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Math Help - proving help

  1. #1
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    proving help

    The roots of the equation x^2+px+1=0 are \alpha and \beta . If one of the roots of the equation x^2+qx+1=0 is \alpha^3 , prove that the other root is \beta^3 .

    I tried this ..

    For this equation x^2+px+1=0 ,

    \alpha+\beta=-p
    \alpha\beta=1

    For the second equation
    x^2+qx+1=0


    \alpha^3+\beta^3=(\alpha+\beta)(\alpha^2-\alpha\beta+\beta^2)
    =(-p)[(-p)^2-2]
    =-p^3+2p


    Obviously , i am wrong .. no idea where my mistake is , or i started wrongly .
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  2. #2
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     x^2+px+1 = (x-\alpha)(x-\beta) = x^2+x(-\beta-\alpha) + \beta \alpha

    Hence we can deduce that:

     \alpha \beta = 1
     \therefore \alpha = \frac{1}{\beta}

    Now let's look at the 2nd equation!
    x^2+qx+1 = x^2+qx+\alpha \beta

    But we also know that  \alpha^3 is a root (and hence  (x-\alpha^3) is a factor), and we need to find the other root, call it y (hence (x-y) is a factor!) Hence:

     x^2+qx+\alpha \beta = (x-\alpha^3)(x-y) = x^2+x(\alpha^3-y) + \alpha^3y

    From this we can deduce that:

    Hence  \alpha^3y = \alpha \beta

    And with a bit of manipulation, and then plugging in the result from the first equation that  \alpha = \frac{1}{\beta}, then we can find y

     y = \frac{\alpha \beta}{\alpha^3}

     y = \frac{ \beta}{\alpha^2}

     y = \frac{\beta}{(\frac{1}{\beta})^2}

     y = \frac{\beta}{\frac{1}{\beta^2}}

     y = \beta \times (\frac{1}{\beta^2})^{-1}

     y = \beta \times \beta^2

     y =  \beta^3
    Last edited by Mush; January 12th 2009 at 06:55 AM.
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  3. #3
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    Re :

    Lots thanks Mush , understooD !
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  4. #4
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    Quote Originally Posted by Mush View Post
     x^2+px+1 = (x-\alpha)(x-\beta) = x^2+x(-\beta-\alpha) + \beta \alpha

    Hence we can deduce that:

     \alpha \beta = 1
     \therefore \alpha = \frac{1}{\beta}

    Now let's look at the 2nd equation!
    x^2+qx+1 = x^2+qx+\alpha \beta

    But we also know that  \alpha^3 is a root (and hence  (x-\alpha^3) is a factor), and we need to find the other root, call it y (hence (x-y) is a factor!) Hence:

    \alpha^3-y) + \alpha^3y" alt=" x^2+qx+\alpha \beta = (x-\alpha^3)(x-y) = x^2+x(\alpha^3-y) + \alpha^3y" />

    should be -\alpha^3-y


    From this we can deduce that:

    Hence  \alpha^3y = \alpha \beta

    And with a bit of manipulation, and then plugging in the result from the first equation that  \alpha = \frac{1}{\beta}, then we can find y

     y = \frac{\alpha \beta}{\alpha^3}

     y = \frac{ \beta}{\alpha^2}

     y = \frac{\beta}{(\frac{1}{\beta})^2}

     y = \frac{\beta}{\frac{1}{\beta^2}}

     y = \beta \times (\frac{1}{\beta^2})^{-1}

     y = \beta \times \beta^2

     y = \beta^3
    should be
    [tex]-\alpha^3-y[tex]
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  5. #5
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    Quote Originally Posted by repcvt View Post
    should be
    [tex]-\alpha^3-y[tex]
    True, that's a typo. Luckily it's irrelevant to the workings of the solution.

    PS: html doesn't tend to work inside latex commands.
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