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Math Help - 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 \frac{\alpha}{2} is a root of 2x^2+x+1. This means:
    2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+  1=0
    \frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0
    \frac{\alpha^2}{2}=-\frac{\alpha}{2}-1
    \alpha^2=-\alpha-2

    Use this, along with \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 \frac{\alpha}{2} is a root of 2x^2+x+1. This means:
    2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+  1=0
    \frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0
    \frac{\alpha^2}{2}=-\frac{\alpha}{2}-1
    \alpha^2=-\alpha-2

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

    --Kevin C.
    I've tried that before,
    \begin{array}{l}<br />
 \alpha ^2  =  - 2 - \alpha  \\ <br />
 \alpha ^4  = \left( { - 2 - \alpha } \right)^2  = \alpha ^2  + 4\alpha  + 4 \\ <br />
 \alpha ^2  + 4\alpha  + 4 = p\alpha  + q \\ <br />
 ????? \\ <br />
 \end{array}
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  4. #4
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    Good job getting that far.
    You still have an \alpha^2. Use again to get the answer
    Last edited by badgerigar; October 11th 2008 at 03:53 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 \alpha^2. Use again to get the answer
    oh ya! stupid me, I didn't realize I can still sub \alpha^2 again!
    Thanks.
    Last edited by NyRychVantel; October 11th 2008 at 04:39 AM.
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