cubic formula
CubicFormula
2002-01-06 06:25:08
2005-03-05 04:47:12
Theorem
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The three roots $r_1, r_2, r_3$ of a cubic polynomial equation $x^3 + ax^2 + bx + c = 0$ are given by
\begin{eqnarray*}
r_1 & = & -\frac{a}{3} + \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
& & {} + \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
r_2 & = & -\frac{a}{3} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
& & {} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
r_3 & = & -\frac{a}{3} + \frac{-1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c + \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3} \\
& & {} - \frac{1+i\sqrt{3}}{2} \left(\frac{-2a^3 + 9ab - 27c - \sqrt{(2a^3-9ab+27c)^2 + 4(-a^2+3b)^3}}{54}\right)^{1/3}
\end{eqnarray*}