Chapter 9. Quadratics

Transcription

1 Chpter 9 Qudrtics Artificil Body Prts 9.1 Solving Qudrtic Equtions by Fctoring 9. Completing the Squre 9.3 The Qudrtic Formul 9.4 Eponentil Functions (Growth nd Decy) Chpter Review Chpter Test 147

2 Section 9.1 Solving Qudrtic Equtions by Fctoring Similr to liner eqution (y m + b), qudrtic eqution hs two vribles (, y), but unlike in liner eqution, one of the vribles is squre: y + b + c. (qudrtic squre) As the nme indictes, liner eqution, when plotted, forms line. On the other hnd, qudrtic eqution, when plotted, forms curve. A qudrtic eqution doesn t form just ny curve, but curve tht is symmetricl (hlf of the curve is the reflection of the other hlf). Clled prbol, it is the shpe of the pth of bsebll (or projectile) in flight. The solution of qudrtic is chieved through fctoring, if we remove the equl sign () nd the y vrible: y + b + c + b + c This is the type of trinomil studied in section 7.5. For emple, 8 cn be fctored into ( 4)( + ) putting bck the equl sign nd the y vrible it becomes: y ( 4)( + ) This is the type of eqution to be solved in this section. HOW TO SOLVE A QUADRATIC EQUATION Becuse these equtions represent curves nd curves come round (unlike the stright line of liner equtions tht crosses es only once), in mny cses curve must cross the sme is twice. The figure to the right shows this. A prbolic curve lwys comes bck like boomerng. Notice how in the grph it comes down from left to right nd crosses the -is twice before going off to the top right. Specificlly, notice how the curve comes down nd crosses the -is t, nd comes bck up crossing the -is gin t +4. Fctoring qudrtic eqution, it is these very points we seek. These points become the roots or zeros of the qudrtic. The eqution plotted in the previous grph is y 8 Becuse the vlue of y is zero (y 0) when the prbol crosses the -is, we now set up the eqution equl to zero to find the specific vlues of when the curve crosses the -is (roots). 0 8 Fctoring the trinomil ( 4)( + ) 0 Becuse it is not known which of the two binomils, ( 4) or ( + ), is equl to zero, setting up ech binomil equl to zero is the wy to find both vlues for : 4 0 or These nswers re clled the roots of the qudrtic nd cn be verified in the grph bove. 9.1 Solving Qudrtic Equtions by Fctoring 149

3 Emple: Solve ( + )( 3) 0 ( + ) 0 or ( 3) 0 3 Emple: Solve or Emple: Solve (fctor trinomil) 0 ( + 3)( + 4) Emple: Solve (move to left-side nd dd to 4) (fctor) ( 6)( + 1) Emple: Solve (perfect trinomil squre) ( 3)( 3) One double, equl root, 3. This curve comes down nd touches the -is only t 3, insted of crossing it. Emple: Solve (find common fctor) ( 6) Emple: Solve ( + 5)( + ) Chpter 9: Qudrtics

4 Emple: Solve ( + 3)(3 4) Prctice: CONTENTS OF THIS PAGE HAVE BEEN REMOVED TO PROTECT COPYRIGHT 9.1 Solving Qudrtic Equtions by Fctoring 151

5 Section 9. Completing the Squre So fr we hve solved qudrtic equtions tht cn be fctored, like: Determine: Which two integers give product of 6 nd difference of 5? ( 6)( + 1) nd 1 But wht hppens when we encounter qudrtic eqution which cnnot be fctored? Emple: Determine: Which two integers give product of 7 nd difference of 4? The differences of 7 nd 1 re 6 or 6, not 4 The fctors of 7 re 1 nd 7 Becuse 7 hs only two fctors (7 nd 1), nd these two fctors subtrcted re 6 or 6 nd not 4, like the middle term bove, the nswers to this qudrtic re not integers, but decimls. Finding which combintion is correct would be very lengthy nd difficult tsk. To solve this qudrtic we turn to process clled, completing the squre, which involves mking perfect trinomil squre out of the left side of the eqution STEP ONE: Move 7 to the right side of the eqution, leving n empty spce where the 7 ws: 4 7 STEP TWO: Tke one-hlf of the middle term ( 4) nd squre it. Complete the trinomil squre with this number by plcing it into the empty spce (one-hlf of 4 is, squred is 4). Becuse 4 ws dded to the left-hnd side, then 4 must be dded to the right-hnd side: Chpter 9: Qudrtics

6 STEP THREE: Fctor the left-hnd side of the eqution s perfect trinomil squre: ( )( ) 11 ( ) 11 STEP FOUR: Find the squre root of both sides: Answer: , HOW TO SOLVE A QUADRATIC WITH A LEADING COEFFICIENT GREATER THAN ONE Emple: Solve Becuse to complete squre the leding coefficient MUST be one, divide the whole eqution by : Move 4.5 to the right: Complete the squre: Fctor: ( )( ) 8.5 Find squre root: Emple: Solve Divide eqution by Move to the right Completing the Squre 153

7 Complete the squre (one-hlf of -- is -- nd -- squred is ) nd dd to both sides of the eqution: Add right-hnd side Fctor left-hnd side s perfect trinomil squre nd find squre root of both sides: Solve for : Prctice: CONTENTS OF THIS PAGE HAVE BEEN REMOVED TO PROTECT COPYRIGHT 154 Chpter 9: Qudrtics

8 Section 9.3 The Qudrtic Formul The qudrtic formul is wy of solving ll qudrtic equtions to find their roots. The qudrtic formul is derived by the generl use of completing the squre. If we tke the generl form of the qudrtic eqution + b + c 0 nd use completing the squre to solve for, then the qudrtic formul is born. STEP ONE: Divide whole eqution by to reduce the leding coefficient to b ā - c b ā - c c c STEP TWO: Move -- to the right side of the eqution, leving n empty spce where the -- ws: + b ā - c -- b STEP THREE: Tke one-hlf of the middle term -- nd squre it. Complete the trinomil squre with b b b this epression by plcing it into the empty spce (one-hlf of -- is -----, nd squred is ). Becuse ws dded to the left-hnd side, then must be dded to the right hnd side: + b ā c STEP FOUR: Fctor the left hnd-side of the eqution s perfect trinomil squre nd dd the right-hnd side: FACTOR LEFT SIDE ADD RIGHT SIDE b c c + 4c b c b STEP FIVE: Find the squre root of both sides c b c The Qudrtic Formul 155

9 b STEP SIX: solve for by moving to the right hnd side: b c This is the formul for qudrtic roots: b 4c Where is the leding coefficient (qudrtic term), b is the coefficient of the middle (liner) term, nd c is the lst (constnt) number. UNDERSTANDING WHAT THE ROOTS MEAN In section 8.1 the shpe of the qudrtic line ws given s the prbolic curve crossing the -is. Below left, the curve crosses the -is twice (two distinct roots), below center the curve does not cross but touches the -is (one double, equl root), below right the curve does not cross the -is (no rel roots). Depending on the, b nd c vlues of the qudrtic formul, the shpe nd position of the curve is determined. no rel roots two distinct roots one double, equl root Emple (for two distinct roots): Solve b 5 c 1 Substitute in the qudrtic formul: b 4c Emple (for one double, equl root): Solve b 5 c 6.5 Substitute in the qudrtic formul: b 4c Chpter 9: Qudrtics

10 Emple (for no roots): Solve b c 6 Substitute in the qudrtic formul: b c Becuse the squre root of negtive number cnnot be determined, the nswer is not rel. The Discriminnt: 4c The discriminnt is tht prt of the qudrtic formul tht llows us to peek into it to see if the qudrtic hs ny vlid roots. If the discriminnt is negtive, no need to continue; if it is positive or zero, continue. Emple: Solve b 13 c 13 Using the discriminnt Rdicl is negtive, zero roots. Prctice: Using the qudrtic formul, find the roots of the qudrtic. CONTENTS OF THIS PAGE HAVE BEEN REMOVED TO PROTECT COPYRIGHT Use the discriminnt to determine if the qudrtic hs roots. 9.3 The Qudrtic Formul 157

11 Section 9.4 Eponentil Functions (Growth nd Decy) Wht shpe would line on grph hve if the eponent is vrible? From pst eperience we know tht n eqution such s y + 3 is stright line (see grph t right). Also from pst eperiences we discovered tht squring the bse, will turn the line into curve, y y specificlly prbol (see grph to the left). Wht shpe would line on grph hve if the eponent is vrible? The shpe is curve tht increses (climbs) rpidly. Nturlly, we cll this type of curve eponentil growth. The prticulr eqution of the grph in figure 1 is: y VARIATIONS OF THE EXPONENTIAL CURVE Becuse the coefficient is negtive, curve such s y (figure ) Figure 1 will send the curve the opposite wy, towrds the negtive side. A deciml bse such s y 0.5 will mke the curve decrese (drop) rpidly. This type of curve is referred to s eponentil decy. See figure 3. Figure Figure Chpter 9: Qudrtics

12 Emple in eponentil growth: On her birthdy, five-yer-old receives $5,000 gift nd her prents plce the money in bnk ccount tht erns interest of 7% per yer (7 cents for very dollr deposited). How much money would be vilble in the ccount when the girl turns 0 (15 yers lter)? After the first yer one dollr would hve grown to $1.07, nd fter the second yer the $1.07 would hve to grow to $ ( ), nd fter the third yer to $1.5, nd so on. This problem fits the eponentil growth pttern becuse the more interest is ccumulted, the higher the interest erned. In other words, the money grows rpidly by compounding the interest mount. The eponentil eqution would be: y (1 + i) Where y is the finl mount is the initil mount i is the rte (percent) of increse per period in deciml form is the number of periods (yers) Solving for y y 5000 ( ) 15 Becuse multiplying 1.07 to the 15th power is tedious, we use the y function of scientific clcultor*: Press 1.07 Press y Press 15 Press ( ) y 5000 ( ) The girl t twenty will hve $13, She erned $8, in interest. Emple in eponentil decy: A chlorine solution decys 50% (0.5) for every week it is left uncovered. If the originl mount of chlorine used is 560 ounces (0 gllons), how much chlorine is vilble in the solution fter 1 weeks? 560 If it decys 50% (hlf the mount) every week, fter the first week there is ounces, fter the second week there is ounces vilble, nd so on. This problem fits the eponentil decy pttern becuse the mount of chlorine drops rpidly. The eponentil eqution would be y (1 i) Compre this eqution to the one bove. For growth, i is dded, for decy, i is subtrcted. Where y is the finl mount is the initil mount i is the rte (percent) of decrese per period in deciml form is the number of periods (weeks) 9.4 Eponentil Functions (Growth nd Decy) 159

13 Solving for y y 560 (1 0.5) 1 Becuse multiplying 0.5 to the 1th power is tedious, we use the y function of scientific clcultor*: Press 0.5 Press y Press 1 Press (1 0.5) 1 (0.5) y 560 ( ) 0.65 The mount of vilble chlorine fter 1 weeks is 0.65 ounces (slightly more thn tblespoon). *Performed on TI-35X clcultor. Prctice: Solve nd find the vlue of y when is 9. Round off nswers to the nerest thousndth. CONTENTS OF THIS PAGE HAVE BEEN REMOVED TO PROTECT COPYRIGHT Solve 5. The yer she retires, Gerri receives severnce check for $135,000. If she leves this mount in 6.5% retirement ccount for 10 yers, how much money would be vilble in the ccount fter ten yers? How much interest hs ccumulted? 6. A lrge fctory hs 5,000 bulbs. If 4% of the bulbs fil every week, how mny originl bulbs remin fter 6 weeks? 7. The popultion of certin city is projected to decrese % per yer. If the city hs pproimtely 1. milli 160 Chpter 9: Qudrtics