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Thread: Help - equation

  1. #16
    TD!
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    Quote Originally Posted by ThePerfectHacker
    You can always use the quadtric formula!!!
    Yes, but that's rather 'nasty' as well, I don't like it
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  2. #17
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    Thanks TD! for help, finally solved!
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  3. #18
    TD!
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    Quote Originally Posted by DenMac21
    Thanks TD! for help, finally solved!
    You're welcome
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  4. #19
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    Quote Originally Posted by TD!
    I haven't really looked into the problem so I haven't been able to come up with a more elegant solution, but here's a go.

    We have a 4th degree polynomial without a 3rd degree term so a possible factorization would be two second degree polynomials with opposite lineair terms.

    $\displaystyle \begin{gathered}
    x^4 - 2ax^2 + x + a^2 - a = 0 \hfill \\
    \left( {x^2 + \alpha x + \beta } \right)\left( {x^2 - \alpha x + \gamma } \right) = 0 \hfill \\
    x^4 + \left( {\alpha ^2 - \beta - \gamma } \right)x^2 + \left( {\gamma - \beta } \right)x + \beta \gamma = 0 \hfill \\
    \end{gathered} $

    Identifying coefficients gives us a 3x3 system

    $\displaystyle
    \left\{ \begin{gathered}
    \alpha ^2 - \beta - \gamma = 2a \hfill \\
    \alpha \left( {\gamma - \beta } \right) = 1 \hfill \\
    \beta \gamma = a^2 - a \hfill \\
    \end{gathered} \right.
    $

    Solving yields $\displaystyle \alpha = 1 \wedge \beta = - a \wedge \gamma = 1 - a$

    So we have the following factorization

    $\displaystyle \left( {x^2 + x - a} \right)\left( {x^2 - x - a + 1} \right) = 0$

    Then it's just solving two quadratics, solutions are

    $\displaystyle x = \frac{{1 \pm \sqrt {4a - 3} }}
    {2} \vee x = - \frac{{1 \pm \sqrt {4a + 1} }}
    {2}$

    Of course, these solutions for x depend on a, so I don't see why a negative solution would be impossible - for certain values of the parameter a.
    Very well done.
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  5. #20
    TD!
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    Thanks ticbol
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