Sample
StudyMath

# Cubic equation

##### Solution of cubic equation using Vieta's formulae. Created by user request
Timur2011-06-17 22:23:12

The canonical form of cubic equation is
$ax^3+bx^2+cx+d=0$

Vieta's formulae are used to solve equations as
$x^3+ax^2+bx+c=0$
thus, first step is to divide all coefficients by "a".

Here is calculator, description of calculation using Vieta's formulae are below

##### Cubic Equation

The only place I was able to find Vieta's formulae tailored for cubic equations is here

First we calculate
$Q=\frac{a^2-3b}{9}$
$R=\frac{2a^3-9ab+27c}{54}$

Then
$S=Q^3-R^2$

If S > 0, then
$\phi = \frac{1}{3}\arccos\left(\frac{R}{\sqrt{Q^3}}\right)$
and we have three real roots:

$x_1=-2\sqrt{Q}\cos(\phi)-\frac{a}{3}$
$x_2=-2\sqrt{Q}\cos\left(\phi+\frac{2}{3}\pi\right)-\frac{a}{3}$
$x_3=-2\sqrt{Q}\cos\left(\phi-\frac{2}{3}\pi\right)-\frac{a}{3}$

If S < 0, trigonometric functions are replaced with hyperbolic. Depending on sign of Q

Q > 0:
$\phi = \frac{1}{3}\,\operatorname{Arch}\left(\frac{|R|}{\sqrt{Q^3}}\right)$
$x_1=-2sgn(R)\sqrt{Q}\,\operatorname{ch}(\phi)-\frac{a}{3}$
(real root)
$x_{2,3}=sgn(R)\sqrt{Q}\,\operatorname{ch}(\phi)-\frac{a}{3} \pm i \sqrt{3}\sqrt{Q}\,\operatorname{sh}(\phi)$
(two complex roots)

Q < 0:

$\phi = \frac{1}{3}\,\operatorname{Arsh}\left(\frac{|R|}{\sqrt{|Q|^3}}\right)$
$x_1=-2sgn(R)\sqrt{|Q|}\, \operatorname{sh}(\phi)-\frac{a}{3}$
(real root)
$x_{2,3}=sgn(R)\sqrt{|Q|}\, \operatorname{sh}(\phi)-\frac{a}{3} \pm i \sqrt{3} \sqrt{|Q|}\,\operatorname{ch}(\phi)$
(two complex roots)

If S = 0, then it is singular equation and has only two roots:

$x_1=-2sgn(R)\sqrt{Q}-\frac{a}{3}=-2\sqrt[3]{R}-\frac{a}{3}$
$x_2=sgn(R)\sqrt{Q}-\frac{a}{3}=\sqrt[3]{R}-\frac{a}{3}$

View all calculators
(467 calculators in total. )