# Help - equation

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• January 10th 2006, 12:18 PM
TD!
Quote:

Originally Posted by ThePerfectHacker
You can always use the quadtric formula!!! :D

Yes, but that's rather 'nasty' as well, I don't like it ;)
• January 10th 2006, 01:29 PM
DenMac21
Thanks TD! for help, finally solved!
• January 10th 2006, 01:33 PM
TD!
Quote:

Originally Posted by DenMac21
Thanks TD! for help, finally solved!

You're welcome :)
• January 10th 2006, 11:52 PM
ticbol
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.

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

Identifying coefficients gives us a 3x3 system

$
\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 $\alpha = 1 \wedge \beta = - a \wedge \gamma = 1 - a$

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

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

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.
• January 11th 2006, 06:03 AM
TD!
Thanks ticbol :)
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