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Thread: Quadratic equation

  1. #1
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    Quadratic equation



    The answers are p=3 q=2

    I don't know the steps, pls help me

    Thanks.
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  2. #2
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    Note that $\displaystyle \frac{\alpha}{2}$ is a root of $\displaystyle 2x^2+x+1$. This means:
    $\displaystyle 2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
    $\displaystyle \frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
    $\displaystyle \frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
    $\displaystyle \alpha^2=-\alpha-2$

    Use this, along with $\displaystyle \alpha^4=(\alpha^2)^2$, to get your result.

    --Kevin C.
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  3. #3
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    Quote Originally Posted by TwistedOne151 View Post
    Note that $\displaystyle \frac{\alpha}{2}$ is a root of $\displaystyle 2x^2+x+1$. This means:
    $\displaystyle 2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
    $\displaystyle \frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
    $\displaystyle \frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
    $\displaystyle \alpha^2=-\alpha-2$

    Use this, along with $\displaystyle \alpha^4=(\alpha^2)^2$, to get your result.

    --Kevin C.
    I've tried that before,
    $\displaystyle \begin{array}{l}
    \alpha ^2 = - 2 - \alpha \\
    \alpha ^4 = \left( { - 2 - \alpha } \right)^2 = \alpha ^2 + 4\alpha + 4 \\
    \alpha ^2 + 4\alpha + 4 = p\alpha + q \\
    ????? \\
    \end{array}$
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  4. #4
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    Good job getting that far.
    You still have an $\displaystyle \alpha^2$. Use again to get the answer
    Last edited by badgerigar; Oct 11th 2008 at 02:53 AM. Reason: typo
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  5. #5
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    Quote Originally Posted by badgerigar View Post
    Good job getting that far.
    You still have an $\displaystyle \alpha^2$. Use again to get the answer
    oh ya! stupid me, I didn't realize I can still sub $\displaystyle \alpha^2$ again!
    Thanks.
    Last edited by NyRychVantel; Oct 11th 2008 at 03:39 AM.
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