**p** and **q** are roots of the equation:-
\[x^{2}-a(x-1)+b\]
Then find out the value of:-
\[\frac{1}{p^{2}-ap}+\frac{1}{q^{2}-bq}+\frac{2}{a+b}\]

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## Comments

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TopNewest@Michael Mendrin @Satvik Golechha @Krishna Ar @Aditya Raut please help......and sorry to disturb you all

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The question is wrong, it should be \(\dfrac{1}{q^2-aq}\)

Then,

\(\dfrac{1}{p^2-ap} + \dfrac{1}{q^2-aq} + \dfrac{2}{a+b}\\ = \dfrac{1}{p(p-a)} + \dfrac{1}{q(q-a)} + \dfrac{2}{pq} \\ = \dfrac{1}{-pq} + \dfrac{1}{-pq} + \dfrac{2}{pq} = 0\).

Seems logical @Aman Sharma ?

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I saw this problem in an old iit question paper........hard to believe that question is wrong.....anyway thank you so much for helping...you know i wasted almost 3 hours on this question and realy got frustrated........you are doing great job as a moderator as well a problem writer...keep up good work (and thank you so much again)

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and thank you very much for the appreciation, it inspires me to work harder :)

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Hi..Write the equation as \(x^2-ax+(b-a)\) and try finding roots using the Shrredaracharya Formula..Tell me if this worked...

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By shrredaracharya's formula calculations became too much long....(i am still struggling with algebra and can't handle such large equations) well i am trying to solve it in this way:- By the vieta's formula:- \[p+q=a,pq=a+b\] Rearranging a bit gives:- \[p-a=-q ,q=\frac{a+b}{p}\] Putting the values of p-a ,a+b and q in give expression and manupulating a bit we get:- \[\frac{p+q-b}{(a+b)(q-b)}\] Putting the value of p+q:- \[\frac{a-b}{(a+b)(q-b)}\] But official answer of this question is 0 i have to somehow prove that a-b=0 .......i think i am doing a mistake somewhere :-( (Thanks for replying)

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Wow @Aman Sharma , you just said that \(2 = \dfrac{1}{2}\).

Shouldn't it be \(q = \dfrac{a+b}{p}\) in place of \(\dfrac{p}{a+b}\) ?

LOL, i can edit that as a moderator, but i want you to know where you actually missed it. So please try it again.

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