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.
\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
 = 1 \hfill \\<br />
\beta \gamma = a^2 - a \hfill \\ <br />
\end{gathered} \right.<br />
)
Solving yields
So we have the following factorization
\left( {x^2 - x - a + 1} \right) = 0)
Then it's just solving two quadratics, solutions are
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.
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About LinkBacksOriginally Posted by ThePerfectHacker
You can always use the quadtric formula!!!
