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Math Help - Sum & Product of the Roots of a Quadratic Equation

  1. #1
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    Sum & Product of the Roots of a Quadratic Equation

    Having a bit of trouble with this one. Can anyone help?

    Many thanks.

    Q.: (i)
    The roots of the equation 2x^2+6x+3=0 are \alpha & \beta. Show that \alpha^2+\beta^2=6. (ii) The roots of 2x^2+px+q=0 are 2\alpha+\beta & \alpha+2\beta. Find the value of p & the value of q.

    Attempt: (i) a = 2, b = 6, c = 3

    \alpha+\beta=\frac{-b}{a} => \frac{-6}{2} => -3

    \alpha\beta=\frac{c}{a} => \frac{3}{2}

    Sum of roots: \alpha^2+\beta^2 => (\alpha+\beta)^2-2\alpha\beta => (-3)^2-2(\frac{3}{2}) => 9-\frac{6}{2} => 9 - 3 => 6

    (ii) a = 2, b = p, c = q

    Sum of roots: 2\alpha+\beta+\alpha+2\beta => 3\alpha+3\beta => 3(\alpha+\beta) => \frac{-b}{a} => \frac{-p}{2} => 3(\alpha+\beta) => 3(\frac{-p}{2}) => \frac{-3p}{2}

    Product of roots: (2\alpha+\beta)(\alpha+2\beta) => 2\alpha^2+5\alpha\beta+2\beta^2 => 2(\alpha^2+\beta^2)+5\alpha\beta => 2((\alpha+\beta)^2-2\alpha\beta)+5\alpha\beta => 2(\alpha+\beta)^2+\alpha\beta => 2(\frac{-p}{2})^2+(\frac{c}{a}) => \frac{2p^2}{4}+\frac{q}{2} => \frac{p^2}{2}+\frac{q}{2} => \frac{p^2+q}{2}

    x^2-(sum of roots)x + (product of roots) => x^2-(\frac{-3p}{2})x+\frac{p^2+q}{2} => 2x^2+3px+p^2+q...

    Ans.: (From text book): p = 18, q = 39
    Last edited by GrigOrig99; June 20th 2012 at 11:14 AM.
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  2. #2
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    Re: Sum & Product of the Roots of a Quadratic Equation

    It seems that more informations are needed. Since if \alpha=0,\beta=1 then the roots will be 1,2. And hence p=-6,q=4.
    Also, if
    \alpha=1,\beta=-1 then the roots will be 1,5. And hence p=0,q=-2.
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  3. #3
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    Re: Sum & Product of the Roots of a Quadratic Equation

    I've made the changes to the question I think you're referring to. There was an earlier piece to the question, but I had already solved it and wasn't sure it was relevant.
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  4. #4
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    Re: Sum & Product of the Roots of a Quadratic Equation

    Yes, you do have to use the \alpha and \beta from part a).

    Sum of roots:
    3 \alpha + 3 \beta = -\frac{p}{2}. From part a), \alpha + \beta = -3, substituting yields -9 = -\frac{p}{2} \Rightarrow p = 18

    Product of roots:
    2 \alpha^2 + 5 \alpha \beta + 2 \beta^2 = \frac{q}{2} \Rightarrow 2(\alpha^2 + \beta^2) + 5 \alpha \beta = \frac{q}{2}. From part a), \alpha^2 + \beta^2 = 6, and \alpha \beta = \frac{3}{2}. Therefore,

    2(6) + 5(\frac{3}{2}) = \frac{q}{2} \Rightarrow \frac{39}{2} = \frac{q}{2} \Rightarrow q = 39.
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  5. #5
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    Re: Sum & Product of the Roots of a Quadratic Equation

    So, from the first part you have
    \alpha^2+\beta^2=6 and also \alpha+\beta=-3.
    We also need \alpha\beta which is from the first equation
    \alpha\beta=\frac{3}{2}.
    Now,
    p=-((\alpha+2\beta)+(2\alpha+\beta))=-6(\alpha+\beta)=-6*-3=18.

    q=(2\alpha+\beta)(\alpha+2\beta)=4(\alpha^2+\beta^  2)+10\alpha\beta=4*6+10\frac{3}{2}=39
    Last edited by Kmath; June 20th 2012 at 01:24 PM.
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