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help!!!!!!!!!!

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:- \[\frac{1}{p^{2}-ap}+\frac{1}{q^{2}-bq}+\frac{2}{a+b}\]

Note by Aman Sharma
2 years, 7 months ago

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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... Krishna Ar · 2 years, 7 months ago

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@Krishna Ar 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) Aman Sharma · 2 years, 7 months ago

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@Aman Sharma 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. Aditya Raut · 2 years, 7 months ago

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@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) Aman Sharma · 2 years, 7 months ago

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@Michael Mendrin @Satvik Golechha @Krishna Ar @Aditya Raut please help......and sorry to disturb you all Aman Sharma · 2 years, 7 months ago

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@Aman Sharma 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 ? Aditya Raut · 2 years, 7 months ago

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@Aditya Raut 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 · 2 years, 7 months ago

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@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 :) Aditya Raut · 2 years, 7 months ago

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