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The **quartic formula** gives the roots of any quartic equation

- ${\begin{array}{l}\displaystyle {ax^{4}+bx^{3}+cx^{2}+dx+e=0,\quad a\neq 0.}\end{array}}$

The four (distinct or not) roots are given by

- $x=\ldots ,\,$

where

- (...)

Equivalently, the four (distinct or not) roots of

- ${\begin{array}{l}\displaystyle {x^{4}+{\tfrac {b}{a}}\,x^{3}+{\tfrac {c}{a}}\,x^{2}+{\tfrac {d}{a}}x+{\tfrac {e}{a}}=0=x^{4}-4Bx^{3}+6Cx^{2}-4Dx+E,\quad a\neq 0,}\end{array}}$

with

, and

, may be written as

- $x=\ldots ,\,$

where

- (...)

## Vieta's formulas for the quartic

Vieta's formulas for the quartic

- $ax^{4}+bx^{3}+cx^{2}+dx+e=0=(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4}),\quad a\neq 0,\,$

gives a system of four equations in four variables (which are the four roots)

- ${\begin{array}{l}\displaystyle {\begin{aligned}x_{1}+x_{2}+x_{3}+x_{4}&=-{\frac {b}{a}}=4B,\\x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{1}\cdot x_{4}+x_{2}\cdot x_{3}+x_{2}\cdot x_{4}+x_{3}\cdot x_{4}&=+{\frac {c}{a}}=6C,\\x_{1}\cdot x_{2}\cdot x_{3}+x_{1}\cdot x_{2}\cdot x_{4}+x_{1}\cdot x_{3}\cdot x_{4}+x_{2}\cdot x_{3}\cdot x_{4}&=-{\frac {d}{a}}=4D,\\x_{1}\cdot x_{2}\cdot x_{3}\cdot x_{4}&=+{\frac {e}{a}}=E.\end{aligned}}\end{array}}$

If

, then at least one root is

:

- if , then is a simple root;
- if , then is a double root;
- if
*E* = 0, *D* = 0, *C* = 0, *B* ≠ 0 |

, then is a triple root;
- if
*E* = 0, *D* = 0, *C* = 0, *B* = 0 |

, then is a quadruple root.

If

, then dividing the fourth equation into the third equation, one obtains a formula for the harmonic sum of the roots, and dividing the fourth equation into the second equation, one obtains a formula for the harmonic sum of the products of root pairs

- ${\begin{array}{l}\displaystyle {\begin{aligned}{\frac {1}{x_{1}}}+{\frac {1}{x_{2}}}+{\frac {1}{x_{3}}}+{\frac {1}{x_{3}}}&=-{\frac {d}{e}}=4{\frac {D}{E}},\\{\frac {1}{x_{1}\cdot x_{2}}}+{\frac {1}{x_{1}\cdot x_{3}}}+{\frac {1}{x_{1}\cdot x_{4}}}+{\frac {1}{x_{2}\cdot x_{3}}}+{\frac {1}{x_{2}\cdot x_{4}}}+{\frac {1}{x_{3}\cdot x_{4}}}&=+{\frac {c}{e}}=6{\frac {C}{E}}.\end{aligned}}\end{array}}$

## See also