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# Quartic formula

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

(...)

## Contents

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
B =  −
 b 4a
, C =
 c 6a
, D =  −
 d 4a
, and
E =
 e a
, 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
 E = 0
, then at least one root is
 0
:
• if  E = 0, D   ≠   0
, then  0
is a simple root;
• if  E = 0, D = 0, C   ≠   0
, then  0
is a double root;
• if  E = 0, D = 0, C = 0, B   ≠   0
, then  0
is a triple root;
• if  E = 0, D = 0, C = 0, B = 0
, then  0
{\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}} 