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Math Help - problem in finding quad. eqn. from the roots

  1. #1
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    problem in finding quad. eqn. from the roots

    In general, if α(alpha) and (beta) are roots of eqn. ax^2 +bx +c=0
    then for finding the equation whose roots are α+2 and +2 can be done by
    addition of roots (α+2++2=-b/a) and product of roots (α+2)(+2)=c/a
    By solving this we get ax^2 -(4a-b)x + (4a-2b+c)=0

    The problem is this that,by replacing x in place of (x-2)in the given equation
    ax^2 +bx +c=0 we get the same answer ax^2 -(4a-b)x + (4a-2b+c)=0
    but this method ( replacing x by (x-2) ..) is not mentioned anywhere
    i am not able to understand this method
    please help .
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  2. #2
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    Re: problem in finding quad. eqn. from the roots

    You can write every quadratic equation \displaystyle ax^2 + bx + c = 0 in terms of its roots, \displaystyle a\left(x - \alpha\right)\left(x- \beta\right) = 0.

    The roots of \displaystyle ax^2 + bx + c = 0 are \displaystyle \alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a} and \displaystyle \beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a}.

    So \displaystyle \alpha + 2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + 2 = \frac{4a - b + \sqrt{b^2 - 4ac}}{2a}

    and \displaystyle \beta + 2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a} + 2 = \frac{4a - b- \sqrt{b^2 - 4ac}}{2a}.

    So for a quadratic which has \displaystyle \alpha + 2, \beta + 2 as its roots, the equation is

    \displaystyle \begin{align*} &\phantom{=} A\left[x - \left(\alpha + 2\right)\right]\left[x - \left(\beta + 2\right)\right] \textrm{ where }A\textrm{ is some other parameter} \\ &= A\left[x - \left(\frac{4a - b + \sqrt{b^2 - 4ac}}{2a}\right)\right]\left[x - \left(\frac{4a - b - \sqrt{b^2 - 4ac}}{2a}\right)\right] \end{align*}

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  3. #3
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    Re: problem in finding quad. eqn. from the roots

    thank you very much i got it
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