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Math Help - problem of quadratic equation with two variables

  1. #1
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    problem of quadratic equation with two variables

    If 3x2+2αxy+2y2+2ax-4y+1 can be resolved into two linear factors, prove that α is the root of the equation x2+ 4ax+2a2+6=0.

    please don't solve the problem
    just hint is expected.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: problem of quadratic equation with two variables

    Quote Originally Posted by sumedh View Post
    If 3x2+2αxy+2y2+2ax-4y+1 can be resolved into two linear factors, prove that α is the root of the equation x2+ 4ax+2a2+6=0.
    Hint : The discriminant of the second degree equation on x : 3x^2+(2\alpha y+2a)x+2y^2-4y+1=0 must be a perfect square.
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  3. #3
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    Re: problem of quadratic equation with two variables

    That is, if you meant "can be resolved into two linear factors" with rational coefficients.
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    Re: problem of quadratic equation with two variables

    Quote Originally Posted by HallsofIvy View Post
    That is, if you meant "can be resolved into two linear factors" with rational coefficients.
    Why?. The problem is equivalent to find where the given conic is degenerated. That is \Delta=\begin{vmatrix}{3}&{\alpha}&{a}\\{\alpha}&{  2}&{-2}\\{a}&{-2}&{1}\end{vmatrix}=0 i.e. \alpha^2+4a\alpha +2a^2+6=0 . Why do we need rational coefficients ?
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  5. #5
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    Re: problem of quadratic equation with two variables

    on solving i got
    α^2 y^2+a^2+2aαy =-6y^2-3+12y [α means alpha]
    α^2 y^2+a^2+2aαy -9 (y-3)^2
    after equating what should i do???

    could you please tell me the concept for solving this????
    Last edited by sumedh; August 17th 2011 at 08:40 AM. Reason: edit
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: problem of quadratic equation with two variables

    The discriminant of the equation p(x,y)=3x^2+(2\alpha y+2a)x+2y^2-4y+1=0 is \Delta=(2\alpha y+2a)^2-12(2y^2-4y+1) then,

    p(x,y)=3\left(x-\dfrac{-(2\alpha y+a)+\sqrt{\Delta}}{6}\right)\left(x-\dfrac{-(2\alpha y+a)-\sqrt{\Delta}}{6}\right)


    \Delta is a perfect square iff the discriminant \Delta_1 of \Delta=0 is 0 . Now, verify \Delta_1=0\Leftrightarrow \ldots \Leftrightarrow \alpha^2+4\alpha a+2a^2+6=0 . Equivalently, \alpha is a root of x^2+4ax+2a^2+6=0 .
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    Re: problem of quadratic equation with two variables

    thank you very much i got it
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