1. ## Algebra problem

I am stuck at a question after trying a lot,plzz help me
Q. If the equation $\dfrac{x^2}{(1+x^2)^2}+\dfrac{ax}{1+x^2}+3=0$
has two real and distinct roots then a lies in the interval $(-\infty,a)\cup(b,\infty)$.
Then find the value of $\displaystyle{\frac{a+b}{13}}$?.

2. ## Re: Algebra problem

An open interval does not contain its end points, so are you saying that $b<a$? If not, then how does $a$ lie in the union of those two intervals? I could see that it is in $[a,b]$, but I don't think that is what you mean.

Anyway, to help you out, let $u = \dfrac{x}{1+x^2}$. Then you have $u^2+au+3 = 0$. Solve for $u$. Now, $ux^2-x+u = 0$, so solve for $x$.

3. ## Re: Algebra problem

Sorry there was a typo so posted a new thread...interval in which a will lie is $(-\infty,-s)\cup(t,\infty)$
find the value of $\dfrac{s+t}{13}$

4. ## Re: Algebra problem

Originally Posted by Sidd07
Sorry there was a typo so posted a new thread...interval in which a will lie is $(-\infty,-s)\cup(t,\infty)$
find the value of $\dfrac{s+t}{13}$
If $s,t$ are just arbitrary numbers, then that value would equal anything you like. If $s,t$ are the roots of the equation, that is a different story. I stand by my recommendation that you should solve for $u$, then solve for $x$. Once you have solutions for $x$ in terms of $a$, you can use that to determine $\dfrac{s+t}{13}$ in terms of $a$.

5. ## Re: Algebra problem

This really is a poorly worded question. Anyway, to give you a few more hints:

$\displaystyle u = \dfrac{-a \pm \sqrt{a^2-12}}{2}$

You know that $a^2 \ge 12$.

$\displaystyle x = \dfrac{1 \pm \sqrt{1-4u^2}}{2u}$

To get real solutions, you know that $4u^2 \le 1$ (if it were equal, then $x$ would have only a single solution in terms of $u$, which would imply $a^2 > 12$ to get two solutions for $u$).