Quartic equation

In mathematics, a quartic equation is one which can be expressed as a quartic function equaling zero. The general form of a quartic equation is

Not to be confused with Quadratic equation.

where a ≠ 0.

The quartic is the highest order polynomial equation that can be solved by radicals in the general case (i.e., one in which the coefficients can take any value).

History[edit]

Lodovico Ferrari is attributed with the discovery of the solution to the quartic in 1540, but since this solution, like all algebraic solutions of the quartic, requires the solution of a cubic to be found, it could not be published immediately.^[1] The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna (1545).

The proof that this was the highest order general polynomial for which such solutions could be found was first given in the Abel–Ruffini theorem in 1824, proving that all attempts at solving the higher order polynomials would be futile. The notes left by Évariste Galois before his death in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.^[2]

M4. The general quartic equation corresponds to this case whenever $p=q=r=0$ , so the four roots of this equation are given as follows: $x_{1}=x_{2}=x_{3}=x_{4}=-{\frac {b}{4a}}$ .

M3. The general quartic equation corresponds to this case whenever $p^{2}=-12r>0$ and $27q^{2}=-8p^{3}>0$ , so the four roots of this equation are given as follows: $x_{1}=x_{2}=x_{3}={\sqrt {-{\frac {p}{6}}}}-{\frac {b}{4a}}$ and $x_{4}=-{\sqrt {-{\frac {3p}{2}}}}-{\frac {b}{4a}}$ , whether $q>0$ ; otherwise, $x_{1}=x_{2}=x_{3}=-{\sqrt {-{\frac {p}{6}}}}-{\frac {b}{4a}}$ and $x_{4}={\sqrt {-{\frac {3p}{2}}}}-{\frac {b}{4a}}$ .

DM2. The general quartic equation corresponds to this case whenever $p^{2}=4r>0=q$ , so the four roots of this equation are given as follows: $x_{1}=x_{3}={\sqrt {-{\frac {p}{2}}}}-{\frac {b}{4a}}$ and $x_{2}=x_{4}=-{\sqrt {-{\frac {p}{2}}}}-{\frac {b}{4a}}$ .

Biquadratic SM2. The general quartic equation corresponds to this subcase of the SM2 equations whenever $p\neq q=r=0$ , so the four roots of this equation are given as follows: $x_{1}=x_{2}=-{\frac {b}{4a}}$ , $x_{3}={\sqrt {-p}}-{\frac {b}{4a}}$ and $x_{4}=-{\sqrt {-p}}-{\frac {b}{4a}}$ .

Non-Biquadratic SM2. The general quartic equation corresponds to this subcase of the SM2 equations whenever $(p^{2}+12r)^{3}=[p(p^{2}-36r)+{\frac {27}{2}}q^{2}]^{2}>0\neq {q}$ , so the four roots of this equation are given by the following formula: $x={\frac {1}{2}}\left[\xi {\sqrt {s_{1}}}\pm {\sqrt {2{\biggl (}s_{2}-{\frac {\xi q}{\sqrt {s_{1}}}}{\biggr )}}}\right]-{\frac {b}{4a}}$ , where $s_{1}={\frac {9q^{2}-32pr}{p^{2}+12r}}>0$ , $s_{2}=-{\frac {2p(p^{2}-4r)+9q^{2}}{2(p^{2}+12r)}}\neq 0$ and $\xi =\pm 1$ .

[4]

Alternative methods[edit]

Quick and memorable solution from first principles[edit]

Most textbook solutions of the quartic equation require a substitution that is hard to memorize. Here is an approach that makes it easy to understand. The job is done if we can factor the quartic equation into a product of two quadratics. Let

Quartic equation

History[edit]

[4]

Alternative methods[edit]

Quick and memorable solution from first principles[edit]

Linear equation

Quadratic equation

Cubic equation

Quintic equation

Polynomial

Newton's method

Ferrari's achievement

Quartic formula as four single equations

Calculator for solving Quartics