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![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 Jan 2007 21:16:17 IST
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If , are the roots of x2 +px +q=0 and also of x2n+pnqn+qn=0 and if / , / are the roots of xn +1 +(x+1)n=0,then n is:
(a) an odd integer
(b) an even integer
(c) any integer
(d) none
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Jan 2007 00:20:40 IST
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hiiiii Neeraj .. The question has not come up properly due to some problem .. u know these crosses.. what is in there place .. it is not clear . I was wondering if u can post this qusetion again .. I hope u will not mind ... cheers
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Puneet Agrawal
IIT Delhi
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Feb 2007 13:41:32 IST
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alpha ,beta, alpha/beta ,beta/alpha
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 16 Feb 2007 13:44:02 IST
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would u plz put it once again i am not gettin u
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ok i got ur problem i think in the second equation it's xn instead of pn  +  = -p   = q the next equation has roots n & n therefore n + n = -p ( let x n =y ) ok n =k and n = m ( disgustin typin all that) km=qn now (  /  )^n + (  /  +1)^n + 1 (since  /  is its root) =k/m + (  +  )^n/  ^n + 1 ( where k and m hav their usual meanings) takin l.c.m = {-p^n + (-p)^n}/ n now if it is a factor then the above term should add upto zero therefore n is even now that was a tough job to pen down all those xponentials huh..... plzzzzzzz give me rates if i deserve. if u didn't get this i am there to sort it out
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Feb 2007 15:40:56 IST
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But why are the roots of the second equation alpha^n and beta^n?
Its roots are given alpha and beta?I've understood the rest of the solution
Your answer is correct
thanks
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Feb 2007 15:51:11 IST
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since and are the roots of the second equation, hence the roots of "y"(=xn) must be equal to (alpha)n and (beta)n, because y=xn, and x=alpha,beta.
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-Ramya |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Feb 2007 16:58:55 IST
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ya i agree with ramya by symmetry if  &  are roots of equation x^2 + ax + c=0, then  ^n &  ^n should b the roots of the other equation.
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Feb 2007 18:21:51 IST
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WHY???
If first polynomial is represented by 'y' ,then the other equation is not
y^n .
then where comes the symmetry??
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Feb 2007 20:47:22 IST
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but u r taking xn=y and when u substitute this in the second equation, u get two roots of y,mind u, not of xn, and u equate this to (alpha)n and (beta)n, bcoz y=xn, and from the first equation, we know, x= alpha and beta
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Feb 2007 21:16:47 IST
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I got it....thanks
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Feb 2007 13:57:04 IST
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ok let me put it the other way see alpha and beta are the roots of the second equation so alpha^2n + p^n*alpha^n +q^n =0 since it satisfies it. now this is a quadratic with roots alpha^n & beta^n . so the roots are alpha^n and beta^n. i think u got my point!!!!!! 
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