Please give a hint how to solve this algebra problem:-
If p and q are roots of the equation:-
$x^{2}-a(x-1)+b$
Then find out the value of:-
$\frac{1}{p^{2}-ap}+\frac{1}{q^{2}-bq}+\frac{2}{a+b}$

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)

@Aman Sharma
–
I got that question is wrong, because it's really obvious from the first term that because $p$ and $q$ are roots of that equation, $p^2-ap +a+b=0$ and $q^2-aq+a+b=0$ . Hence the first term is simply $\dfrac{1}{p^2-ap} = \dfrac{1}{-(a+b)}$ . If you want there the zero, then only thing needed is making the 2nd term equal to$\dfrac{1}{-(a+b)}$ and that is done by replacing $b$ in second term by $a$.

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

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)

@Aditya Raut
–
:-) ..lol.. i typed it wrong sorry....but you can see that after this wrong step each step is correct(i am not still able to solve this problem please help)

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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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$p$ and $q$ are roots of that equation, $p^2-ap +a+b=0$ and $q^2-aq+a+b=0$ . Hence the first term is simply $\dfrac{1}{p^2-ap} = \dfrac{1}{-(a+b)}$ . If you want there the zero, then only thing needed is making the 2nd term equal to$\dfrac{1}{-(a+b)}$ and that is done by replacing $b$ in second term by $a$.

I got that question is wrong, because it's really obvious from the first term that becauseand 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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