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.