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Thread: roots of quadratic equation

  1. #1
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    roots of quadratic equation

    Question: a)Prove that, if the sum of the reciprocals of the roots of the equation
    $\displaystyle ax^2+bx+c=0 $
    is 1 then $\displaystyle b+ c = 0. $
    If, in addition, one root of the equation is twice the other use the result of $\displaystyle 2b^2=9ac $ to find one set values of a, b, c. Solve the equation.


    My attempt


    a) $\displaystyle ax^2+bx+c=0 $
    $\displaystyle \alpha + \beta = -\frac{b}{a} $
    $\displaystyle \alpha \beta = \frac{c}{a} $
    $\displaystyle \frac{1}{\alpha} +\frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = -\frac{b}{c} = 1 $
    $\displaystyle b+c=0 $
    b)
    $\displaystyle \alpha_{1} = \frac{1}{\alpha} $
    $\displaystyle \beta_{1}=\frac{1}{\beta} =\frac{2}{\alpha} $
    sum of roots
    $\displaystyle sum =\frac{1}{\alpha} +\frac{2}{\alpha} = \frac{3}{\alpha} $
    product of roots
    $\displaystyle product =\frac{1}{\alpha} \times \frac{2}{\alpha} = \frac{2}{\alpha^2} $


    $\displaystyle \frac{3}{\alpha} =1 $


    $\displaystyle \alpha =3 $
    therefore


    $\displaystyle \beta_{1} = \frac{2}{3}$
    since

    sum of roots = $\displaystyle 1 $



    products of roots = $\displaystyle \frac{2}{\alpha^2} = \frac{2}{3^2} = \frac{2}{9} $


    $\displaystyle x^2 - (1)x + \frac{2}{9} = 0 $
    $\displaystyle 9x^2 - 9x + 2 =0 $
    from this; a=9, b = -9, c = 2 and x = 2/3 or 1/3


    the answer given by the book is a =2, b= -9 c= 9 x = 3, 3/2
    I know how to find $\displaystyle ax^2+bx+c=0 $ but i dont know how to use it to find the answer

    Please help me
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  2. #2
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    Re: roots of quadratic equation

    I think your error is at the start of part b) It says one root is twice the other, so $\alpha = 2\beta$, $\frac1\alpha = \frac1{2\beta}$, $\alpha + \beta = 3\beta = 1$, and $\alpha\beta = 2\beta^2$
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