We can write the left side as a perfect square, and the right side as a single fraction:
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1 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios. QUADRATIC EQUATIONS The followig equtio is clled qudrtic equtio: + + c = 0, where,, d c re rel umers with 0... Qudrtic Formul: We will derive the qudrtic formul elow. The method used is clled completig the squre method. The method works for ll qudrtic equtios. + + c = 0 c Sice 0, we c divide oth sides : 0. Now we complete the squre: c = +. c We c write the left side s perfect squre, d the right side s sigle frctio: c c. Tke the squre root of ech side:. Solve for : Simplif:,, c c. c.. Emple. 999 NC Alger II Solve for : c = c. A. =, c B. =, c C. =, c + D. =, c E. oe of these Solutio: D. c + c = 0 c c c c c Usig the qudrtic formul, c,
2 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios Emple. 00 NC Alger II If the discrimit of c 0 is zero, the which of the followig sttemets is true out,, d c? A. the form rithmetic progressio B. the re uequl C. ol is egtive D. the re ll egtive umers E. the form geometric progressio Solutio: E. The discrimit of + + c = 0 is c, which fctors s c. If this is zero, the = c, or / = c/, mkig the geometric me etwee d c. Emple. 999 NC Alger II For which vlue of c will the equtio c = c hve ectl oe solutio? A. c = 0 B. c = C. c = D. c = 0.5 E. oe of these Solutio: C. I qudrtic A +B+C, ol oe solutio eists if B AC = 0. For this equtio, c c = 0 c = 0 c =. Emple. 999 NC Alger II For which vlue of is there ol oe itersectio etwee the lie = + d the prol = + 5? A. 0 B. C. D. 5 E. oe of these Solutio: B. Itersectios occurs whe + = + 5 or + 5 = 0. There is ol oe itersectio poit whe the ove qudrtic hs ol oe solutio, i.e. it is perfect squre. This occurs whe 5 = 0. =... Viet s Theorem If d re two roots of qudrtic equtio c 0, 0, the.. c.. Emple NC Alger II If m is positive rel umer, determie the sum of the roots of the equtio m 0.
3 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios A. B. C. 5 D. 7 E. 9 Solutio: B. m = 0 9 m + 9 m = 0 0, so the sum of the roots is =. Emple NC Alger II Fid the sum of the rel roots of the equtio, k = 0, if is fctor of k. A. 7 5 B. 7 C. 5 D. E. 6 Solutio: E k = k = 0. Therefore /6 is the sum of the roots of the equtio.... Useful Forms Of Viet s Theorem.. c.. [ ]..5 c c c c..9 c I ll the ove formuls, d represet the two roots of the qudrtic equtio c 0. Emple UNC- Chrlotte Alger II Let d e the rel solutios of the equtio + + c = 0 with 0. If = d + = 0, the must e equl to
4 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios A. 8 B. C. D. 8 or E. 8 or 8 Solutio: E. B sustitutio of the gives, we hve + = 0 d = 6,. The = =, 6. Sice = + d 0, we must hve either = 6 + = 8 or = 6 = 8. Emple UNC- Chrlotte Alger II Let, e the two solutios to the equtio = 0. Fid the vlue of / + /. A. B. C. D. / E. /. Solutio: D. Note tht / + / = + / = // = /.... A Differet Form Of Viet Theorem c..0 If > 0, the..0 c e simplified s c.. Emple Alm Alger II Cosider the equtio: + p + q = 0. If the roots of this equtio differ, the q equls p p p p A. p B. C. D. E. Solutio: D. B..0, c p q p q Squrig oth sides of : p q q =. QUADRATIC FUNCTIONS.. The Qudrtic Fuctio: p.
5 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios The followig fuctio is clled the qudrtic fuctio: = f = + + c. where,, d c re rel umers with 0. = c f... Vertices h, d c k. > 0, mi = k < 0, m = k Emple NC Alger II Fid the geerl epressio of qudrtic fuctio tht psses through 0, d 8,. A. f = 8 + B. f = 8 + C. f = 8 + D. f = + E. oe of these Solutio: B. Iitill, let f = + + c. f0 = c =. f8 = c = = 0 = 8. f = = 8 +. Emple NC Alger II Fid the verte of qudrtic fuctio tht hs for the coefficiet d itercepts:, 0 d 0, 0. A. 6, 6 B. 6, C., 0 D. 6/, 6 E. oe of these Solutio: A. Kowig the roots d the costt multiplier, we c s tht the form of the qudrtic fuctio is f = 0 This epds to + 0. The -vlue of the verte is /, which is / = 6. f6 = 6, mkig the verte 6, 6. 5
6 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios Emple. 005 NC Alger II Ech sprig meter meter rectgulr grde hs its legth icresed meters ut its width decresed 50 cetimeters. Wht will e the mimum ttile re of the grde? A. m B. 76 m C. 89 m D. 00 m E. 5 m Solutio: E. The re will e = + 6 = + 8, where is the umer of ers. This qudrtic epressio hs mimum t its verte, which occurs whe = / = 8/ = 9. The re whe = 9 is = 5... Distce Betwee Two X-Itercepts Of A Prol The qudrtic fuctio 0. d stisf c 0. The distce etwee A d B is: c meets the -es t two poits A, 0 d B, AB c. Emple. 999 NC Alger II Fid the vlues of m so tht the differece etwee the roots of + m is 5. A. 0 B. C. 7 D. 9 E. oe of these Solutio: D. Method officil solutio: For this qudrtic, we wt the followig equtio to e true B B AC B A m 5 B AC A 5 m = 9. B AC 5 A Method : B the formul, we hve 5 c m 5 m m 8 m 9. 6
7 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios 7. SOME USEFUL FORMULAS.. Perfect squre triomil z z z z z w z w z w z w zw Emple. 008 Idi Alger II + c = A. + c B. + + c C. + c + c c D. + + c + c c E. oe of these Solutio: D. c c c c = + + c + c c... Differece d sum of two squres.. Differece d sum of two cues for ll.... for ll eve.... for ll odd. Emple. 00 NC Alger II If d 6, fid the umericl vlue for. A. 7/ B. 8 C. 9 D. 9/ E. Solutio: A. + = + = 9, so + + = 9, ut we re lso told tht + = 6, so =. = + = + + +, so 7 = = + + +, d + = 7 + = 7 / = 7/.
8 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios PROBLEMS Prolem. 000 NC Alger II For wht vlues of k does the equtio k 5 0 hve two imgir roots? A. k B. k C. k D. k E. k Prolem. 000 NC Alger II Fid ll rel umer solutios of 6 80 = 0. A. 5 B. 5, C. D. 0, E. 0 Prolem. 000 NC Alger II A qudrtic equtio = + + c is kow to pss through the poits 0, 5,,, d, 5. Fid the sum of d. A. 7 B. C. D. E. Prolem. 00 NC Alger II The grph of two prols = d = itersect i two poits. A equtio for the lie tht psses through these two poits is A. +8 = 0 B. 8 = 0 C. + = 0 D. + = 0 E. + = 0 Prolem NC Alger II If 0, equls A. B. C. D. or E. or Prolem NC Alger II The product of the vlues of,, d c, tht require the grph of = + + c to pss through the poits 0,,,6 d,9 is A. 08 B. 56 C. D. 6 E. oe of these Prolem NC Alger II Let e positive iteger d cosider f = As icreses, how does the grph of f chge? A. The verte chges with the -coordite decresig t slower rte th the - coordite. B. The -coordite of the verte icreses wheever the -coordite of the verte decreses 6. C. The verte chges with the -coordite decresig while the -coordite remis costt. D. The -coordite of the verte decreses wheever the -coordite of the verte decreses. E. Noe of these. 8
9 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios Prolem NC Alger II A prol with verticl is of smmetr psses through the poits 0,7,,5, d,7. Fid the two -itercepts. A. {, } B.{, } C. { 6, 6 } D. { 6 56, 6 56 } E. The prol does ot cross the -is. Prolem NC Alger II Cosider the equtio + k + = 0. A sigle fir die is rolled to determie the vlue of the middle coefficiet, k. The vlue of k is the umer of dots o the upper fce of the die. The proilit tht the equtio will hve rel, uequl roots is: A. / B. / C. / D. / E. Noe of these. Prolem NC Alger II Jerem hs iccle repir shop d eeds to kow how much to chrge for lor. If he chrges too much he will lose customers; if he chrges too little he wo t mke much moe. At the preset, he chrges $0 per hour d hs 5 hours of work week. He kows tht for ever $5 icrese i the hourl rte his worklod drops hours, d for ever $5 decrese his worklod goes up hours. How much should Jerem chrge per hour to mimize his profits? A. $.50 B. $0.00 C. $5.00 D. $.50 E. $0.00. Prolem. 009 UNC- Chrlotte Alger II Give tht, stisfies + = 9, wht is the lrgest vlue of + +? A. B. C. 6 D. 7 E. 9 Prolem. 009 UNC- Chrlotte Alger II If m = p + 5 +, wht is m + p? A. 0 B. 50 C. 80 D. E. 50 Prolem. 008 Idi Alger II Wht is the lrgest possile vlue of if = + d is rel umer? A. 0 B. 6 C. D. 6 E. oe of these Prolem. 008 Idi Alger II If the sstem hs ectl oe k solutio, the k = A. 6 B. C. or 6 D. or 6 E. oe of these 9
10 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios Prolem 5. 0 Alm Alger II Suppose is comple umer stisfig the equtio + /=. Wht is the vlue of + /? A. B. C. 0 D. E. Prolem 6. 0 Alm Alger II If = d + = 5, the A. 0 B. 5 C. D. 5 E. 5 is equl to Prolem Alm Alger II If the equtio + c = 0 hs two rel solutios, = p d = q, d p q, the p q =? A. c B. c C. c D. c E. c Prolem Alm Alger II If the verte of the grph of = + 7 is h, k, the wht is h + k? A. B. C. D. E. 5 Prolem NC Alger II The lie = itersects the prol = t two poits. Fid the distce etwee these two poits. A. 6 5 B. 0 C. 5 D. 8 E. Prolem 0. 0 Illiois Alger II The sum of the cues of the roots for of the equtio 60 + k = 0 is 8,0. Fid the lrger of the two roots for. 0
11 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios SOLUTIONS Prolem. Solutio: A. For qudrtic + + c = 0, the roots re imgir ol if c < 0. Here, 5 k < 0. k < 5/6. Prolem. Solutio: A. Let =. The 6 80 = = 0. 0 is etreous. So 5. Prolem. Solutio: B. Pluggig the first poit, we get = 5 = c. The we get from the et two poits + + c = + c = 5 : = = =. + = = Prolem. Solutio: C. = d = = = 0 + = 0 = or =. = = 8 or = = 9. The prols itersect t, 8 d, 8. The lie through, 8 d, 8 hs slope m = d equtio + = 0. Prolem 5. Solutio: B Prolem 6. Solutio: A. All three poits stisf the equtio = + + c, so =, 6 = +, 9 = + + = 9, + = 6 =, =. c = = 08. Prolem 7. Solutio: A. f 0 0 0, so whe = 0 the verte is 0, 0 d s icreses, the verte moves to the left t rte of while it lso moves dow t rte of, which, whe > is fster rte th the.
12 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios Prolem 8. Solutio: A. Sice the prol hs verticl is, it will e of the form = + + c. Whe we plug the poits ito this equtio, we get the followig sstem of equtios: 7 = c 5 = + + c 7 = + + c d the solutio to this sstem is, =, c = 7, so the prol re d. Prolem 9. Solutio: B. For rel, uequl roots c > 0. So k > 0 k =,,5,6. So the proilit tht + k + = 0 will hve rel, uequl roots is /6 or /. Prolem 0. Solutio: D. Let e the mout he chrges per hour. Hours worked = 5 /5 0. Thus Profits = = Its mimum is t = / 9/0.6 = $.5. Prolem. Solutio: E. Replce with 9 to get = 9 + = 9, which must e t lest 9. It s vlue t = is 9. Prolem. Solutio: C. Sice m = 5m m + 7, it follows tht = d 5 + m = 5. Thus m = 0 d 5m = p so p = 50. Hece m + p = = 80. Prolem. Solutio: D. c 0 B., the lrgest possile vlue of is k 6. Prolem. Solutio: C. k k 0. Sice the ssem hs ectl oe solutio, 0 k 0 k or k 6. Prolem 5. Solutio: A.
13 Alger II Through Competitios Chpter Qudrtic Equtios d Fuctios. Prolem 6. Solutio: B. 5. Prolem 7. Solutio: C. c c B.., p q c. Prolem 8. Solutio: C. c B., h, d k. h k c 7. Prolem 9. Solutio: A. First we should fid the two poits. So = 5 + 6, which simplifies to = 0, which fctors to 7= 0 =, 7 d the poits of itersectio re,0 d 7,. So the distce etwee these poits is Prolem 0. Solutio:. c B the Viet s Theorem, we hve 60 k k = 7. Solvig 60 + k = = 0 7 = 0 =, = 7. The lrger root is., or
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