 Page 3 of 4 2 I 3a gtzn 23116 quot 3x6218 1a in L 3W 4 V W quot2 3 C5W rf zy f7 y 3 I 3wct 42 x 1 J In 6an c f ILv Jung aJt 3x HE A 4 L g 6 l 391 c3J 7 3 5t 7272 sic7V 7L 5quot 7 m g 3 47739 L L6 kaJ 5 4 LQJ L WL 1 1 7 i L H39CL 7quot 1139ch 39 5 S 5 CLwK L 1 lt 2 47374 77 7w1 81 I L 6ft quart l 67 on FIJI ki rftxb L M nTH39 C39 C L L66 A Page4of4 W Slug 54 Mr Simonds MTH 65 The quadratic formula Key Concepts The quadratic formula MartinGave sections and practice 93 1 55 odd The quadratic formula If a 72 0 Then The soluTions To The equaTion ax2 bx c 0 are b i ibz 4610 2a x Use The quadraTic formula To solve each equaTion 11x2 10x20 222z3 Page 1 of 4 Mr Simonds MTH 65 The quadratic formula 10 x22x 5t2 2t10 Page 2 of 4 Mr Simonds MTH 65 The quadratic formula 2 y 3 1 2 y 511 41122 2 Page 3 of 4 Mr Simonds MTH 65 The quadratic formula 4x2 4x 1 36x2 15 Page 4 of 4 MTH 65 Mr Simonds class Key Concepts Polynomials Definitions Addition and Subtraction Multiplication part1 52 1 81odd89 111odd MartinGave sections and practice 53 135 19 21 23 29 31 33 35 37 45 47 55 57 65 efinitions A Polynomial Term is a constant or a product of variables raised to whole number powers or a product of a constant and variables raised to whole number powers The constant factor of a polynomial term is called the coefficienfof the term The sum of the exponents on the variable factors is called the degree of the term A constant term has degree 0 A single polynomial term or the sum of two or more polynomial terms is called a polynomial A polynomial with one term is called a monomial A polynomial with two terms is Wed a binomial A polynomial with three terms is called a frinomial Example kq Yy List the terms of the polynomial 4x2 y7 x4 xy2 7y 9 Also state the coefficient and degree of each term 3 i 4 I 4M f wL quot73 Coefficient IL I I I 7 0 Degree Cl quot 3 l O Page 1 of 7 MTH 65 Mr Simonds class Addition and Subtraction of Polynomials 1 Distribute the subtraction sign if subtracting Remove parentheses regardless 2 Combine the like terms Like terms have the same variable factors including exponents Examgles Find each sum or difference x27x 25x2 7 kwm 4 4 KS XL 7X v 3XL7X 2 rx 77x 35 4 134 3 S x za 2XL39 u Q u k a Y x1 lx7 3787X 4 2KXLk1x l 2 L1XL ExtMIX 7 Page 2 of 7 x1 xr kl MTHBS MrSimonds class FindThesumof 3xy2 4xy 10xy2 and yx2 8yx 9y2x 3791 ij ngtltquotI 9xe 47x V L ZXJL LXj 0ij XJ 4 75 quot lxj a x f 44 44 Find The difference of x 9 and y 9 mm 4 WW 1 X jl lf SubTr39acT Twice The polynomial a2 Zab b2 from The polynomial Zaz 2ab sz L L 1 26 5 i gal155 IQ a L lax 4ampL 39l39l l 2 12 LP15 T L7 K cal Page 3 of 7 MTH 65 Mr Simonds class Multiplying Polynomials 1 Use the distributive property of multiplication to eliminate all grouping symbols 2 Simplify each polynomial term 3 Combinetheliketerms Examples Find each product Ix lxquot 4KL2 K4x 4Hquot W l l l f 393 I3 x5 x 4343 V lx1 4x23x3 4x29x11 vy jx 558 ch f1 L 4 5X44 uv22u3 v2 11uv 7 uv UJ i WW 7 1 1 Au Page 4 of 7 MTH 65 Mr Simonds class 39s7x3 and x2 5x2 7x44 J sxuj xl 39xl 1x339 4 7y 4 4 L Spo L xcx 43x 7x5 3 454 1 3 J 3 My 1 L 7y3x1 sx 736 393Sgtc 4 va 4 331 Ark 4 7X3 3LLL39 ltlC Multiply The polynomials 6 x and 2x2 x 2 60 1x1 gtlt ll ll listx w quotlxj Wk Rx 39 9343 4 3 XL le L ll Page 5 of 7 MTH 65 Mr Simonds class Multiply The polynomials 2x 2y and 3x y 5 Mama 35 Llxj Wquot J 43 quotU 39 Cx 7 fo HM 39Jjquot 03 CO Kjx YXKJL Multiply The polynomials x y and xy2 7xy 8xZ y L y L y xjl xj i bcj gt85 XL jL 7X j D jj Xf 7xy 1 W y L 7x2yquot r7342 39gxijij3 7Xj1 Page 6 of 7 MTH 65 Mr Simonds class Find The producT 6 x2 x 4 x2 x 7 cf x K XL X 7 T 6yT x3 71quot X3 it WYLI39 Lx39 lt 1 f Laq 7x3 y quotx A 5 cgrrw MK P xv Ly llx gt4 L 7 1vquot Ply quotlf Lxquot quotY Hf 1H Ll Find and simplify The volume and surface area of a cube if The heighT of The cube is Twice The widTh and The lengTh is 2 more Than The widTh Use The variable w for The widTh 1 xQVJL Lam qu LM 1 UT Q In 1LJL U QNjJ VmL f t bn S 5 If H La SP 1LJLJL 4 1U391U i JUH i 391 2 lu uiA l QU JU 4 L U 39i QU 4amp1 4 fuz39 421 4 YD mufpllw Page 7 of 7 Mr Simonds MTH 65 graphing parabola Key Concepts Graphing parabolas MartinGaye sections and practice problems 65 71 76 all 95 1 23 odd 33 35 37 l I l Complete Table 1 and then graph onto Figure 1 the quadratic equation y x2 2x 3 Table 1 x 4 3 2 1 0 1 3 4 5 6 2 yli 11 5 03 444 0 JL2 39 Lulka 39IJ lbJ 4Lj 4393wa 0 3 L a X i lquotquot quot1J at ebb A Lol 3 9 11 1 WWW quotJ Keptr Ur Figure 1 Hour A L6 State the intercepts vertex and axisofsymmetry for the parabola y 62 2x 3 W a UL g Mrs at If an 5 72 Page 1 of 8 Mr Simonds MTH 65 graphing parabola CompIeTe Table 1 and Then graph onTo Figure 1 The quadraTic equaTion y x2 2x 10 Table2 x 7 6 5 4 3 2 1 0 1 2 3 y 0 Cf 15 0 ll ll IL ll 139 0gt L 5 CDHCNLB Qquot Q tuud f Figure2 STaTe The inTercest verTex and axisofsymmeTry for The parabola y 2 Ex2 2x 10 3 1JTMCUT1 O n X39lurmw X Jrampcamp r 5 1n j o lt ad Q l 3 D i x1 x 10 39 o 39 Vzr 391 Q x1 uh 10 4310 L 12 c lx 10 139 i S 139 S at 4quot I x f lrmm39 FM 0 j j 1 Q t J JO C 2 1 riL X Page of8 Z 3 L LI a 7 l JLY lt am Mr Sim onds MTH 65 graphing parabola x For each parabola use The formula x 7 To find The xcoordinaTe of The verTex Then make a a Table showing 7 poinTs cenTered aT The verTex if possible ThaT lie on The parabola Find and sTaTe The inTercest of The parabola Finally graph The parabola y74x212x727 j 1 43 f KJ LII L 3917 l I L 5 lLI It 1 A l39 UL39 X 3 7 3 Utr l lt39 L 39 7 ll lug ML quot X I quot 39 J I 73 Page 3 of 8 Mr Sim onds MTH 65 graphing parabola 2 y 7 x 7 8 x 1 2 3 x y a l gtL 5 7quot 7 gtlt 3 3 gt1 1 H W sz l L Y S lt L JLIL L 27 A4 er X 3k x Fa er gt K LIquot 1 Jr 1 LA m7 W AI A R C T U 4 M 3 D X 39LCI 7ltL1XIL b th L A o 39x 7X IL 3 L 1i qr 1 Lquot C I Page4of8 L X wig Eh 4 15 t W Kr bl NLAHI rQMI jdm4 0 Iv Mr Sim onds MTH 65 graphing parabola yx279x25 Mil 41 3 Hquot W5 L qr 34 39 1quot 1 i3 IluJuhxi 39 10 If Lan N quot X 015 lL IA 4 hr l a W quot391cJ bahcw 3 9 w J Jquot lszu 7 Mn N 41 L1X I k1 1 J KJ a l l l Page 5 of 8 Mr Sim onds MTH 65 graphing parabola X m A 3 ix7w 1 39K L iqq 39 7 h 5 f E 1I 71 lLl 441 V x x at h y 1x1 I L37 139 06 1 L 7 1 iquot 391L 1 4 T 201 on w A I ILu j i r a 011 Page 6 of 8 Mr Simonds MTH 65 graphing parabola DieTer39 swiped his sisTerquots lacrosse sTick and ball and climbed To The roof of his gr39anny39s condo Tower39 DieTer39 sTuck The ball in The neT of The sTick and flung The ball sTr39aighT up wiTh The sTick exachy aT r39oof level and The ball 3 feeT pasT The edge of The roof The heighT above The ground of The ball fT 1 seconds afTer39 DieTer39 did his flinging is modeled by The funcTion 111 161 2 381 86 Find each of The following The heighT of gr39anny39s condo Tower39 a b The maximum heighT reached by The ball c The Time To The near39esT 10 of a second iT Took for The ball To hiT The ground once DieTer39 had flung The ball a L bi 7 rio calm1 W 7LT huh L 11 vulur 0quot uJ LN L L Lia1quot 9 I LJ f 1 Q J 11 Ill L1 d39un l k 4 U 4 L 2 fwsLud Ll Lg 21 u m LL 1 T rudxl u 4 nix L QL l y7q A Ul quotUj Fquot kH l b Tch F I 3 4 3 1 SvLl L 1L 4 4L LJV G k 1U H 1 Jun 39 JL w Jcnr vl w 4 L W gtL quot 393 3 kamead 39mb emWAM 50 k 6 of mummm foc n nmm ek fl JIM b 39 J H r thj 4quot b 501 1 Page7of8 ij ld k Mr Simonds MTH 65 graphing parabola GI OUQ work questions 1 lquot 0quot 5 Find the vertex of each parabola a yx2 7x11 b y 2x2 12x 50 c yx2 8x 9 d y3x82 Find the intercepts of each parabola a yx2 7x11 b y 2x2 12x 50 c yx2 8x 9 d y3x82 Graph each parabola Make sure that you clearly show the symmetry of the parabola and that your axes are well labeled and scaled a yx2 7x11 b y 2x2 12x 50 c yx2 8x 9 d y3x82 The next day Ricky ran into Abdou in the Subway shop on Capitol Highway Ricky said whaddup Abdou said Check it out fool I took the tofu pita pocket my moms packed me for lunch and tossed it up in the air from the bleachers of the Wilson High athletic field The height above the ground ft of the pita pocket 1 seconds after I tossed the nasty thing can be modeled by the function 111 161 2 321 48 I bet you can39t figure out how high the pita was flung nor how long it took for the pita to hit the groundquot Lucky Ricky just happened to have a notebook and pencil with him so he sat down in a booth and did some figuring Ricky once again impressed Abdou by coming up with the correct answer to both of his questions Recreate Ricky39s calculations and find the correct answers to each of Abdou39s stumpers The hypotenuse of a certain right triangle is 3 inches longer than twice the length of one of the legs The other leg of the triangle is 11 inches long Find to the nearest 10 of an inch the length of the hypotenuse of the triangle Page 8 of 8 Mr Simonds MTH 65 paraboa group work answers 1 a and b x coor dinafe ycoor dinafe Ver fex a x2 a a1 7 y352 73511 35125 b 7 E 125 35 b b x 2a 2 a 2 12 y 2 3 12 3 50 3532 b 12 22 32 3 c i 2a a1 8 y42849 425 b 8 E 25 4 d y3x82 xi 9x248x64 2quot 33 2 48 y s W a9 29 0 b48 Please nofe Fr acfional answers would be fine and dandy for ques rion 1 a Decimal answers would no be accep rable for 1 d because The decimals would no be exact Find The in rer cep rs of each parabola 0 x2 7x110 a1 b 7 c11 The xinfer cepfs are H Em and 7 0 2 2 The yin l er cepf is 011 Page 1 of 4 Mr Simonds MTH 65 rparaboia group Work answers 72 712x7500 1 1 734 712x750750 x 125 There are no xinTer cest i 6x 25 0 a1 2 Theyintercept is 011 b 6 c 25 These are not real numbers x2 7 8x 7 9 0 x 7 9X 1 0 The ximercepfs are 9 0 and 710 x7900rx10 Theyintercept is 079 x 9 orx 71 3x 82 0 3x80 The onlyximercepf is 70 x8 A Whenx0y08z64 Theyintercept is 064 r Page 2 of 4 3 Mr Simonds MTH 65 rparaboia group Work answers The maximum heighT reached by The piTs can be deTermined by finding The verTex of The parabola ht716tZ 32t 48 a 716 and b 32 so The t coordinaTe of The verTex is t 7 1 This Tes us ThaT 32 2716 iT only Took one second for The piTa To reach iTs maximum heighT The maximum heighT is deTermined by 111 64 So The maximum heighT reached by The piTa was 64 feeT The piTa hiT The ground when ht 0 716 32t480 ii716t2 32t487i0 16 16 t2 721730 I73t10 t3ort71 Since The nasTy piTa didn39T hiT The ground before Abdou Tossed iT in The air iT musT have Take 3 seconds for The piTa To reach The ground once iT had been Tossed by Abdou Page 3 of 4 Mr Simonds MTH 65 parabola group work answers 5 LeT x represenT The lengTh of The unknown leg measured in inches Then The lengTh of The hypoTenuse is 2x 3 and from The PyThagorean Theorem we geT x2 112 2x 32 x21122x32 a3 x2 12i4122 4l3ll 112 x21214x212x9 b12 23 03x212x 112 c 112 12iJ1488 6 121488 6 m 44 an 84 d 12 J1488 N 6 Since The lengTh of a righT Triangle leg surely can39T be negaTive The unknown leg lengTh musT be abouT 44 inches This makes The hypoTenuse lengTh abouT 118 inches Check 442 112 14036 and 1182 13924 The answer checks the numbers are not exactly the same because of our rounding Page 4 of 4 Mr Simonds MTH 65 1 0 8iJ44 8r 411 Scratch work divide 2 ou r of each ferm 6 6 8iJZJ11 6 8211 6 4iJ 3 b39 10im10i1 93 C39 12iV100 12i10 2 2 4 4 10iJ J 2 1210land 12 10E 10i3J 4 2 4 2 2 2 0 5w 22 88 6 6t102 72 5w 2iJ 6t10i 5W2izm 6t10i6J2 Swzzim 6t710r6x 2im tM W 5 6 I775i3x5 T 3 The solufions are 5 i 3J5 5 The solufions are b 3x12 40 3x12 4 The equa rion has no real number solu rions Page 1 of 2 Mr Simonds MTH 65 Problem 3 LeT x represenT The lengTh fT of each side of The original pen Then The lengThs of The sides of The new pen are shown in The diagram below I 16xx7744 I A 7744 2 16 x2 7744 I 2 x I I x 484 I M x r 484 hquotquotquotquot xi22 x 06x 16x Since The lengTh cannoT be negaTive each side of The pen was originally 22 fT Problem 4 LeT h represenT The heighT of The TV screen in and w represenT The widTh of The screen in From The aspecT raTio we have g 3 w 11 From The PyThagorean Theorem we have 2 h2 Eh 402 9 h 40 in if Eh2 1600 81 gh2 Eh2 1600 16 81 81 3h 112 1600 81 h2 1600 i 337 hi 1600 i 337 h m 196 Since The heighT cannoT be negaTive The heighT musT be abouT 196 inches This makes The widTh abouT 196 in m 348 in Page 2 of 2

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