# Cubic and Quartic Equations

### Cubic Equation x3 + a1x2 + a2x + a3 = 0

Let $Q = \frac{3a_2 - a_1^2}{9}$

$R = \frac{9a_1a_2 - 27a_3 - 2a_1^3}{54}$

$S = \sqrt[3]{R+\sqrt{Q^3+R^2}}$

$T = \sqrt[3]{R-\sqrt{Q^3+R^2}}$

Solutions:
$\begin{cases}x_1 = S + T - \frac{1}{3}a_1 \\ x_2 = -\frac{1}{2}(S+T) - \frac{1}{3}a_1+\frac{1}{2}i\sqrt{3}(S-T) \\ x_3 = -\frac{1}{2}(S + T) - \frac{1}{3}a_1-\frac{1}{2}i\sqrt{3}(S-T) \end{cases}$

If a1, a2, a3 are real and if D = Q3 + R2 is the discriminant, then
(i)     one root is real and two complex conjugate if D > 0
(ii)     all roots are real and at least two are equal if D = 0
(iii)     all roots are real and unequal if D < 0.
If D < 0, computation is simplyfied by use of trigonometry.

Solutions if D < 0 :
$\begin{cases} x_1 = 2 \sqrt{-Q} \cos \left( \frac{1}{3} \theta \right) \\ x_2 = 2 \sqrt{-Q} \cos \left( \frac{1}{3} \theta + 120^{\circ} \right) \\ x_3 = 2 \sqrt{-Q} \cos \left( \frac{1}{3} \theta + 240^{\circ} \right) \end{cases}$

where $\cos \theta = -\frac{R}{\sqrt{-Q^3}}$

x1 + x2 + x3 = -a1,      x1x2 + x2x3 + x3x1 = a2,      x1x2x3 = - a3
where x1, x2 , x3 are the three roots.

### Quartic equation x4 + a1x3 + a2x2 + a3x + a4 = 0.

Let $y_1$ be a real root of the cubic equation
(1)   $y^3 - a_2y^2 + (a_1a_3 - 4a_4)y + (4a_2a_4 - a_3^2 - a_1^2a_4) = 0$

Solutions:    The roots of
$z^2+\frac{1}{2}(a_1 \pm \sqrt{a_1^2-4a_2+4y_1})z+\frac{1}{2}(y_1 \pm \sqrt{y_1^2 - 4a_4})=0$

If all roots of (1) are real, computation is simplified by using that particular real root which produces all real coefficients in the quadratic equation.

where x1,x2,x3,x4 are the four roots.

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