THEORY OF EQUATIONS & FUZZY SETS

Transcription

1 THEORY OF EQUATIONS & FUZZY SETS B. Sc. MATHEMATICS SIXTH SEMESTER ADDITIONAL COURSE IN LIEU OF PROJECT UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION CALICUT UNIVERSITY P.O., THENJIPALAM, MALAPPURAM

2 UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION B.Sc. MATHEMATICS STUDY MATERIAL SIXTH SEMESTER ADDITIONAL COURSE IN LIEU OF PROJECT THEORY OF EQUATIONS & FUZZY SETS Prepared by: Viod Kumar P. Assistat Professor P. G.Departmet of Mathematics T. M. Govermet College, Tirur Sathosh P.K. Assistat Professor P. G.Departmet of Mathematics Govt.BreeCollege,Thalassery. Scrutiised by ; Dr. Ailkumar. V. Associate Professor, Departmet of Mathematics, Uiversity of Calicut. This This study study otes otes are are oly oly a aabridged abridgedversio versioofofthe the topics give i i the syllabus. syllabus. For For further further details details like like proofs proofsof theorems, of theorems, illustratios illustratios etc., etc., studets studets are advised are to advised go through to go the through prescribed the prescribed Text Books Text. Books. Reserved

3 Module-I THEORY OF EQUATIONS Prepared by: Viod Kumar P. Asst. Professor Dept. of Mathematics T. M. Govt. College, Tirur.

4 T H E O R Y O F E Q U A T I O N S.0 Itroductio I this module, we will study about polyomial fuctios ad various methods to fid out the roots of polyomial equatios. Solvig equatios was a importat problem from the begiig of study of Mathematics itself. The otio of complex umbers was first itroduced because equatios like x + 0 has o solutio i the set of real umbers. The fudametal theorem of algebra which states that every polyomial of degree > has at least oe zero was first proved by the famous Germa Mathematicia Karl Fredrich Gauss. We shall look at polyomials i detail ad will discuss various methods for solvig polyomial equatios... Polyomial Fuctios Defiitio: A fuctio defied by f ( x) a x + a x a, where a 0, is a o egative 0 o iteger ad a i (i 0,.,) are fixed complex umbers is called a polyomial of degree i x. The umbers a, a,..., a are called the coefficiets of f. o polyomial. If α is a complex umber such that f(α) 0, the α is called zero of the.. Theorem ( Fudametal Theorem of Algebra) Remark: Every polyomial fuctio of degree > has at least oe zero. Fudametal theorem of algebra says that, if f(x) a + 0 x + ax +... a,

5 School of Distace Educatio, Uiversity of Calicut where a 0 is the give polyomial of degree >, the there exists a complex 0 umber α such that a α + a α a 0. 0 We use the Fudametal Theorem of Algebra, to prove the followig result... Theorem Proof: Every polyomial of degree has ad oly zeroes. Let f ( x) a x + a x a, where a 0, be a polyomial of degree >. 0 o By fudametal theorem of algebra, f(x) has at least oe zero, let α be that zero. The x α ) is a factor of f(x). ( Therefore, we ca write: f(x) x α ) Q ( ), where Q(x) is a polyomial fuctio of degree -. ( x If - >, agai by Fudametal Theorem of Algebra, Q (x) has at least oe zero, say α. Therefore, f x) ( x α )( x α ) Q ( x) where Q ( ) is a polyomial fuctio of degree -. ( x Repeatig the above argumets, we get f ( α x x) ( x α )( x α )...( x ) Q ( ), where Q (x) is a polyomial fuctio of degree - 0, i.e., Q (x) is a costat. Equatig the coefficiet of x o both sides of the above equatio, we get Q ( x). a o Therefore, f x) a ( x α )( x α )...( x α ). ( o If α is ay umber other tha α, α,... α, the f (x) 0 α is ot a zero of f(x). Hece f(x) has ad oly zeros, amely α...,, α α. Note: Let f ( x) a x + a x a ; a 0 be a th degree polyomial i x. o o The, a x + a x a () o is called a polyomial equatio i x of degree. A umber α is called a root of the equatio () if α is a zero of the polyomial f(x).

6 School of Distace Educatio, Uiversity of Calicut So theorem (..)ca also be stated as : Every polyomial equatio of degree has ad oly roots... Theorem If the equatio a o x + a x a 0, where a o, a,... a are real umbers ( a o 0), has a complex root α + iβ, the it also has a complex root α iβ. (i.e., complex roots occur i cojugate pairs for a polyomial equatio with real coefficiets). Proof: Let f(x) a x + a x a, a 0 o o Give that Cosider α + iβ is a root of f(x) 0. x ( α + iβ ) ( x ( α iβ ) ( x α) +. ( β Divide f(x) by ( x α ) + β. Let Q(x) be the quotiet ad Ax + B be the remaider. The, f ( x) [( x α ) + β ] Q( x) + Ax + B ( α β α β [ x ( + i ))( x ( i ))] Q( x) + Ax + B f ( α + iβ) 0 + A( α + iβ) + B A( α + iβ) + B (Aα + B) + iaβ But f ( α + iβ) 0. Equatig real ad imagiary parts, we see that A α + B 0 ad Aβ 0 But β 0 A 0 ad so B 0 The remaider Ax + B is zero. i.e., [( x ( α + iβ ))( x ( α iβ ))] i.e., α iβ is a root of f(x) 0. is a factor of f(x)..4. Theorem I a equatio with ratioal coefficiets, the roots which are quadratic surds occur i cojugate pairs. Proof: Let f ( x) a x + a x a, a 0, be a th degree polyomial with ratioal coefficiets. o o Let α + β is a root of f(x) 0. ( x ( + ))( x ( )) ( x ). Divide f(x) by [ α β α β ] α β

7 Let Q(x) be the quotiet ad Ax + B be the remaider. School of Distace Educatio, Uiversity of Calicut Proceedig exactly as i the above theorem, we get Ax + B 0. Thus we coclude that α β is also a root of f(x) Theorem If the ratioal umber p q, a fractio i its lowest terms (so that p, q are itegers prime to each other, q 0) is a root of the equatio a x + a x... + a 0 o where a a,..., a 0, are itegers ad o 0, a the p is a divisor of a ad q is a divisor of a o. Proof: Sice p q is a root the give polyomial equatio, we have a o p q Multiplyig by q, we get Dividig by p, we have + a p a p + a 0 q q a p + a p q a pq + a q () o a o p + a p q a q aq p Now, the left side of the above equatio is a iteger ad therefore aq p is also must be a iteger. Sice p ad q have o commo factor, p must be a divisor of a. Also, from (), a p q a pq + a q a o p Dividig this expressio by q, we get a p a pq + a q ao p q Sice the left side is a iteger ad sice q does ot divide p, q must be a divisor of a 0. This completes the proof. 4

8 Corollary School of Distace Educatio, Uiversity of Calicut Every ratioal root of the equatio x + a x a 0,where each a i is a iteger must be a iteger. Moreover, every such root must be a divisor of the costat a. Proof: This follows from the above theorem, by puttig a 0. Multiple Roots If a root α of f(x) 0 repeats r times, the α is called a r-multiple root. A - multiple root is usually called a double root. For example, cosider f(x) (x- ) (x- 5) (x + ). Here is a - multiple root, 5 is a double root, ad - is a sigle root of the equatio f(x) Theorem If α is a r - multiple root of f(x) 0 the α is a (r ) multiple root of f (x) 0, where f (x) is the derivative of f(x). Proof: Give that α is a r - multiple root of f(x) 0. r The f ( x) ( x α) φ( x) where φ( α) 0. Now, by applyig product rule of differetiatio, we obtai: f ( x) r ( x α ) φ ( x) + φ( x) r.( x α) r r ( x α ) [( x α) φ ( x) + rφ( x)] Whe x α, ( x α) φ ( x) + rφ( x) rφ( α) 0 α is a (r ) multiple root of f (x) 0. Remark: If α is a (r ) -multiple root of f (x) 0, similarly as above, we ca see that α will be a (r ) multiple root of f (x) 0 ;(r ) - multiple root of f (x) 0, ad so o. 5

9 Solved Problems School of Distace Educatio, Uiversity of Calicut. Solve x 4 4x + 8x + 5 0, give + i is a root. Solutio : Give that + i is a root of x 4 4x + 8x + 5 0; sice complex roots occurs i cojugate pairs i is also a root of it. polyomial. [ x ( + i )][ x ( i )] ( x ) + x 4x + 7 is a factor of the give Dividig the give polyomial by this factor, we obtai the other factor as x + 4x ± 6 0 The roots of x + 4x are give by ± i. Hece the roots of the give polyomial are + i, i, + i ad i.. Solve x 4-5x + 4x + 8x 8 0, give that oe of the roots is 5. Solutio: Sice quadratic surds occur i cojugate pairs as roots of a polyomial equatio, + 5 is also a root of the give polyomial. [ x ( 5)][ x ( + 5)] ( x ) 5 x x 4is a factor. Dividig the give polyomial by this factor, we obtai the other factor as x x +. Also, x x + (x ) (x ) Thus the roots of the give polyomial equatio are + 5, 5,,.. Fid a polyomial equatio of the lowest degree with ratioal coefficiets havig ad i as two of its roots. Solutio: Sice quadratic surds occur i pairs as roots, is also a root. Sice complex roots occur i cojugate pairs, + i is also a root of the required polyomial equatio. Therefore the desired equatio is give by ( x ( + i) ) 0 ( x ) (x + )(x ( i) 4 i.e., x x + x + 6x Solve 4x 5 + x + x x + 0, give that it has ratioal roots. Solutio: Let f(x) 4x 5 + x + x x +. 6

10 By theorem (..5.), ay ratioal root School of Distace Educatio, Uiversity of Calicut p q (i its lowest terms) must satisfy the coditio that, p is divisor of ad q is positive divisor of 4. So the possible ratioal roots are ±, ± ½, ± ¼. Note that f(-) 0, f (½ ) 0. But f() 0, f( -½) 0, f(¼) 0 ad f( - /4) 0. Sice f(-) 0 ad f (½) 0, we see that (x + ) ad (x ½) are factors of the give polyomial. Also by factorizig, we fid that f(x) (x ½) (x + ) (4x x + 4x ) Note that x ½ is a root of the third factor, if we divide 4x x + 4x by x ½, we obtai f(x) (x ½) (x + ) (4x + 4) 4 (x ½) (x + ) (x + ) Hece the roots of f(x) 0, are ½, ½, -, ± i. 5. Solve x x 8x + 0, give that has a double root. Solutio: Let f(x) x x 8x + Differetiatig, we obtai: f (x) x x 8. Sice the multiple roots of f(x) 0 are also the roots of f (x) 0, the product of the factors correspodig to these roots will be the g.c.d of f(x) ad f (x). Let us fid the g.c.d of f(x) ad f (x). x x x 8 x 6x 4 4x 8 4x 8 x x 8x + x x 4x + 6 x x 8x x x 6x x 48x x + x x + 00 x Therefore, g.c.d (x ) f(x) has a factor (x ). 7

11 Also, f(x) (x ) (x + ) Thus the roots are,, -. School of Distace Educatio, Uiversity of Calicut 6. Show that the equatio x + qx + r 0 has two equal roots if 7r + 4q 0. Solutio: Let f(x) x + qx + r () Differetiatig, we obtai: f (x) x + q () Give that f(x) 0 has two equal roots, i.e., it has a double root, say α. The α is a root of both f(x) 0 ad f (x) 0. From the d equatio, we obtai α -q / Now the first equatio ca be writte as: α ( α + q) + r 0 r i.e., α (-q/ + q) + r 0 α q Squarig ad simplifyig, we obtai: 7r + 4q Relatio betwee the Roots ad Coefficiets of a Polyomial Equatio Let Cosider the polyomial fuctio f(x) aox + ax a, a 0 0 α, α be the roots of f(x) 0. α,..., The we ca write f x) a ( x α )( x α )...( x α ) ( o Equatig the two expressios for f(x), we obtai: a o x + a x Dividig both sides by a 0, x a + a a ao ( x α)( x α )...( x α ) o x a ( x α)( x α )...( x α ) a o x Sx + S x... + ( ) S where Sr stads for the sum of the products of the roots time. Comparig the coefficiets o both sides, we see that a S, a o S a a o,... S ( ) a a o. α,..., α take r at a 8

12 Special Cases School of Distace Educatio, Uiversity of Calicut b If α ad β are the roots of ax + bx + c 0, ( a 0), the α + β ad αβ a If α ad β ad γ are the roots of ax + bx +cx +d 0, ( a 0), the c d ad αβ + βγ + αγ ad αβγ. a a c a b α + β + γ, a Illustrative Examples:. If the roots of the equatio x + px + qx + r 0 are i arithmetic progressio, show that p 9 pq + 7r 0. Solutio: Let the roots of the give equatio be a d, a, a + d. The S a d + a + a + d a -p a Sice a is a root, it satisfies the give polyomial p + p. p p + q. p + 0 r O simplificatio, we obtai p 9pq + 7r 0.. Solve 7x + 4x 8x 8 0, give that its roots are i geometric progressio. Solutio: Let the roots be The, Sice a, a, ar r a 8. a. ar a a r 7 a is a root, x is a factor. O divisio, the other factor of the polyomial is 7x + 60x ± Its roots are or 7 9 Hece the roots of the give polyomial equtio are, 9. Solve the equatio 5x x + 9x 0 whose roots are i harmoic progressio.,. 9

13 School of Distace Educatio, Uiversity of Calicut Solutio: [Recall that if a, b, c are i harmoic progressio, the /a, /b, /c are i arithmetic ac progressio ad hece b ] a + c Let α, β, γ be the roots of the give polyomial. 9 The αβ + βγ + αγ..() 5 αβγ.() 5 αγ Sice α, β, γ are i harmoic progressio, β α + γ αβ + βγ αγ Substitute i (),.9 αγ + αγ αγ αγ. 5 Substitute i (), we obtai β 5 5 β is a root of the give polyomial. Proceedig as i the above problem, we fid that the roots are,, Show that the roots of the equatio ax + bx + cx + d 0 are i geometric progressio, the c a b d. Solutio: Suppose the roots are The k d. k. kr r a i.e., k d a Sice k is a root, it satisfies the polyomial equatio, k r ak + bk + ck + d 0, k, kr d a a + bk + ck + d 0 0

14 School of Distace Educatio, Uiversity of Calicut bk + ck 0 bk ck 6 ( bk ) ( ck) i. e., b k c k b d a d c a b d c b d c a. a 5. Solve the equatio x - 9x + 4x + 4 0, give that two of whose roots are i the ratio :. Solutio: Let the roots be α, α, β The, α + α + β 5α + β 9.() α.α + α. β + α. β 4 i.e., 6α + 5αβ 4 () ad α.α. β 6α β 4 α β 4 () From (), β 9 5α. Substitutig this i (), we obtai 6α + 5α (9 5α ) 4 Whe i.e., 9α 45α O solvig we get α α or. 9 6, from (), we get β. But these values do ot satisfy (). 9 So, α, the from (), we get β Therefore, the roots are 4, 6, Symmetric Fuctios of the Roots Cosider the expressios like α + β + γ,( β γ ) + ( γ α) + ( α β ), ( β + γ )( γ + α)( α β ). Each of these expressios is a fuctio of α, β, γ with the property that if ay two of α, β, γ are iterchaged, the fuctio remais uchaged. Such fuctios are called symmetric fuctios.

15 School of Distace Educatio, Uiversity of Calicut Geerally, a fuctio f α, α,..., α ) is said to be a symmetric fuctio of ( α, α,..., α if it remais uchaged by iterchagig ay two of α, α,..., α. Remark: The expressios S, S,.., S where Sr is the sum of the products of, α α take r at a time, are symmetric fuctios. These are called elemetary α,..., symmetric fuctios. Now we discuss some results about the sums of powers of the roots of a give polyomial equatio.... Theorem The sum of the r th powers of the roots of the equatio f(x) 0 is the x f ( x) coefficiet of x r i the expasio of i descedig powers of x. f ( x) Proof: Let f(x) 0 be the give th degree equatio ad let its roots be, α α the, f(x) a0 ( x α)( x α )...( x α ) α,..., Takig logarithm, we obtai log f x) log a + log( x α ) log( x α ) where a 0 is some costat. ( 0 Differetiatig w.r.t. x, we have: f (x) f (x) x α x α Multiplyig by x, x f (x) f (x) x x α x x α α x α x α α α α x x x x + ( α i ) x + ( α i ) x Therefore α r i is the coefficiet of x r i the expasio of descedig powers of x. x f ( x) f ( x) i

16 School of Distace Educatio, Uiversity of Calicut... Theorem (Newto s Theorem o the Sum of the Powers of the Roots) If α, α,..., α are the roots of the equatio x + P x + P x P 0, ad Sr r r + α. The, Sr + S r P SPr + rpr 0, if r. α... + ad S S P + S P S P 0 if r >. r + r r r Proof: We have x + P x + P x P ( x α)( x α )...( x α ) Put x y P P P ( )( )...( ) α α α y y y y y y ad the multiplyig by y, we obtai:, + P y + P y P y ( α y)( α y)...( α y) Takig logarithm ad differetiatig w.r.t y, we get P + P y + P y + P y + P y P y P y α α α + y α α y α y α α α α ( y) ( y)... ( y) α α α ( + α y + α y +...) α ( + α y + α y +...)... α ( + α y + α y +...) r S S y S y... S r + y... Cross multiplyig, we get P + P y + P y P y ( + P y + P y P r [S + S y Sr + y +...] y ) Equatig coefficiets of like powers of y, we see that P +.P 0 S S SP S P + S P + P 0 S SP SP S p + S P + S P + P 0, ad so o. S If r <, equatig coefficiets of y r o both sides, rp r S r S r P S r P... S P r

17 S r + S r P + Sr P S P r + School of Distace Educatio, Uiversity of Calicut rp 0 r If r >, the r >. Equatig coefficiets of y r o both sides, 0 Sr S r P Sr P... Sr i.e., S S P + S P S P 0 r + r r r P Remark: To fid the sum of the egative powers of the roots of f(x) 0, put x y ad fid the sums of the correspodig positive powers of the roots of the ew equatio. Illustrative Examples. If α, β, γ are the roots of the equatio x + px + qx + r 0, fid the value of the followig i terms of the coefficiets. ( i ) ( βγ ii ) α ( iii ) α β Solutio: Here α + β + γ p, αβ + βγ + αγ q, αβγ r (i) βγ + + αβ βγ αγ α + β + γ αβγ p r p r (ii) α + + α β γ αβ + βγ + αγ αβγ q r q r (iii) α β α β + β α + γ α + γ β + α γ + β γ ( αβ βγ + αγ )( α + β + γ ) αβγ + (q. p) (- r ) r pq.. If α is a imagiary root of the equatio x 7 0 form the equatio whose roots are α + α, α + α, α + α. Solutio: Let a α + α 6 b α + α 5 c α + α 4 4

18 School of Distace Educatio, Uiversity of Calicut The required equatio is (x a) (x b) (x c) 0 i.e., x ( a+b+c )x + ( ab+bc+ac )x abc 0 () a + b + c α + α + α + α 4 + α 5 + α α( α ) α α α α α α ( Sice α is a root of x 7 0, we have α 7 ) Similarly we ca fid that ab + bc + ac, abc. Thus from (), the required equatio is x + x x 0. If α, β, γ are the roots of x + x + x + 0, fid α ad α. Solutio: Here α + β + γ -, αβ + βγ + αγ, αβγ Usig the idetity a +b +c abc (a+b+c)( a + b + c ab bc ac ), we fid that [ α + β + γ ( αβ + βγ + αγ )] + αβγ α ( α + β + γ ) ( α + β + γ ) [( α + β + γ ) ( αβ + βγ + αγ )] ( αβ + βγ + αγ ) + αβγ [(9 4 ) ] 9 β γ + α γ + β α Also, α + + α β γ α β γ ( αβ + βγ + αγ) α βγ..() α β γ We have: α βγ ( α + β + γ) αβγ. ( ) α Fid the sum of the 4 th powers of the roots of the equatio x 4 5x + x 0. Solutio: Let f(x) x 4 5x + x 0 The f (x) 4x 5x + 5

19 Now, xf ( x) f ( x) ca be evaluated as follows : School of Distace Educatio, Uiversity of Calicut Therefore, xf (x) f (x) x x x x Sum of the fourth powers of the roots coefficiet of x If α + β + γ, α + β + γ, α + β + γ. Fid α 4 + β 4 + γ 4. Solutio: Let x + Px + Px + P 0 be the equatio whose roots are α, β, γ, the α + β + γ By Newto s theorem, P P Agai, by Newto s theorem S + SP + P 0 i.e., +. ( ) + P 0 P / S + SP + SP + P 0 i.e., / +.P 0 P /6 Also S4 + SP + SP + SP 0 (By Newto s theorem for the case r < ) Substitutig ad simplifyig, we obtai S4 5 /6 Thus α β + γ Calculate the sum of the cubes of the roots of x 4 + x + 0 6

20 School of Distace Educatio, Uiversity of Calicut Solutio: Let the give equatio be x 4 + Px + Px + Px + P4 0 Here P P 0, P ad P4 By Newto s theorem, S + SP + SP + P 0 i.e., S S 6 i.e., sum of the cubes of the roots of x 4 + x + 0, is Trasformatios of Equatios Let f(x) 0 be a polyomial equatio. Without explicitly kowig the roots of f(x) 0, we ca ofte trasform the give equatio ito aother equatio whose roots are related to the roots of the first equatio i some way. Now we discuss some importat such trasformatios.. To form a equatio whose roots are k-times the roots of a give equatio. Let f (x) a ox + ax a () Suppose that α, α are the roots of f(x) 0 α,..., The f (x) a (x α )(x α )...(x α )..() o Put y kx i (), we obtai: y f a k o y k α y k α y... k α Thus the roots of f ( y / k) 0, are kα,..., kα Therefore the required equatio is y y y f a o + a a 0 k k k i.e., a y + ka y + k a y k a 0 o Thus; to obtai the equatio whose roots are k times the roots of a give equatio, we have to multiply the coefficiets of k, k - ad k respectively. x, x,..., x ad the costat term by, k, 7

21 School of Distace Educatio, Uiversity of Calicut Remark: To form a equatio whose roots are the egatives of the roots of a give equatio of degree, multiply the coefficiets of x, x -,... by, -,, -, respectively.. To form a equatio whose roots are the reciprocals of the roots of a give equatio. Cosider, f (x) a x + a x a 0 o.. () Let α, α be the roots of the equatio. The, α,..., f (x) a (x α )(x α )...(x α )..() o I (), put y x i. e., x y The f a y o α y α y The roots of this equatio are... y α,,... α α But from (), f a o + a a 0 y y y i.e., a + a y + a y a y 0 o o Therefore, the required equatio is a y + a y a y + a 0 α o. To form a equatio whose roots are less by h the the roots of a give equatio. ( i.e., Dimiishig the roots by h ) Let f (x) a x + a x a 0 o. () Suppose that α, α are the roots of f(x) 0 α,..., Therefore, f (x) a ( x α )( x α )...( x α ) Put y x h so that x y + h o. () From (), f (y + h) a ( y + h α )( y + h α )...( y + h α ) a o o ( y ( α h) )( y ( α h) )...( y ( α h) ) The roots of f (y + h) 0 are α h,..., α h. By (), we obtai, a ( y + h) + a (y + h) a 0 o 8

22 School of Distace Educatio, Uiversity of Calicut Expadig usig biomial theorem ad combiig like terms, we get a equatio of the form b o y + by b 0. () Replacig y x h, we get b ( x h) + b ( x h) b 0 o. (4) Now, equatio () ad (4) represets the same equatio. Dividig equatio (4) cotiuously by (x h), we obtai the remaiders as b, b,..., b o Substitutig these i (), we obtai the required equatio. Remark: Icreasig the roots by h is equivalet to decreasig the roots by h. 4. To form a equatio i which certai specified terms of the give equatio are abset. Cosider the equatio a x + a x a 0 o () Suppose it is required to remove the secod term of the equatio (). Dimiish the roots of the give equatio by h. For this, put y x h i.e., x y + h i (), we obtai the ew equatio as a ( y + h) + a ( y + h) a 0 o ie a y + ( a h + a ) y a 0 o o Now to remove the secod term of the equatio (), we must have a o h + a 0 i.e., we must have a h. a o Thus to remove the secod term of the equatio (), we have to dimiish its roots by a h a o Remarks: If α, α are the roots of the polyomial equatio f(x) 0. α,..., Formatio of a equatio whose roots are φ α ), φ( α ),..., φ( ) is kow as a geeral trasformatio of the give equatio. 9 ( α I this case, the relatio betwee a root x of f(x) 0 ad a root y of the trasformed equatio is that y φ(x). Also, to obtai this ew equatio we have to elimiate x betwee f(x) 0 ad y φ(x).

23 School of Distace Educatio, Uiversity of Calicut Solved Problems. Form a equatio whose roots are three times those of the equatio x x + x + 0. Solutio: To obtai the required equatio, we have to multiply the coefficiets of x, x, x, ad by,,, ad respectively. Thus x x + 9x is the desired equatio.. Form a equatio whose roots are the egatives of the roots of the equatio Solutio: By multiplyig the coefficiets successively by, -,, - we obtai the required equatio as x + 6x + 8x Form a equatio whose roots are the reciprocals of the roots of x 4 5x + 7x 4x Solutio: We obtai the required equatio, by replacig the coefficiets i the reverse 4 order, as 5x 4x + 7x 5x Fid the equatio whose roots are less by, tha the roots of the equatio x 5 4 x x + 5x + 0x Solutio: To fid the desired equatio, divide the give equatio successively by x

24 Thus the required equatio is School of Distace Educatio, Uiversity of Calicut x 5 + 7x 4 + 4x + x + 40x Solve the equatio x 8x x + 68x by removig its secod term. Solutio: To remove the secod term, we have to dimiish the roots of the give a 8 equatio by h. a 4. o Dividig the give equatio successively by x, we obtai the ew equatio as x 4 5x O solvig, we get x - 4, 4, -,. Thus the roots of the origial equatio are, 6, - ad If α, β, γ are the roots of the equatio x + ax + bx + c 0. Form the equatio whose roots are αβ, βγ, γα. Solutio: Note that αβ αβγ γ c γ Put y c x x c y Hece the give equatio becomes c y c + a y c + b + c 0 y i.e., y by + acy c 0, which is the required equatio. + α + β + γ 7. If α, β, γ are the roots of x x + 0, show that + + α β γ Solutio: We have to form the equatio whose roots are + α + β + γ,, α β γ. For this, put y + x x i.e., x y y + y y Therefore the required equatio is + 0 y + y + O simplifyig, we obtai y y + 7y + 0

25 School of Distace Educatio, Uiversity of Calicut + α + β + γ The sum of the roots of this equatio is. i.e., + + α β γ.5. Reciprocal Equatios If Let f(x) 0 be a equatio with roots,,..., α α α α, α. α,..., are also roots of the same equatio, the such equatios are called reciprocal equatios. Suppose that a x + a x a 0 o... () is a reciprocal equatio with roots The α, α,..., α.,,..., α α α are also roots of the same equatio. The equatio with roots,,..., α α α is : a x + a x a 0 () o Sice () ad () represets the same equatio, we must have a a o a a a a o k Takig the first ad last terms i the above equality, we obtai k i.e., k + whe k, we have ao a, a a-... Such equatios are called reciprocal equatios of first type. Whe k, we have ao a, a a-, These type of equatios are called reciprocal equatios of secod type. A reciprocal equatio of first type ad eve degree is called a stadard reciprocal equatio. Note:. If f(x) 0 is a reciprocal equatio of first type ad odd degree, the x is always a root. If we remove the factor x + correspodig to this root, we obtai a stadard reciprocal equatio.. If f(x) 0 is a reciprocal equatio of secod type ad odd degree, the x is always a roots. If we remove the factor x correspodig to this root, we obtai a stadard reciprocal equatio.

26 School of Distace Educatio, Uiversity of Calicut. If f(x) 0 is a reciprocal equatio of secod type ad eve degree, the x ad x are roots. If we remove the factor x correspodig to these roots, we obtai a stadard reciprocal equatio. Solved Problems 4. Solve the equatio 60x 76x + a4x 76x Solutio: The give equatio is a stadard reciprocal equatio. Dividig throughout by x, we obtai, x 76x x x 60 x + x 76 x+ x Puttig y x + ad simplifyig, we obtai x O solvig, we get 60y 76y y or Whe y, x + 0x 0x x 0 i.e., Similarly whe x 0, 0 y, we get 6 x, Thus the roots of the give equatio are. Solve : Solutio: 0, 0 This is a secod type reciprocal equatio of odd degree. So x is a root. O divisio by the correspodig factor x, we obtai the other factor as x 4 4x x + 5x 5 x + 9 x 9 x + 5 x 0,, 4x + 0, which is a stadard reciprocal equatio. Proceedig exactly as i the above problem, we may fid that ± i x or ± x 5

27 Hece the roots of the give equatio are School of Distace Educatio, Uiversity of Calicut, ± i,. Show that o dimiishig the roots of the equatio 6x 4 4x + 76x + 5x 00 0 ± 5 by, it becomes a reciprocal equatio ad hece solve it. Solutio: To dimiish the roots of the give equatio by, divide it successively by (x ), we obtai: x + 5x 8x + 5x + 6 0is the required equatio, which is a stadard reciprocal equatio. It ca be writte as Whe Puttig Whe 6 x + x + 5 ( x + ) 8 0 x x + y ad solvig for y, we get x y 5, we have 0 y, we have x + 0x + 0 or Thus the roots of the origial equatio are above roots) 0 y or 5 x +. O solvig we get: x x or 5 x, 4, 5,, 5 ( by addig to each of the 4

28 School of Distace Educatio, Uiversity of Calicut.6 Carda s Method of Solvig a Stadard Cubic Equatio Kowledge of the quadratic formula is older tha the Pythagorea Theorem. Solvig a cubic equatio, o the other had, was doe by Reaissace mathematicias i Italy. I this sectio we describe some methods to fid oe root of the cubic equatio ax bx cx d () so that other two roots (real or complex) ca the be foud by polyomial divisio ad the quadratic formula. The solutio proceeds i two steps. First, the cubic equatio is depressed; the oe solves the depressed cubic..6. Depressig the cubic equatio This trick, which trasforms the geeral cubic equatio ito a ew cubic equatio with missig x -term is due to Italia mathematicia Nicolo Fotaa Tartaglia ( ). To remove the secod term of Eq. (), we dimiish the roots of a () by h with (i.e., the degree of the polyomial equatio), a0 a, a b; a 0 b b so that h. Set x h y or x y + h y. a a We apply the substitutio b x y to the cubic equatio (), ad obtai a b b b a y b y c y d 0. a a a Multiplyig out ad simplifyig, we obtai + 0, a + + 7a a b b bc ay c y d a cubic equatio i which y -term is abset..6. Solvig the Depressed Cubic Method to solve a depressed cubic equatio of the form y + Ay B () had bee discovered earlier by Italia mathematicia Scipioe dal Ferro (465-56). The procedure is as follows: First fid s ad t so that st A... () ad s t B... (4) 5

29 School of Distace Educatio, Uiversity of Calicut The y s t will be a solutio of the depressed cubic. This ca be verified as follows: Substitutig for A, B ad y, equatio () gives ( s t) + st( s t) s t. This is true sice we ca simplify the left side usig the biomial formula to obtai s t. Now to fid s ad t satisfyig () ad (4), we proceed as follows: From Eq.(), we have A s ad substitutig this ito Eq.(4), we obtai, t A t t B Simplifyig, this turs ito the tri-quadratic equatio, t A + Bt 0, 7 6 which usig the substitutio u t becomes the quadratic equatio, u A + Bu 0. 7 From this, we ca fid a value for u by the quadratic formula, the obtai t, afterwards s. Hece the root s t ca be obtaied. Illustrative Examples:. Usig the discussio above, fid a root of the cubic equatio Solutio: x 0x + 6x Comparig with ax bx cx d ,... (5) we have a, b 0, c 6, ad d 50. b Hece substitutig x y y + 5 i (5), expadig ad simplifyig, we obtai a the depressed cubic equatio y + 6y 0 0. Now to fid the solutio of depressed equatio y + 6y 0, we proceed as follows: We eed s ad t to satisfy st 6 (6) ad s t 0. (7) Solvig for s i (6) ad substitutig the result ito (7) yields: 8 t t 0, which multiplied by t becomes, t 6 + 0t 8 0. Usig the substitutio u t the above becomes the quadratic equatio u + 0u

30 Usig the quadratic formula, we obtai that u 0 ± 08. School of Distace Educatio, Uiversity of Calicut We take the cube root of the positive value of u ad obtai By Equatio (7), s t ( ) t ad hece s Hece a root y for the depressed cubic equatio is y s t Hece a root of the origial cubic equatio x 0x + 6x 50 0 is give by x y Fid oe real root of the cubic equatio x x 5 0. Solutio. Sice the term x is abset, the give equatio is i the depressed form. Here x x 5, ad hece st ad s t 5. Now substitutig s t i s t 5, we obtai t t 5 or 8 7t t 5 or t + 5t Take u t. The the above becomes the quadratic equatio u + 5u + 0 with ± ± 4.88 u 4.94 or We take the cube of root of the value with largest absolute value, ad obtai t / u.70. Puttig this value i s t, we obtai s Hece oe of the roots of the give equatio is s t Carda s Solutio of the Stadard cubic Italia Reaissace mathematicia Girolamo Cardao (50 576) published the solutio to a cubic equatio i his Algebra book Ars Maga. 7

31 Usually we take the cubic equatio as School of Distace Educatio, Uiversity of Calicut a x + a x + a x + a. 0 0 But it has bee foud it is more advatageous to take the geeral cubic as ax bx cx d (8) This method of writig is referred to as the cubic with biomial coefficiets. Takig the form (8) ad puttig y ax + b or x ( y b) / aad multiplyig throughout by a, we obtai : ( y b) b( y b) ac( y b) a d i.e., i.e., y + ( ac b ) y + ( a d abc + b ) 0 y Hy G + + 0, (9) where H ac b ad G a d abc b +. The equatio (9), where the term i y abset, is the stadard form of the cubic. Now to solve (9) usig Cardao s method, assume that the roots are of the form p + q ; where p ad q are to be determied. / / Puttig / / y p + q, we get / / / / y p + q + p q ( p + q ) / / p + q + p q y. Hece, / / y p q y p q ( + ) 0. (0) Comparig the coefficiets i (9) ad (0), we have p + q G, ad / / p q H. i.e., p + q G () ad Now, i.e., p q G H pq H. p q [( p + q) 4 pq] / + 4 () Usig () ad (), solvig for p ad q, we get G + G + 4 H 4 p, q G G + H The the solutio is give by Remark / / y p q. We otice that / p has three values, viz., m, mω ad mω where m is a cube root of p (i.e., G + G + 4H m ) ad ω is oe of the imagiary cube roots of uity. But we caot take the three values of / q idepedetly, for we have the 8

32 relatio / / p q H. Thus if, ω, School of Distace Educatio, Uiversity of Calicut ω are the three values of / q where is a cube root of q ad ω is oe of the imagiary cube roots of uity, we have to choose those pairs of cube roots of p ad q such that the product of each pair is ratioal. Hece the three admissible roots of equatio (9) are Illustrative Examples m +, m +, m + ω ω ω ω. Solve the cubic x 9x by Carda s method. Solutio. Let / / x p + q be a solutio. The / / / / x p + q + p q ( p + q ) Hece ( + ) 0 / / x p q x p q Comparig this with the give cubic equatio, we get p + q 8 () / / p q Hece pq 7. Now, ( ) ( ) 4 p q p + q pq Hece p q 6 (4) From () ad (4), we get p ad q 7. Hece / p, ω, ω ad / q, ω, ω ; where ω is oe of the imagiary cube roots of uity. Hece the roots of the give cubic are, ω ω ad ω ω. i.e., 4, ω ω, ω ω. Aother Method Sice 4 is a root of the give cubic, x + 4 is a factor of the polyomial i the give cubic equatio. Removig the factor x + 4, the cubic equatio yields the quadratic equatio Hece x 4x ± x ± i Hece the roots of the give cubic are 4, + i, i. Solve x x x 9

33 Solutio. School of Distace Educatio, Uiversity of Calicut To reduce to stadard form [otig that h b / a 6/ ], put x y. i.e, x y + ad obtai ( y + ) 6( y + ) + ( y + ) 0 i.e., y 9y 0, which is the stadard form of the cubic. Puttig / / y p + q, takig the cube ad a rearragemet yields / / y p q y p q ( + ) 0. Comparig this with the stadard form of the cubic, we obtai p + q (5) ad / / p q Hece pq 7 ad p q p q pq ( + ) 4 6 (6) From (5) ad (6), p 9 ad q. Hece / / y + 9 or / / / / ω + 9 ω or ω + 9 ω Hece the roots of the give equatio are 9, 9, 9 / / / / / / ω + ω + ω + ω..6.4 Nature of the roots of a cubic Let α, β, γ be the roots of the cubic y Hy G (7) The the equatio whose roots are ( β γ ), ( γ α) ad ( α ) β is z + 8Hz + 8H z + 7( G + 4 H ) 0. Hece ( β γ ) ( γ α) ( α β ) 7( G 4 H ). + (8) The ature of the roots α, β, γ of Eq. (7) ca be obtaied by a cosideratio of the product i Eq. (8). Sice imagiary roots occur i pairs, equatio (7) will have either all real roots, or oe real ad two imagiary roots. The followig cases ca occur. Case : The roots α, β, γ are all real ad differet. I this case ( β γ ) ( γ α) ( α β ) is positive. Therefore by Eq. (8), G + 4H is egative. Case : Oe root, say α, is real ad the other two imagiary. Let β ad γ be m ± i. The ( β γ ) ( γ α) ( α β ) ( i) ( m i α) ( m + i α) 4 {( m ) } α +, 0

34 School of Distace Educatio, Uiversity of Calicut which is egative, whatever α, m, may be. Therefore by Eq.(8), positive i this case. G + 4H is Case : Two of the roots, say β, γ are equal. The ( β γ ) ( γ α) ( α β ), ad therefore G + 4H, is zero. Case 4: α, β, γ are all equal. I this case all the three roots of equatio (7) are zero. This will be so if H G 0. Coversely, it is easy to see that (i) whe (ii) whe (iii) whe G G G + 4H < 0, the roots of the cubic i Eq. (7) are all real; + 4H > 0, the cubic i Eq. (7) has two imagiary roots; + 4H 0, the cubic i Eq. (7) has two equal roots; ad Remark (iv) whe G H 0, all the roots of the cubic i Eq. (7) are equal. O substitutig the values of G ad H, it ca be see that G + 4 H a { a d 6abcd + 4ac + 4b d b c }. The expressio i brackets is called the discrimiat of the geeral cubic i Eq. (7), ad is deoted by..7. Quartic (or Biquadratic) Equatio A quartic fuctio is a polyomial of degree four ad is of the form 4 f ( x) ax + bx + cx + dx + e, where a is ozero. Such a fuctio is sometimes called a biquadratic fuctio, but the latter term ca occasioally also refer to a quadratic fuctio of a square, havig the form 4 ax + bx + c, or a product of two quadratic factors, havig the form ( ax + bx + c)( dy + ey + f ). Settig f ( x ) 0 results i a quartic equatio (or biquadratic equatio) of the form 4 ax bx cx dx e where a 0. Quartic equatio is some times called biquadratic equatio. Solutio of Quartic Equatios (Ferrari s method) Shortly after the discovery of a method to solve the cubic equatio, Lodovico Ferrari (5-565), a studet of Cardao, foud a similar method to solve the

35 School of Distace Educatio, Uiversity of Calicut quartic equatio. I this method the solutio of the quartic depeds o the solutio of a cubic. We ow describe the Ferrari s method. Writig the quartic equatio 4 u ax bx cx dx e (9) We assume that au ax + bx + s mx + ( ) ( ). Equatig coefficiets of like powers of x ad the elimiatig m ad we will obtai a cubic equatio i s. The correspodig values of m ad will be obtaied. Usig these, roots of the quartic equatio will be obtaied. The method is illustrated i the followig problems. Illustrative Examples. Solve x x x x Solutio: Let u x x x x ad The, ( + + ) ( + ) (0) u x px s mx x x x + 0x + ( x + px + s) ( mx + ) 4 Equatig coefficiets of like powers of x, we obtai p, s p m, + 0 ps m, s. Hece, p s + m,, 0 s m, s. Thus p m s +,, m s s 5,. To elimiate m ad, we ote that ( m) m which gives ( s 5) (s )( s ) + or s + 0s + 5 s 6s + s 9 which is the cubic equatio s 6s + s 64 0 or s s s () ( Remarks : Equatio (), which is a cubic i s, is kow as the reducig cubic. The reducig cubic gives three values of s. These do ot however lead to three differet sets of roots for the quartic equatio. They oly give three differet methods of factorizig the left had side of the quartic. Hece it is eough to fid ay oe root of the reducig cubic. ) By ispectio, s is a root of the cubic equatio ().

36 Hece, m School of Distace Educatio, Uiversity of Calicut s + ( ) + 9 s ( ). m s 5 5. We take m. The, as m ad, we take. We take (Referrig Remark above), s, m,. Notig [by Eq.(0)] that u x + p + m x + s + x + p m x + s { ( ) }{ ( ) } we have u x + + x x + x + { ( ) }{ ( ) } or u x x x x { + }{ 4 }, or u x x x x ( )( + ){ 4 } () As the quadratic equatio x 4x has roots + 5ad 5, by (), we have the roots of the give quartic equatio are x,, + 5, 5.. Solve 4 x x x 6 0 Solutio. Let 4 u x x x 6 0. ad u x px s mx ( + + ) ( + ) () The, x x 6x ( x + px + s) ( mx + ) 4 Equatig coefficiets of like powers of x, we obtai p m s + 0,, m s +,. To elimiate m ad, we ote that ( m) m which gives 9 (s + )( s + ) i.e., we obtai the cubic equatio, s s 4s (4) I order to fid a root of (4) we proceed as follows: If α, β, γ are the roots of (4), the the equatio with roots α, β, γ is give by y y y Or y y y (5) y is evidetly a root of equatio (5). If we let α the i.e., s is a solutio of the reducig cubic (4). Hece α is a root of (4). m + 4,

37 Take m. Hece, as 9 School of Distace Educatio, Uiversity of Calicut m, ad, we have to take 4. Puttig p 0, s, m,, i u x p m x s x p m x s { + ( + ) + + }{ + ( ) + }, we get u x x x x { + + }{ }. As the quadratic equatio x + x + have roots + i, i ad x x have roots +, ; hece the roots of the give quartic equatio are +,, + i, i.. Solve 4 x + x + x 0. Solutio. Let u x x x ad u x px s mx ( + + ) ( + ) (6) The, x + x + x ( x + px + s) ( mx + ) 4 Equatig coefficiets of like powers of x, we obtai 5 p m s + 4,,,. m s s + To elimiate m ad, we ote that ( m) m, which gives 5 s s + s + 4 ( ) ad is the cubic equatio 4s s 8s (7) I order to fid a root of (7) we proceed as follows: If α, β, γ are the roots of (4), the the equatio with roots 4 α, 4 β, 4γ is give by y y y Or y y y (8) y is evidetly a root of equatio (8). If we let 4α the α is a root of (7). i.e., reducig cubic (7). Hece s is a solutio of the 5 5 m s s

38 School of Distace Educatio, Uiversity of Calicut We take m. Sice m s, ad sice 4 9, we are forced to take 4, Hece u x p m x s x p m x s { + ( + ) + + }{ + ( ) + } gives u x + x x + x + { }{ } Hece the roots of the give quartic equatio are x + + ω ω,,,, where ωis a imagiary cube root of..6. Isolvability of the Quitic. A quitic fuctio is a polyomial fuctio of the form where a, b, c, d, e ad f are ratioal umbers, real umbers or complex umbers, ad a is ozero. I other words, a quitic fuctio is defied by a polyomial of degree five. If a is zero but oe of the coefficiets b, c, d, or e is o-zero, the fuctio is classified as either a quartic fuctio, cubic fuctio, quadratic fuctio or liear fuctio. As we oted above, solvig liear, quadratic, cubic ad quartic equatios by factorizatio ito radicals is fairly straightforward, o matter whether the roots are ratioal or irratioal, real or complex; there are also formulae that yield the required solutios. However, there is o formula for geeral quitic equatios over the ratioals i terms of radicals; this is kow as the Abel Ruffii theorem, first published i 84, which was oe of the first applicatios of group theory i algebra. This result also holds for equatios of higher degrees. This meas that ulike quadratic, cubic, ad quartic polyomials, the geeral quitic or all polyomials of degree greater tha 5 caot be solved algebraically i terms of a fiite umber of additios, subtractios, multiplicatios, divisios, ad root extractios. A example quitic whose roots caot be expressed by radicals is Some fifth-degree equatios ca be solved by factorizig ito radicals; for example,, which ca be writte as, or, as aother example,, which has as solutio. 5

39 School of Distace Educatio, Uiversity of Calicut Évariste Galois developed techiques for determiig whether a give equatio could be solved by radicals which gave rise to Galois theory..7. Descarte s Rule of Sigs ad Sturm s Theorem Nature of Roots - Descarte s Rule of Sigs To determie the ature of some of the roots of a polyomial equatio it is ot always ecessary to solve it; for istace, the truth of the followig statemets will be readily admitted.. If the coefficiets of a polyomial equatio are all positive, the equatio has o positive root; for example, the equatio caot have a positive root. x + x If the coefficiets of the eve powers of x are all of oe sig, ad the coefficiets of the odd powers are all of the opposite sig, the equatio has o egative root; thus for example, the equatio caot have a egative root x + x + x x + x x + x 7 0. If the equatio cotais oly eve powers of x ad the coefficiets are all of the same sig, the equatio has o real root; thus for example, the equatio caot have a real root. 8 4 x x x 0 4. If the equatio cotais oly odd powers of x, ad the coefficiets are all of the same sig, the equatio has o real root except x 0; thus the equatio 7 5 x x x x has o real root except x 0. Suppose that the sigs of the terms i a polyomial are ; here the umber of chages of sig is 7. We shall show that if this polyomial is multiplied by a biomial (correspodig to a positive root) whose sigs are +, there will be at least oe more chage of sig i the product tha i the origial polyomial. 6

40 School of Distace Educatio, Uiversity of Calicut Writig dow oly the sigs of the terms i the multiplicatio, we have the followig: ± ± + ± ± Here i the last lie the ambiguous sig ± is placed wherever there are two differet sigs to be added. Here we see that i the product (i) a ambiguity replaces each cotiuatio of sig i the origial polyomial; (ii) the sigs before ad after a ambiguity or set of ambiguities are ulike; (iii) a chage of sig is itroduced at the ed. Let us take the most ufavourable case (i.e., the case where the umber of chages of sig is less) ad suppose that all the ambiguities are replaced by cotiuatios; the the sig of the terms become ad the umber of chages of sig is , We coclude that if a polyomial is multiplied by a biomial (correspodig to a positive root) whose sigs are +, there will be at least oe more chage of sig i the product tha i the origial polyomial. If the we suppose the factors correspodig to the egative ad imagiary roots to be already multiplied together, each factor x a correspodig to a positive root itroduces at least oe chage of sig; therefore o equatio ca have more positive roots tha it has chages of sig. Agai, the roots of the equatio f ( x) 0 are equal to those of f ( x ) 0 but opposite to them i sig; therefore the egative roots of f ( x ) 0 are the positive roots of f ( x) 0; but the umber of these positive roots caot exceed the umber of chages of sig i f ( x); that is, the umber of egative roots of f ( x ) 0 caot exceed the umber of chages i sig i f ( x). 7

41 School of Distace Educatio, Uiversity of Calicut All the above observatios are icluded i the followig result, kow as Descarte s Rule of Sigs. I ay polyomial equatio f(x) 0, the umber of real positive roots caot exceed the umber of chages i the sigs of the coefficiets of the terms i f(x), ad the umber of real egative roots caot exceed the umber of chages i the sigs of the coefficiets of f( x). Example: Cosider the equatio f(x) x 4 + x 0 This a polyomial equatios of degree 4, ad hece must have four roots. The sigs of the coefficiets of f(x) are + + Therefore, the umber of chages i sigs By Descarte s rule of sigs, umber of real positive roots <. Now f(-x) x 4 - x 0 The sigs of the coefficiets of f( x) are + Therefore, the umber of chages i sigs. Hece the umber of real egative roots of f(x) 0 is <. Therefore, the maximum umber of real roots is. If the equatio has two real roots, the the other two roots must be complex roots. Sice complex roots occur i cojugate pairs, the possibility of oe real root ad three complex roots is ot admissible. Also f (0) < 0, ad f () > 0, so f(x) 0 has a real roots betwee 0 ad. Therefore, the give equatio must have two real roots ad two complex roots. Problem. Discuss the ature of roots of the equatio Solutio. 9 8 x x x x With f ( x) x + 5x x + 7x +, there are two chages of sig i f ( x ) 0, ad therefore there are at most two positive roots. 9 8 Agai f ( x) x + 5x + x 7x +, ad there are three chages of sig, therefore the give equatio has at most three egative roots. Obviously 0 is ot a root of the give equatio. Hece the give equatio has at most real roots. Thus the give equatio has at least four imagiary roots. 8

42 School of Distace Educatio, Uiversity of Calicut.8. Exercises 4. Solve the equatio x + x x x 0give that oe root is.. Form a ratioal quartic whose roots are,, + 5. Solve x x + 4x x + 0 give that it has multiple roots Solve the equatio x x x + x whose roots are i A.P Solve the equatio x x + 4x + 6x 0 give that two of its roots are equal i magitude ad opposite i sig. 6. Fid the coditio that the roots of the equatio x + px + qx + r 0 may be i geometric progressio. 7. Fid the coditio that the roots of the equatio x lx + mx 0 may be i arithmetic progressio. 8. If α, β, γ are the roots of x + px + 0, prove that α α α. 9. If α, β, γ are the roots of x + qx + r 0, the fid the values of ( ). ad β γ β + γ 0. Prove that the sum of the ith powers of the roots of x + x is zero.. If α, β, γ are the roots of x 7x + 7 0, fid the value of α + β + γ.. Fid the equatio whose roots are the roots of the equatio x 4 + 7x 5x + x 0, each icreased by 7.. Remove the secod term of the equatio x 6x + 4x Solve the equatio x 8x + 9x x + 0 by removig its secod term. 5. If α, β, γ are the roots of x + px + q 0, form the equatio whose roots are α + βγ, β + γα, γ + αβ. 6. If α, β, γ are the roots of the equatio x + px + q 0, fid the equatio whose β γ γ α α β roots are +, +, +. γ β α γ β α Solve 6x 5x + x x + 5x Solve x 5x + 5x Solve x 9x 0 usig Carda s method. 9

43 0. Solve. Solve. Solve x x x usig Carda s method. 4 x x x x School of Distace Educatio, Uiversity of Calicut usig Ferrari s method. 4 x + 6x + 4x 0 usig Ferrari s method.. Fid the greatest possible umber of real roots of the equatio x 5 6x 4x Fid the umber of real roots of x x x 6x Show that 5 x x has at least two imagiary roots. 6. Determie the ature of the roots of the equatio 4 x x x Prescribed Text Book (As per Syllabus) Barard ad Child, Higher Algebra, Macmilla. 40

44 Module-II FUZZY SETS Prepared by: Sathosh P.K. Asst. Professor Dept. of Mathematics Govt. Bree College, Thalassery.

45 From Crisp Sets to Fuzzy Sets: A Grad Paradigm Shift Sice its iceptio i 965, the theory of fuzzy sets has advaced i a variety of ways ad i may disciplies. Applicatios of this theory ca be foud, for example, i artificial itelligece, computer sciece, medicie, cotrol egieerig, decisio theory, expert systems, logic, maagemet sciece, operatios research, patter recogitio, ad robotics. Mathematical developmets have advaced to a very high stadard ad are still forthcomig to day. I this course material, the basic mathematical framework of fuzzy set theory is described. Applicatios of fuzzy set theory to real problems are aboud.. Crisp Sets: A Over View The cocept of Sets caot be itroduced without rememberig the great mathematicia George Cator, the fouder of Set Theory. A ordiary set or a crisp set is a well defied collectio of distict objects. This meas

46 . From Crisp Sets to Fuzzy Sets: A Grad Paradigm Shift that elemets are o repeated ad give a elemet, it must be possible to determie whether it belogs or does ot belogs the give collectio. The followig are some examples of crisp sets.. Collectio of boys i a class. That is, {x : x is a boy i the class}.. Collectio of positive itegers betwee ad 0. That is, {,,, 4, 5,6, 7, 8, 9, 0}. The followig are some o example of sets as they caot be uiquely represeted.. Collectio of beautiful girls i a class.. The collectio of real umbers much greater tha 8. To idicate a idividual object x is a member or elemet of a set A, we write x A, else x / A. A set ca be represeted usig three ways, amig its elemets, called List Method; defied by a property of its elemets, called Rule Method ad the last by, usig a fuctio, called Characteristic Fuctio. A set A X is defied by its characteristic fuctio χ A as follows, ; for x A χ A (x) 0 ; for x / A That is, a characteristic fuctio maps elemets of X to its elemets of the set {0, }, which is formally deoted by χ A : X {0, }. Geeral Priciple of Duality i Set Theory: For each valid equatio i set theory that is based o the uio ad itersectio operatio, there correspods a dual equatio, also valid, that is obtaied by replacig φ,, by X, ad i order. School of Distace Educatio, Uiversity of Calicut