help!!!!!!!!!!

Please give a hint how to solve this algebra problem:- If p and q are roots of the equation:- x2a(x1)+bx^{2}-a(x-1)+b Then find out the value of:- 1p2ap+1q2bq+2a+b\frac{1}{p^{2}-ap}+\frac{1}{q^{2}-bq}+\frac{2}{a+b}

Note by Aman Sharma
4 years, 10 months ago

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

Aman Sharma - 4 years, 10 months ago

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The question is wrong, it should be 1q2aq\dfrac{1}{q^2-aq}

Then,

1p2ap+1q2aq+2a+b=1p(pa)+1q(qa)+2pq=1pq+1pq+2pq=0\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 - 4 years, 10 months ago

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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)

Aman Sharma - 4 years, 10 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 pp and qq are roots of that equation, p2ap+a+b=0p^2-ap +a+b=0 and q2aq+a+b=0q^2-aq+a+b=0 . Hence the first term is simply 1p2ap=1(a+b)\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 to1(a+b)\dfrac{1}{-(a+b)} and that is done by replacing bb in second term by aa.

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

Aditya Raut - 4 years, 10 months ago

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Hi..Write the equation as x2ax+(ba)x^2-ax+(b-a) and try finding roots using the Shrredaracharya Formula..Tell me if this worked...

Krishna Ar - 4 years, 10 months ago

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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+bp+q=a,pq=a+b Rearranging a bit gives:- pa=q,q=a+bpp-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:- p+qb(a+b)(qb)\frac{p+q-b}{(a+b)(q-b)} Putting the value of p+q:- ab(a+b)(qb)\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 - 4 years, 10 months ago

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Wow @Aman Sharma , you just said that 2=122 = \dfrac{1}{2}.

Shouldn't it be q=a+bpq = \dfrac{a+b}{p} in place of pa+b\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 - 4 years, 10 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 - 4 years, 10 months ago

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