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Math Help - 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.

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

    Identifying coefficients gives us a 3x3 system

    <br />
\left\{ \begin{gathered}<br />
  \alpha ^2  - \beta  - \gamma  = 2a \hfill \\<br />
  \alpha \left( {\gamma  - \beta } \right) = 1 \hfill \\<br />
  \beta \gamma  = a^2  - a \hfill \\ <br />
\end{gathered}  \right.<br />

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

    So we have the following factorization

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

    Then it's just solving two quadratics, solutions are

    x = \frac{{1 \pm \sqrt {4a - 3} }}<br />
{2} \vee x =  - \frac{{1 \pm \sqrt {4a + 1} }}<br />
{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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