Cardano's Formula (with some help from Tartaglia)
(from mathworld.wolfram.com)
$
z^{3}+a_{2}z^{2}+a_{1}z+a_{0}=0
$
(1)
Put $z \equiv x-a_{2}/3$
and
$
\begin{eqnarray}
Q &\equiv& \frac{ 3a_{1}-a_{2}^{2}}{9} & \;\;\;\; (2)
\\
R &\equiv& \frac{ 9a_{1}a_{2}-27a_{0}-2a_{2}^{3}}{54}
& \;\;\;\; (3)
\end{eqnarray}
$
Then compute
$
\begin{eqnarray}
D &\equiv& Q^{3}+R^{2}
& \;\;\;\; (4)
\\
S &\equiv& (R+\sqrt{D})^{1/3}
& \;\;\;\; (5)
\\
T &\equiv& (R-\sqrt{D})^{1/3}
& \;\;\;\; (6)
\end{eqnarray}
$
Note:
(a) as the cubic root of a real, negative number, use
the real, negative result;
(b) if the argument in $S$ is complex, take the phase angle in
$T$ as the negative of the ph. a. in $S$, such that
$T=S^{*}$.
Finally,
$
\begin{eqnarray}
z_{1}&=&-\frac{1}{3}a_{2}+(S+T)
& \;\;\;\; (7)
\\
z_{2}&=&-\frac{1}{3}a_{2}-\frac{1}{2}(S+T)+\frac{1}{2}i\sqrt{3}(S-T)
& \;\;\;\; (8)
\\
z_{3}&=&-\frac{1}{3}a_{2}-\frac{1}{2}(S+T)-\frac{1}{2}i\sqrt{3}(S-T)
& \;\;\;\; (9)
\\
\end{eqnarray}
$
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