Repeted Root d Commo Root 1 (Method 1) Let α, β, γ e the roots of p(x) x + x + 0 (1) The α + β + γ 0, αβ + βγ + γα, αβγ - () (α - β) (α + β) - αβ (α + β) [ (βγ + γα)] + [(α + β) + γ (α + β)] +γ (α + β) + [(α + β+ γ) (α + β)] +γ (α + β) + 0 +γ ( γ) γ Similrly, (β - γ) α d (γ - α) β (α - β) (β - γ) (γ - α) ( γ )( α )( β ) ( + γ )( + α )( + β ) [6 + 8 (α + β + γ ) + 6(α β + β γ + γ α ) + 7α β γ ] {6 + 8 [(α+ β + γ) (αβ + βγ + γα)] + 6 [(αβ + βγ + γα) - αβγ(α + β + γ)+ 7α β γ ] {6 + 8 [0 ] + 6[ (-)(0)] + 7(-) } { + 7 } (1) hs distict roots (α - β) (β - γ) (γ - α) < 0 + 7 < 0 (Method ) Let p(x) x + x +, p (x) x +, p (x) 6x Sice lim p(x) +, lim p(x), p(x) 0 hs t lest oe rel root x + x For sttiory poits, p (x) 0, x + 0 x ± (1) If > 0, the there is o sttiory poit, ut the poit (0, ) is rel poit of iflexio p(x) 0 hs oly oe rel root d two complex roots If 0 d 0, there is sttiory poit whe x 0 d (0, ) is rel iflexio p(x) 0 hs oly oe rel root d two complex roots If 0 d 0, the origi is t the iflexio d p(x) 0 hs roots 0, 0, 0 If < 0, the y (1), y mi d + x + is mi sice p " > 0 + x is mx sice p " < 0, y mx p(x) 0 hs distict roots y mi < 0 d y mx > 0 + < 0 d > 0 < 0 + 7 + < 0 (Note : If < 0, For two differet rel roots, oe eig repeted, the ecessry d sufficiet coditio is 7 + 0 For oe rel roots, two complex roots : 7 + > 0) 1
(Method 1) Let P(x) x 5 + x + x 6x 5x, the P (x) 15x + 8x + x 1x 5 By the multiple root theorem, P(ω) ω 5 + ω + ω 6ω 5ω 0, P (ω) 15ω + 8ω + ω 1ω 5 0 By Euclide lgorithm, (workig steps ot show here), HCF (P(x), P (x)) x + x + 1 P(x) (x + x + 1) (x ), y divisio (workig steps ot show here) The roots re 1± (doule roots) d (Method ) Sice P(x) 0 hs complex root ω, its cojugte ω is lso root Thus P(x) 0 hs four roots, ω, ω, ω, ω As result, P(x) hs fctor : (x + x + c) Sice deg [P(x)] 5, there is lier fctor left Sice the ledig coefficiet of P(x) is, which is ot complete squre, 1 Similrly, from the costt term of P(x), c ±1 or ± From the costt term 5 of P (x), c ± is rejected As result, P(x) hs fctor : (x + x ±1) d the lier fctor is therefore (x ±) O testig usig the fctor theorem P(/) 0, d the lier fctor is (x ) P(x) (x + x + c) (x ) O comprig coefficiets, we c fid 1, c 1 Result follows z z z Put p(z) 1+ z + + + +!!! Sice p(0) 0 z 0 is ot root of p(z) p(z) 0 1 z z z z z 1+ z + + + + 0 p' (z) 0!!!!! By the multiple root theorem, p(z) 0 hs o repeted roots Let p(x) (x 1 ) (x ) (x 5 ) + k (x ) (x ) (x 6 ) p( 1 ) k ( 1 ) ( 1 ) ( 1 6 ) < 0, p( ) ( 1 ) ( ) ( 5 ) > 0 p( ) ( 1 ) ( ) ( 5 ) < 0, p( 6 ) ( 6 1 ) ( 6 ) ( 6 5 ) > 0 Sice deg [p(x)] d there re chges of sig s x icreses from 1 to 6, there re distict rel roots 5 Let p(x) x x + 6x + 1, p (x) x 1 x 1x Sice p(0) 1 0 q 0 is ot root of p(x) 0 By the multiple root theorem, sice q is the repeted root, p (q) q 1 q 1q 0 q (q q + ) 0 Sice q 0, q q + 0 q + q (1) p(q) 0 q q + 6q + 1 0 ()
q + (1) (), q q + 6q + 1 0 q 6q 0 q 6q + q From (1), 6 (Method 1) q + q ( q + 6q + 9) ( 6q + + 6q + 9) 81q 81( 6q + ) 16 6q + + q + 1 16 8q + 6 7 q + 1 7 q + 1 7 Let p(x) x + p 1 x -1 + p x - + + p -1 x + p [ 1( q + 1) ] 1 q + q 81( 6q + ) q + 1 Sice α is multiple root of p(x) 0, 1/α is multiple root of p(1/y) 0 1 1 1 1 1 or + p1 + + p 1 + p 0, g(y) py + p 1y + + p1 + 1 0 y y y 1 By Multiple Root Theorem, 1/α is root of g (y) 0, ie, '(y) p y + ( 1)p y + + p 0 1 1 1 α is lso root of p + ( 1)p 1 + + p1 0, or x x p 1 x -1 + p x - + p x - + + ( 1)p -1 x + p 0 (Method ) Let p(x) x + p 1 x -1 + p x - + + p -1 x + p p (x) x -1 + ( 1) p 1 x - + ( )p x - + + p -1 α is multiple root p(α) p (α) 0 p(α) - α p (α) 0 α is root of p 1 x -1 + p x - + p x - + + ( 1)p -1 x + p 0 7 () (h) 0 (x) (x h) m q(x), where q(h) 0 + 1 g 1 1 (x) (x h) m q (x) + m(x h) m-1 q(x) (x h) m-1 [(x h) q (x) + m q(x)] (x h) m-1 g(x) where g(x) (x h) q (x) + m q(x) d g(h) m q(h) 0 But (x) (x h) s f(x), where g(h) 0, sice h is s-multiple root of (x) (x h) m-1 g(x) (x h) s f(x) If s m 1, without lost of geerlity, we ssume s > m 1 the g(x) (x h) s-(m-1) f(x) d therefore g(h) (h h) s-(m-1) f(h) 0, cotrdictig to g(h) 0 s m 1, m s + 1, d (x) (x h) s+1 q(x), where q(h) 0 h is (s + 1)-multiple root of (x) Coverse: h is (s + 1)-multiple root of (x) (x) (x h) s+1 q(x), q(h) 0 Oviously, (h) 0 Also, (x) (x h) s+1 q (x) + (s + 1) (x h) s q(x) (x h) s [(x h)q (x) + (s + 1) q(x)] (x h) s g(x) g(h) (h h)q (h) + (s + 1) q(h) (s + 1) q(h) 0 h is s-multiple root of (x)
() Let f(x) x + x + c, f (x) x + By (), h is doule root of f(x) 0 f(h) 0 d f (h) 0 h + h + c 0 (1) d h + 0 () From (), h () () (1), + + c 0 Δ c 0 Δ c 0 h is doule root of f(x) 0 (c) If h is triple root of f(x) x + x + cx + d, the f(h) h + h + ch + d 0 () f (h) h + 6h + c 0 (5) f (h) 6h + 6 0 (6) From (6), h, sust i (), h () - (5), h + ch + d 0 c + + c 0 (7) c c c d + c + d 0 + c + d 0 (8) c Result follows from (7) d (8) (d) (i) x + x + cx + d (x e) (x f), where e f, e 0 (x ex + e ) (x f) x (e + f) x + (e + ef) x e f Comprig coeff (e + f) (9) c (e + ef) (10) d e f (11) (9) (10), 9c e(e + f)(e + f) From (11), 9d 9 e f d c 9d 9c e(e + f)(e + f) 9 e f (e + f)(e + f) 9ef e + 5ef + f 9ef e ef + f 0 (e f) 0 e f Cotrdictio, d c (ii) If k is doule root of f(x) 0, the 1/k is doule root of p(1/y) 0, tht is, g(y) dy + cy + y + 0 g (1/k) 0 g (y) (dy + cy + d) [d(1/k) + c(1/k) + ] 0 k + ck + d 0 (1) (iii) If f(x) x + x + cx + d hs doule root k, the: f(k) k + k + ck + d 0 (1) f (k) k + 6k + c 0 k + k + c 0 (1) (9), k + ck + d 0 (15)
(1), k + k + c 0 (16) (15) (16), (c )k (c d) 0 1 c d k (17) c (iv) (1) c, ck + ck + c 0 (18) (1), k + ck + d 0 (19) (18) (19), (c )k (d c ) 0 (17) (0), d move terms, (c d) (c )(d c ) k d c (0) c 8 Let p(x) x + + 1 x + 0 0, If α is repeted root of p(x) 0, the p(x) (x - α) g(x), where g(x) is polyomil p (x) (x - α) g(x) + (x - α) g (x) (x - α) [g(x) + (x - α) g (x)] p (α) 0 α is root of p (x) x -1 + + + 1 0 Let p(x) x 0x 6x + 9x, p (x) 96x 60x 1x + 9 (x 0x x + ) p (x) (96x 0x ) 1(x 10x 1) Now, p (x) 0 (x 1)(1x + 1) 0 x 1/ or -1/1 Sice p(x) 0 hs triple root, this root must e root of p (x) 0 By tril, p(1/) 0, therefore 1/ is the triple root By divisio, we get p(x) (x 1) (x + ) d hece the roots re 1/ (triple root) d -/ 9 First prt is omitted () Let f(x) Ax +1 + Bx + 1 Sice f(x) is divisile y (x 1), x 1 is doule root of f(x) 0 f(1) A + B + 1 0 (1) f (x) ( + 1) Ax + Bx -1 f (1) ( + 1)A + B 0 () From (1), A + B - () () (), - + A 0 A () () (), B -1 f(x) x +1 ( + 1)x + 1 () See umer 10 Let f(x) x + 1x + x x + the f (x) x + 6x + 6x (x + 9x + 16x 6) (x + 6x ) (x + ) By tril f(-) 0 x + 6x is repeted fctor of f(x) [irrtiol roots occur i pirs] f(x) (x + 6x ) f(x) 0 hs roots ± 11 (repeted roots) 11 Let f(x) x 6 5x 5 + 5x + 9x 1x x + 8 the f (x) 6x 5 5x + 0x + 7x 8x, f (x) 0x 100x + 60x + 5x 8 By tril f() f () f () 0 x is the multiple roots of multiplicity of f(x) 0 By divisio f(x) (x ) (x + x x 1) (x ) [x (x + 1) (x + 1)] (x ) (x 1)(x + ) The roots re (triple root), 1 d -1 (doule roots) 5
1 Let f(x) x ( 0), f (x) x -1 f(x) hs doule root t x r if f(r) 0 d f (r) 0 r - 0 (1) r -1 0 () From (), r 0 is the oly possile multiple root Su r 0 i (1), we get 0 0 cotrdictig to the give 0 f(x) 0 cot hve repeted roots 1 Let f(x) x + x -1 + ( 1)x - + +!, f (x) x -1 + ( - 1)x - + +! f(x) hs doule root t x r if f(r) 0 d f (r) 0 f(r) r + r -1 + ( 1)r - + +! (1), f (r) r -1 + ( - 1)r - + +! () (1) (), r 0 r 0 But f(0)! 0 r 0 is ot root of f(x) 0 Cotrdictio f(x) 0 cot hve repeted roots 1 Let f(x) x + px + q 0 (1) 15 By multiple root theorem, f (x) x + px 0 () f (x) 1x + p 0 () () x, 1x + px 0 () () (), 8x 0 x 0 my e the oly possile repeted root But, If x 0, f(0) q 0 f(x) x + px x (x + p) If p 0, f(x) 0 hs x 0 of multiplicity If p 0, f(x) 0 hs x 0 of multiplicity d ± p s roots I either cse, the equtio x + px + q 0 cot hve exctly three equl roots x + x + 0 x x + x 1 0 x + ( / )x + 1 0 (1) 0() ( x 1)( x x + 1) From (), x 1 is root The other two roots re complex d re cojugte roots () If (1) d () hve exctly oe root i commo The the root i commo must e 1 Su x 1 i (1), 1 + (/)(1) + 1 0 / d the equtio (1) is x x + 1 0 (x 1) 0 x 1 () If (1) d () hve exctly two roots i commo, the x x + 1 0 from () Compre coefficiets with (1), / 1 The commo roots re 1± i x 16 x x + 6x + 0 (1), x x + 0 () (1) () x, (6 )x - x /(6 ) () From (1), x x + 6x + (x + )(x x + 1) 0 Sice the roots of x x + 1 0 re complex d must occur i pirs, (1) d () hve o complex commo roots d the oly commo root is x - Put x - i (), -6 6
17 x + 5x 6x 9 0 (1), x + 7x 11x 15 0 () Let α, β, γ e the roots of (1) d α, β, δ e the roots of () The α + β + γ -5/ () αβγ 9/ () α + β + δ -7/ (5) αβδ 15/ (6) () (5), γ - δ -1/6 (7) ()/(6), γ 9δ/10 (8) (8) (7), δ 5/ (9) (9) (5), α + β - (10) (9) (6), αβ (11) Solvig (10), (11), α -, β -1 18 Let f(x) x x x + 1 0 d g(x) x 7x + 1 0 1 1 +0-7 +0 +1 1 - - +1 1 Apply Euclide Algorithm, the HCF of f(x) d g(x) is x x + 1 1 - - +1 1 - +1 x x + 1 0-5 -1 +1 1 - +1 1 ± 5 - - + 1 - +1 x re the commo roots -1 + -1 19 Let f(x) 6x + 7x x 0 & g(x) 6x + 19x + 17x x 6 0 6 +7-1 - 6 +19+17 - -6 1 Apply Euclide Algorithm, the HCF of f(x) d g(x) is x + x - 1 6 + - 6 +7-1 - x + x - 1 0 (x + 1)(x + 1) 0 x 1/ or -1 + - 1 +18+0-6 x 1/ d -1 re the commo roots + - 1 +1 - - +- 0 () -1/ (doule root), () (doule root), i, - i (c) 1/ (triple root), -/ 1 () f (x) x + x + 0 g(x) x + ' x + ' 0 hs commo root f (x) g(x) ( ' )x + ( ' ) 0 g(x) ' f (x) ( ' )x + (' ' ) 0 hs commo root ' x d ' ' ' x ' ' ' ' ' ' ( ) + ( )( ) 0 (1) () x x (k + )x + 1 0 x (k )x + 6 0 x k + k + x + 6 0, 6 k k + 9 0 ', ' 9 Sustitute i (1) d solve for k, k 9 7