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

This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .

Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.

Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold

- bulleted - list

bulleted

list

1. numbered 2. list

numbered

list

Note: you must add a full line of space before and after lists for them to show up correctly

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)

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 :)

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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)

Log in to reply

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.

Log in to reply

Log in to reply

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

Log in to reply

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 ?

Log in to reply

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)

Log in to reply

$p$ 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 :)

Log in to reply