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Thread: proving help

  1. #1
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    proving help

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

    I tried this ..

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

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

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


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


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

    Hence we can deduce that:

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

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

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

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

    From this we can deduce that:

    Hence $\displaystyle \alpha^3y = \alpha \beta $

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

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

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

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

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

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

    $\displaystyle y = \beta \times \beta^2$

    $\displaystyle y = \beta^3 $
    Last edited by Mush; Jan 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
    $\displaystyle x^2+px+1 = (x-\alpha)(x-\beta) = x^2+x(-\beta-\alpha) + \beta \alpha $

    Hence we can deduce that:

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

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

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

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

    should be -\alpha^3-y


    From this we can deduce that:

    Hence $\displaystyle \alpha^3y = \alpha \beta $

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

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

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

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

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

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

    $\displaystyle y = \beta \times \beta^2$

    $\displaystyle 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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