Let $W_0$, $W_1$, $W_2$, and $W_3$ be known real numbers.
I have to solve a biquartic equation:
\begin{equation} z^8+W_3z^6+W_2z^4+W_1z^2+W_0=0 \notag \end{equation}
Of course I could solve the quartic for $z^2$ and take the square roots of the solutions.
However, I'm looking for an analytical expression, and I guess that extracting the real and imaginary parts of the square roots of the huge expressions of the quartic solutions would be quite a pain.
Thus I tried to factor the biquartic as in $8$ System of equation , so I will only have to change some signs in the expressions of the quartics solutions.
Let $A$, $B$, $C$, $E$, $F$, and $G$ be complex numbers to be determined:
\begin{equation} (z^4+Az^3+(B+C)z^2+(AC+E)z+(F+G))(z^4-Az^3+(B-C)z^2+(AC-E)z+(F-G))=0 \notag \end{equation}
Expanding it and equating the coefficients of the two equations, I found:
\begin{equation} \begin{split} 2B&=W_3+A^2\\ 8F&=4W_2-(W_3+A^2)^2+8AE+4C^2\\ \end{split} \notag \end{equation}
Inserting these equations in the four remaining ones leads to:
\begin{equation} \begin{split} AC(W_3+A^2)-2AG-2CE&=0\\ AC(4W_2-(W_3+A^2)^2+8AE+4C^2)-8EG&=0\\ (W_3+A^2)(4W_2-(W_3+A^2)^2+8AE+4C^2)+8A^2C^2-8E^2-16CG&=8W_1\\ (4W_2-(W_3+A^2)^2+8AE+4C^2)^2-64G^2&=64W_0\\ \end{split} \notag \end{equation}
And this is where I'm stuck. Thus the question is: what are the expressions of $A$, $C$, $E$, and $G$ as functions of the $W_j$ coefficients?
Thanks in advance for your answers.