Quadratic Equations. Math 99 N1 Chapter 8


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1 Qudrtic Equtions Mth 99 N1 Chpter 8 1 Introduction A qudrtic eqution is n eqution where the unknown ppers rised to the second power t most. In other words, it looks for the vlues of x such tht second degree polynomil will tke the vlue zero. The most generl form is going to be fter possibly rerrnging terms) x + bx + c = 0 1) where, b, c re rel numbers. One wy of expressing this is to look t where the grph of the function intersects the x xis. 1.1 Exmples f x) = x + bx + c All of the following re qudrtic equtions: 1. x + 3x 1 = 0 This is in stndrd form. Referring to 1), here = 1, b = 3, c = x x = 3 Sme, fter shifting the constnt to the right, i.e., dding 3 to both sides no one sid coefficients should be integers): = 4 5, b = 7 3, c = x 3 3 x + 8 = 0 This is stndrd form no one sid coefficients should be rtionl), with =, b = 3 3, c = x 1) x) = 1 Once we expnd the prentheses nd do the lgebr, we end up with stndrd looking eqution: 3x 1) x) = 1 1
2 9x 6x x + x ) = 1 besides opening the lst prenthesis, we lso dd 1 to both sides) so, = 5, b =, c = 4. 9x x 6x + 4x = 1 1 5x x 4 = 0 1. Equtions Tht Led to Qudrtic Equtions Sometimes, though the eqution is not qudrtic, simple mnipultions will led to qudrtic eqution whose solutions re or my be) solutions to our originl problem, or tht will led to the sought for solutions fter simple dditionl step. The cutionry note reminds us tht some mnipultions cn led to the ppernce of spurious solutions, i.e., solutions of the modified problem tht re not solutions to the originl. Exmples of problems of this type re: 1. x 1 = 1 squring the two sides leds to qudrtic eqution). 5x +x+1 = multiplying both sides by the denomintor of the left hnd side leds to qudrtic eqution) 3. x 3 3x = 0 solving for the new unknown y = x 1 3leds to qudrtic eqution in y. We will find the solutions we need, by tking cubes of the solutions: x = y 3 ). 4. x 8 +5x 4 1 = 0 solving for the new unknown y = x 4 leds to qudrtic eqution in y y + 5y 1 ) We will find the solutions we need, by tking fourth roots of the solutions: x = ±y ssuming the solutions re positive). Note tht exmple is rtionl eqution, nd tht we need to mke sure tht the solutions to the qudrtic eqution this leds to re not such s to mke the denomintor tke the vlue 0 too! Here we would hve 5x + x + 1 5x + x + 1 ) = 5x + x + 1 ) = 10x + 4x + 0 = 10x + 4x This is esily solved by noticing tht 10x + 4x = x 10x + 4), so tht to be zero we need either x = 0 or 10x+ 4 = 0, i.e. x = 4 10 = 5. We cn then check tht neither of these two vlues will cuse the denomintor to be zero. As for exmple 4, we will quickly lern how to see without even solving the eqution) tht eqution ) hs two rel solutions, one positive, nd the other negtive. For our purposes, only the first will work, becuse we will need to tke fourth root.
3 Solving Algebric Equtions.1 Introduction An lgebric eqution is n eqution of the form Polynomil=0. The eqution my t first look different, but it should reduce to this form fter dding nd subtrcting ppropritely to both sides. For exmple, 4x 3 = x + 1 is n lgebric eqution, since we cn dd x 1 to both sides, nd get polynomil on the left side, nd 0 on the right side: 4x 3 x 1 = 0. On the other hnd, x x 3 +x+1 = 3x is not it is rtionl eqution), becuse we don t hve n equlity of polynomils. If we multiply both sides by x 3 + x + 1, we turn the eqution into n lgebric one, but we need to remember tht this mkes sense only if the solutions we will find do not mke x 3 + x + 1 tke the vlue zero. There would be no such cutionry note if we hd fced n lgebric eqution from the beginning. other exmples of nonlgebric equtions re x 3 +5x 3 = 0 or 7x x 1 = 0 we wnt polynomil on the left hnd side, nd neither is one). The degree of the polynomil the highest power with non zero coefficient) is clled the degree of the eqution. So, for exmple, x 7 + 3x 4 + = 0 is 7th degree eqution, nd 8x 9 10x 8 3x 5 + x 4 x + 4x = 0 is 9th degree eqution One wy to solve n lgebric eqution in principle) is to fctor the polynomil s fr s we cn: this reduces the problem to equting to zero ech of the fctors in turn, nd since these re of lower degree, the problem is, t lest, simplified. A very deep nd remrkble theorem ctully sttes tht we cn lwys fctor polynomil so tht ech fctor is either of degree one, or of degree two in which cse, the nd degree polynomil cnnot be further fctored). Since we get exctly one solution when we equte first degree polynomil to zero, nd no solutions when we equte non fctorisble nd degree polynomil to zero, we hve number of remrkble consequences, such s 1. An nth degree eqution hs t most n solutions. An eqution of even degree cn hve no solution t ll, but n odd degree eqution lwys hs t lest one solution. 3. If the number of solutions is less thn n, sy k, then n k is even it is the sum of the degrees of the irreducible nd degree polynomils). Unfortuntely, the theorem does not give us prcticl nd generl wy to find this fctoring, so we still need wys to compute our solutions. At lest, we hve some indiction of wht to expect in this serch. 3
4 . The Polynomil Equtions We Cn Alwys Solve In generl, the one type of polynomil eqution we cn ssume we know how to solve is one of the form x n = 0 where n is n integer, nd is ny rel number. This eqution hs none, one, or two solutions depending on the sign of, nd whether n is odd or even. 1. If n is odd, there is one solution for every : x = 1 n. If nis even, there is no solution if < 0 3. If n is even, there re two solutions if > 0: 1 n, nd 1 n 4. If = 0 the only solution is x = 0. Note tht it my extremely hrd to compute 1 n in prctice, but we ssume tht this is lwys fesible, s there re lgorithms tht re pretty good t pproximting this number to very high precision. We re not concerned with them, s we will rely on clcultor or computer to do this, unless hppens to be the exct nth power of some number we cn identify. 3 Solving Qudrtic Equtions We know how to solve liner equtions lgebric equtions where the unknown ppers only t degree 1). Given the theorem bove, it mkes specil sense to study how to solve if possible) equtions of degree. Hence, from now on, we restrict to the cse n =. This cse is known since ncient times, nd is reltively esy to hndle. There re wys to mke similr rguments for the cses n = 3 nd n = 4 these were crcked in the 16th Century), but the rguments nd the results re enormously more complicted. In fct, nobody expects nybody else to remember the solution formuls while everybody expects you to remember the solution formul for qudrtic equtions), nd these re very rrely discussed even in dvnced mth clsses. Incidentlly, it is remrkble result of the 19th Century tht there is no wy we cn construct nlogous rguments nd formuls when n 5. This mens tht while some higher order equtions cn be solved with formul e.g.: x 5 = hs solution x = 1 5 ), this does not work for ll such equtions. To strt our investigtion, we recll, from sec.., tht we lredy know how to solve n eqution of the form x = 0: if > 0, we hve two solutions,, nd, if = 0 we hve only the solution = 0, nd if < 0 there is no solution. 4
5 3.1 More Generl Equtions Equtions Tht Reduce To The Simple Cse Even if our qudrtic eqution is not of the form x = 0 or, equivlently, x = ), it is often close enough tht we will be ble to quickly mnipulte it to tht form. Here re few exmples. 1. x = 8: 1 x = 1 8; x = 4, x =, x =. 3x 5 = 7: 3x = 7 + 5; x = 1 3 = 4; x =, x = 3. x 5) = 4: this mens tht we hve one of two cses: x 5 = or x 5 =. We cn solve ech of these, nd get x = 7, x = 3 4. x + x + 1 = 9; we just notice tht x + x + 1 = x + 1), so we solve x + 1) = 9, i.e. x + 1 = 3 or x + 1 = 9, giving x =, x = 10. Note lso tht n eqution like s n exmple) x 5) = 3 x) 3) mens tht either hence x 5 = 3 x or i.e. 3x = 8, x = 8 3 x 5 = 3 x) x 5 = 3 + x x = x = You will note the similrity with type of eqution we met in the bsolute vlue chpter!. In fct, given the fct tht =, you could lso write eqution 3) s x 5 = 3 x we took the squre root of both sides), nd get bck to tht type of exmple. 5
6 3.1. Equtions Tht Are Esy To Fctor If the nd degree polynomil hs on obvious fctoring, our solutions re immeditely found. So, for exmple, since we my notice tht x x 3 = x 3)x + 1), the eqution x x 3 = 0 is equivlent to x 3)x + 1) = 0, nd so hs solutions given by x 3 = 0 i.e., x = 3), nd x+1 = 0 i.e., x = 1). Unfortuntely, in most cses it s hrd to find the right vlues for fctoring, nd, in ny cse, this pproch requires some good intuition, except in the simplest cses. Since the logic of generl theorems in Mthemtics is to provide tools tht llow us to solve problems without or, t lest, with miniml) recourse to good intuition, we will ctully use different pproch in order to find wy to fctor nd degree polynomils utomticlly  in fct, using this technique bckwrds. The one cse where the fctoring is so esy it would be shme to neglect it is when no constnt term is present: x + bx = 0. In this cse x is common fctor tht cn be pulled out, nd the eqution is equivlent to xx + b) = 0, or x = 0 nd x + b = 0, i.e., x = b. For exmple, 1. 4x + x = 0: x 4x + ) = 0, x = 0 nd 4x + = 0, i.e., x = 4 = 1. 6x = 3x: 6x 3x = 0, 3xx 1) = 0, x = 0 nd x 1 = 0, i.e. x = 1 3. Reducing The Most Generl Cse To The Simplest There is simple, but extremely useful technique tht cn be used to reduce the most generl nd degree eqution to n esy vrition of the simple cse i.e., the one we treted in generlity in sec..). First, we recll the ide put forwrd in sec : n eqution tht cn be rewritten s x + b) = c hs solution, provided c 0, given by the observtion tht, either x + b = c 1, or x + b = c 1. Hence, x = b + c, x = b+ c, nd x = b c, x = b c re the solutions. If c < 0, there re no solutions, nd we could see it by pplying these formuls nd relize tht our negtive c hs no squre root. Also, if c = 0, both solutions coincide, since 0 is the only squre root of 0. One often writes x = b± c to shorten the writing down of the solutions. For exmple, 1. x ) = 4: x =, x = 4 or x =, x = 0. The shortened expression would be x = ± 4. x +6x+9 = 5: we notice tht the left hnd side is equl to x + 3), nd so the eqution cn be written s x + 3) = 5, x = 3 ± 5 = 3 ± x 3 ) 1 = 4 9 : x 3 1 = 4 9 = 3, or x 3 1 = 3. Hence, x 3 = = 7 6, x = = 7 4, or x 3 = 1 3 = 1 6, nd x = ) = 1 4 6
7 The second problem leds us to bright ide: fter ll, it is the fct tht perfect squre ws explicitly isolted on the left hnd side tht led us to the esy solution. But, in itself, if we hd looked t the eqution in its stndrd form, x + 6x 16 = 0, the equivlence to the form in point would not hve been obvious t ll. Still, we could hve rgued like this. 1. Look t the first term  it s the squre of x. The second term, 6x, cn lwys be written s x i.e. times the squre root of the first term), times something else  in this cse, to get 6, the something else would be 3. So we rewrite the second term s x 3 3. Now, to hve perfect squre pper we only need 3 = 9. We hve 16. Well, let s just force 9 on it: 16 = 9 5. So we rewrite the eqution s x + x = 0, or x + 3) 5 = 0. Presto! We hve reduced ourselves to the simple cse. It s esy to see tht the strtegy we discovered in point 3 cn lwys be pplied! For exmple, x 5x 3 = 0 ) x = x 5x = 5 x = x 5 = x ) = = = x x 5 ) 5x 3 = nd our eqution is equivlent to x 5 ) = tht is nd 5 49 x = 4 8 = 7 = x = 4 8 = 7 4 x = 5 ± 7 4 7
8 x = 5 ± 7 4 = 5 ± 7 4 i.e. x = 1, or x = 3 much of the clcultions mde sure we eliminted rdicls from the denomintor, s this is prerequisites for ccurte computtions) Sme Clcultion, Less Rdicls If ll the terms we found seem scry they did dispper, in the end, but they might hve tripped you long the wy), we cn work the sme mgic in slightly different wy just s the book does): x 5x 3 = x 5 3 ) Now, we try to mnipulte the function within the prenthesis: x 5 ) ) = x 5 x 4 3 = x 5 x ) = x 5 x = x 5 ) Now, going bck to our eqution, we cn rewrite it s [ x 5 ) ] 49 = nd, since 0, this requires x 5 ) = x 5 4 = ±7 4 leding us to the sme results we obtined before. 4 The Qudrtic Formul We cn repet the steps in the previous section, in generl. Strt from x +bx+c nd ssume > 0 1 if it is not, we look t the qudrtic function x bx c  fter ll, the two equtions ± x + bxx + c ) = 0 hve the sme solutions). We write it s ) b x + x + c = x ) b + x + b ) ) + c b = 4 1 Incidentlly, in this section we will hve to ssume tht 0 we will do lot of dividing by ). But if tht ws not the cse, we wouldn t be deling with qudrtic eqution in the first plce, so this is trivil ssumption. 8
9 = x + b ) ) + c b 4 This trick is known s completing the squre, nd it is extremely productive in so mny different contexts, it is worth lerning. Now the problem of solving the generl eqution is esy: we hve rewritten our eqution s x b ) + = b 4 c = b 4c 4 Hence s long s b 4c 0, we lredy rrnged for > 0) b x + = ± b 4c 4 x = b ) ± b 4c = b 4 ± b 4c 4 = b ± b 4c which is the fmed qudrtic formul, which you re urged to lern very thoroughly. 4.1 You d Like Not To See Tht Mny s We cn deduce the formul by first tking out fctor of, just s we did in 3..1, nd dispense with most of the rdicls but not relly with the issue with the sign of ). This is how the book proceeds s well. Let us ssume tht > 0, just s we did bove. Now, we first observe tht x + bx + c = x + b x + c ) Now we strt mnipulting the prenthesis only: x + b x + c = b x + x + c ) ) = b b b x + x + + c = x + b ) b 4c 4 If you delight in the following kind of wizrdry, we could hve ignored the requirement > 0, if we pretended tht, somehow, there is something we cn cll 1. Let s cll this nonexistent) something i. Then, if < 0, we cn write =, nd = i. As we proceed s bove, we end up with left hnd side which will be positive if > 0, nd b 4c > 0, or if both re negtive. Proceeding with our brzen strtegy, we eventully rech exctly the sme finl formul, except tht the squre root in the numertor my or my not be defined, depending on whether b 4c 0, or not. Tht s no different from wht we got here! If you wish, you could cll this ugly trick n exmple much esier thn the historiclly importnt one, which occurs when solving cubic equtions) of how pretending tht we cn mke sense of 1 nd hence of the squre root of ny negtive number), cn led us to meningful conclusions, s long s these kind of objects dispper in the end. For little more of this, you cn check the compnion file MoreQud.pdf. 9
10 Going bck to the originl eqution, we hve [ x + b ) ] b 4c 4 = 0 Now, since s discussed in previous footnote, 0, this requires x + b ) b 4c 4 = 0 i.e. x + b ) = b 4c 4 Assuming 3 now tht b 4c 0, we cn tke squre roots. Since we ssumed > 0, the roots of re ±, those of b 4c re, obviously, ± b 4c, so the roots of the left hnd side re, combining ll possibilities, ± b 4c Similrly, we should be wre tht the roots of the left hnd side will be ± x + b 5) which would be even more wkwrd, since we hve no explicit vlue for x yet. However, note tht there re only two possible combintions of one 4) nd 5) for instnce, choosing + on both sides, or choosing on both sides, leds to exctly the sme formul). Since of the two vlues of 5) will be equl to x+ b, we cn write the two possible combintions s or x + b = b 4c x = b ± b 4c Actully, by creful blncing ll the signs ppering in expressions of type A = ± A, we could hve mde without our originl ssumption tht > 0. It just seemed tht tking prt ll possibilities in tht cse would hve been bit too much tedious. 3 The book does not mke ny fuss bout the problem in tking the squre root of squre nd neither hd we before now), nd good reson for this sloppiness is tht, in the end, things work out nicely nywy. Even if this degree of fussiness is not needed in the sense tht no hrm is done by little crelessness) here, ignoring the fct tht A = A is, in generl, dngerous mistke. So, since it is only fter the fct tht we my note tht we don t need to be extrcreful in this section, we ll do it right, this time. On the other hnd, if we were willing to use squre roots of negtive numbers without flinching, we could ignore ll cution, nd proceed, rigorously, just s the book does  even dispensing with the requirement tht > 0  nd end up with the right formul. Compre with the previous footnote. 4) 10
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