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# Quartic equation solution

This calculator produces quartic equation solution using resolvent cubic.

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The calculator below solves quartic equation with single variable. A quartic equation formula: $ax^4+bx^3+cx^2+dx+e=0$, where a,b,c,d,e - coefficients, and x is unknown. The equation solution gives four real or complex roots. The formulas to solve a quartic equation follow the calculator.

#### Quartic equation

Digits after the decimal point: 2
x1

x2

x3

x4

As a first step we divide all the quartic coefficients by a to obtain the equation:
$x^4+a_3x^3+a_2x^2+a_1x+a_0=0$
Next we solve the resolvent cubic:
$u^3-a_2u^2+(a_1a_3-4a_0)u-(a_1^2+a_0a_3^2-4a_0a_2)=0$
We can solve it with the method described here: Cubic equation.
A single real root u1 of this equation we'll use further for quadratic equation roots finding. If the cubic resolvent has more than one real roots, we must choose a single u1 root in the way that gives real p and q coefficients in the formulas:
$p_{1,2}=\frac{a_3}{2}\pm\sqrt{\frac{a_3^2}{4}+u_1-a_2}$
$q_{1,2}=\frac{u_1}{2}\pm\sqrt{\frac{u_1^2}{4}-a_0}$
Then we substitute p1, p2,q1,q2, in quadratic equations in the right side of the following equation:
$x^4+a_3x^3+a_2x^2+a_1x+a_0=(x^2+p_1x+q_1)(x^2+p_2x+q_2)$1

Four roots of the two quadratic equations are the roots of original equation if the pi and qi signs are chosen to satisfy the following conditions:

# Condition
1 $p_1+p_2=a_3$
2 $p_1p_2+q_1+q_2=a_2$
3 $p_1q_2+p_2q_1=a_1$
4 $q_1q_2=a_0$

Actually we may check only third condition, and if it is not satisfied — just swap q1 and q2.
The solution can be verified with the calculator: Complex polynomial value calculation

1. M. Abramovitz and I. Stegun Handbook of Mathematical Functions With Formulas, Graphs and Mathematical Tables, 10th printing, Dec 1972, pp.17-18

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