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Thread: Algebra problem

  1. #1
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    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 $\displaystyle{\frac{a+b}{13}}$?.
    Last edited by Sidd07; May 3rd 2018 at 01:28 AM.
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  2. #2
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    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 $x$.
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  3. #3
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    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 $\dfrac{s+t}{13}$
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  4. #4
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    Re: Algebra problem

    Quote Originally Posted by Sidd07 View Post
    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 $a$.
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  5. #5
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    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 $u$).
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