# 四次方程

${\displaystyle y=7x^{4}+9x^{3}-24x^{2}-28x+48}$的圖形
${\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0\,}$ 其中 ${\displaystyle a\neq 0\,}$

## 特殊情况

### 名义上的四次方程

${\displaystyle ax^{3}+bx^{2}+cx+d=0\,}$

### 双二次方程

${\displaystyle ax^{4}+cx^{2}+e=0\,\!}$

${\displaystyle az^{2}+cz+e=0\,\!}$

${\displaystyle z={{-c\pm {\sqrt {c^{2}-4ae}}} \over {2a}}\,\!}$

${\displaystyle x_{1}=+{\sqrt {z_{1}}}\,\!}$
${\displaystyle x_{2}=-{\sqrt {z_{1}}}\,\!}$
${\displaystyle x_{3}=+{\sqrt {z_{2}}}\,\!}$
${\displaystyle x_{4}=-{\sqrt {z_{2}}}\,\!}$

## 费拉里的方法

### 转变成减少次数的四次方程

${\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0\qquad \qquad (1')}$
${\displaystyle x^{4}+{b \over a}x^{3}+{c \over a}x^{2}+{d \over a}x+{e \over a}=0.}$

${\displaystyle x=u-{b \over 4a}}$ .

${\displaystyle u^{4}+\left({-3b^{2} \over 8a^{2}}+{c \over a}\right)u^{2}+\left({b^{3} \over 8a^{3}}-{bc \over 2a^{2}}+{d \over a}\right)u+\left({-3b^{4} \over 256a^{4}}+{b^{2}c \over 16a^{3}}-{bd \over 4a^{2}}+{e \over a}\right)=0.}$

${\displaystyle \alpha ={-3b^{2} \over 8a^{2}}+{c \over a},}$
${\displaystyle \beta ={b^{3} \over 8a^{3}}-{bc \over 2a^{2}}+{d \over a},}$
${\displaystyle \gamma ={-3b^{4} \over 256a^{4}}+{b^{2}c \over 16a^{3}}-{bd \over 4a^{2}}+{e \over a}.}$

${\displaystyle u^{4}+\alpha u^{2}+\beta u+\gamma =0\qquad \qquad (1)}$

### 费拉里的解法

${\displaystyle (u^{2}+\alpha )^{2}-u^{4}-2\alpha u^{2}=\alpha ^{2}\,}$

${\displaystyle (u^{2}+\alpha )^{2}+\beta u+\gamma =\alpha u^{2}+\alpha ^{2}.\qquad \qquad (2)}$

${\displaystyle {\begin{matrix}(u^{2}+\alpha +y)^{2}-(u^{2}+\alpha )^{2}&=&2y(u^{2}+\alpha )+y^{2}\ \ \\&=&2yu^{2}+2y\alpha +y^{2},\end{matrix}}}$

${\displaystyle 0=(\alpha +2y)u^{2}-2yu^{2}-\alpha u^{2}\,}$ 两式相加，可得
${\displaystyle (u^{2}+\alpha +y)^{2}-(u^{2}+\alpha )^{2}=(\alpha +2y)u^{2}-\alpha u^{2}+2y\alpha +y^{2}\qquad \qquad }$ ${\displaystyle y\,}$ 的插入）

${\displaystyle (u^{2}+\alpha +y)^{2}+\beta u+\gamma =(\alpha +2y)u^{2}+(2y\alpha +y^{2}+\alpha ^{2})\,}$

${\displaystyle (u^{2}+\alpha +y)^{2}=(\alpha +2y)u^{2}-\beta u+(y^{2}+2y\alpha +\alpha ^{2}-\gamma ).\qquad \qquad (3)}$

${\displaystyle (su+t)^{2}=(s^{2})u^{2}+(2st)u+(t^{2})\,}$

${\displaystyle (2st)^{2}-4(s^{2})(t^{2})=0\,}$

${\displaystyle (-\beta )^{2}-4(2y+\alpha )(y^{2}+2y\alpha +\alpha ^{2}-\gamma )=0\,}$

${\displaystyle \beta ^{2}-4[2y^{3}+5\alpha y^{2}+(4\alpha ^{2}-2\gamma )y+(\alpha ^{3}-\alpha \gamma )]=0\,}$ 两边除以${\displaystyle 4\,}$ ，再把${\displaystyle -{\frac {\beta ^{2}}{4}}\,}$ 移动到右边，
${\displaystyle 2y^{3}+5\alpha y^{2}+(4\alpha ^{2}-2\gamma )y+\left(\alpha ^{3}-\alpha \gamma -{\beta ^{2} \over 4}\right)=0\qquad \qquad }$

${\displaystyle y^{3}+{5 \over 2}\alpha y^{2}+(2\alpha ^{2}-\gamma )y+\left({\alpha ^{3} \over 2}-{\alpha \gamma \over 2}-{\beta ^{2} \over 8}\right)=0.\qquad \qquad (4)}$

#### 转化嵌套的三次方程为降低次数的三次方程

${\displaystyle y=v-{5 \over 6}\alpha .}$

${\displaystyle \left(v-{5 \over 6}\alpha \right)^{3}+{5 \over 2}\alpha \left(v-{5 \over 6}\alpha \right)^{2}+(2\alpha ^{2}-\gamma )\left(v-{5 \over 6}\alpha \right)+\left({\alpha ^{3} \over 2}-{\alpha \gamma \over 2}-{\beta ^{2} \over 8}\right)=0.}$

${\displaystyle \left(v^{3}-{5 \over 2}\alpha v^{2}+{25 \over 12}\alpha ^{2}v-{125 \over 216}\alpha ^{3}\right)+{5 \over 2}\alpha \left(v^{2}-{5 \over 3}\alpha v+{25 \over 36}\alpha ^{2}\right)+(2\alpha ^{2}-\gamma )v-{5 \over 6}\alpha (2\alpha ^{2}-\gamma )+\left({\alpha ^{3} \over 2}-{\alpha \gamma \over 2}-{\beta ^{2} \over 8}\right)=0.}$

${\displaystyle v^{3}+\left(-{\alpha ^{2} \over 12}-\gamma \right)v+\left(-{\alpha ^{3} \over 108}+{\alpha \gamma \over 3}-{\beta ^{2} \over 8}\right)=0.}$

${\displaystyle P=-{\alpha ^{2} \over 12}-\gamma ,}$
${\displaystyle Q=-{\alpha ^{3} \over 108}+{\alpha \gamma \over 3}-{\beta ^{2} \over 8}.}$

${\displaystyle v^{3}+Pv+Q=0.\qquad \qquad (5)}$

#### 解嵌套的降低次数的三次方程

${\displaystyle U={\sqrt[{3}]{{-Q \over 2}\pm {\sqrt {{Q^{2} \over 4}+{P^{3} \over 27}}}}}}$
（由三次方程
${\displaystyle v=U-{P \over 3U}}$

${\displaystyle y=-{5 \over 6}\alpha -{P \over 3U}+U\qquad \qquad (6)}$

#### 配成完全平方项

${\displaystyle y\,}$ 的值已由${\displaystyle \left(6\right)\,}$ 式给定，现在知道等式${\displaystyle \left(3\right)\,}$ 的右边是完全平方的形式

${\displaystyle s^{2}u^{2}+2stu+t^{2}=\left({\sqrt {s^{2}}}u+{2st \over 2{\sqrt {s^{2}}}}\right)^{2}}$

${\displaystyle (\alpha +2y)u^{2}+(-\beta )u+(y^{2}+2y\alpha +\alpha ^{2}-\gamma )=\left[{\sqrt {\alpha +2y}}u+{(-\beta ) \over 2{\sqrt {\alpha +2y}}}\right]^{2}}$ .

${\displaystyle (u^{2}+\alpha +y)^{2}=\left({\sqrt {\alpha +2y}}u-{\beta \over 2{\sqrt {\alpha +2y}}}\right)^{2}\qquad \qquad (7)}$ .

${\displaystyle (u^{2}+\alpha +y)=\pm \left({\sqrt {\alpha +2y}}u-{\beta \over 2{\sqrt {\alpha +2y}}}\right)\qquad \qquad (7')}$ .

${\displaystyle u\,}$ 合并同类项，得

${\displaystyle u^{2}+\left(\mp _{s}{\sqrt {\alpha +2y}}\right)u+\left(\alpha +y\pm _{s}{\beta \over 2{\sqrt {\alpha +2y}}}\right)=0\qquad \qquad (8)}$ .

${\displaystyle u={\pm _{s}{\sqrt {\alpha +2y}}\pm _{t}{\sqrt {(\alpha +2y)-4(\alpha +y\pm _{s}{\beta \over 2{\sqrt {\alpha +2y}}})}} \over 2}.}$

${\displaystyle u={\pm _{s}{\sqrt {\alpha +2y}}\pm _{t}{\sqrt {-\left(3\alpha +2y\pm _{s}{2\beta \over {\sqrt {\alpha +2y}}}\right)}} \over 2}.}$

${\displaystyle x=-{b \over 4a}+{\pm _{s}{\sqrt {\alpha +2y}}\pm _{t}{\sqrt {-\left(3\alpha +2y\pm _{s}{2\beta \over {\sqrt {\alpha +2y}}}\right)}} \over 2}.\qquad \qquad (8')}$

#### 费拉里方法的概要

${\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e=0\,}$

${\displaystyle \alpha =-{3b^{2} \over 8a^{2}}+{c \over a},}$
${\displaystyle \beta ={b^{3} \over 8a^{3}}-{bc \over 2a^{2}}+{d \over a},}$
${\displaystyle \gamma ={-3b^{4} \over 256a^{4}}+{b^{2}c \over 16a^{3}}-{bd \over 4a^{2}}+{e \over a},}$
${\displaystyle \beta =0\,}$ ，求解 ${\displaystyle u^{4}+\alpha u^{2}+\gamma =0\,}$  并代入 ${\displaystyle x=u-{b \over 4a}}$ ，求得根
${\displaystyle x=-{b \over 4a}\pm _{s}{\sqrt {-\alpha \pm _{t}{\sqrt {\alpha ^{2}-4\gamma }} \over 2}},\qquad \beta =0}$ .
${\displaystyle P=-{\alpha ^{2} \over 12}-\gamma ,}$
${\displaystyle Q=-{\alpha ^{3} \over 108}+{\alpha \gamma \over 3}-{\beta ^{2} \over 8},}$
${\displaystyle R={Q \over 2}\pm {\sqrt {{Q^{2} \over 4}+{P^{3} \over 27}}},}$ （平方根任一正负号均可）
${\displaystyle U={\sqrt[{3}]{R}},}$ （有三个复根，任一个均可）
${\displaystyle y={\displaystyle -{5 \over 6}\alpha +{\begin{cases}U=0&\to -{\sqrt[{3}]{Q}}\\U\neq 0,&\to U-{P \over 3U},\end{cases}}\quad \quad \quad },}$
${\displaystyle x=-{b \over 4a}+{\pm _{s}{\sqrt {\alpha +2y}}\pm _{t}{\sqrt {-\left(3\alpha +2y\pm _{s}{2\beta \over {\sqrt {\alpha +2y}}}\right)}} \over 2}.}$

${\displaystyle x^{4}+6x^{2}-60x+36=0\,}$

### 笛卡兒方法

${\displaystyle (x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})=0\,}$

${\displaystyle (x-x_{1})(x-x_{2})=0\qquad \qquad (9)}$

${\displaystyle (x-x_{3})(x-x_{4})=0.\qquad \qquad (10)}$

${\displaystyle x_{2}=x_{1}^{\star }}$

${\displaystyle {\begin{matrix}(x-x_{1})(x-x_{2})&=&x^{2}-(x_{1}+x_{1}^{\star })x+x_{1}x_{1}^{\star }\qquad \qquad \qquad \quad \\&=&x^{2}-2\,\mathrm {Re} (x_{1})x+[\mathrm {Re} (x_{1})]^{2}+[\mathrm {Im} (x_{1})]^{2}.\end{matrix}}}$

${\displaystyle a=-2\,\mathrm {Re} (x_{1}),}$
${\displaystyle b=[\mathrm {Re} (x_{1})]^{2}+[\mathrm {Im} (x_{1})]^{2}\,}$

${\displaystyle x^{2}+ax+b=0.\qquad \qquad (11)}$

${\displaystyle x^{2}+wx+v=0.\qquad \qquad (12)}$

${\displaystyle x^{4}+(a+w)x^{3}+(b+wa+v)x^{2}+(wb+va)x+vb=0.\qquad \qquad (13)}$

${\displaystyle a+w={B \over A},}$
${\displaystyle b+wa+v={C \over A},}$
${\displaystyle wb+va={D \over A},}$

${\displaystyle vb={E \over A}.}$

${\displaystyle w={B \over A}-a={B \over A}+2\mathrm {Re} (x_{1}),}$
${\displaystyle v={E \over Ab}={E \over A\left([\mathrm {Re} (x_{1})]^{2}+[\mathrm {Im} (x_{1})]^{2}\right)}.}$

${\displaystyle x_{3}={-w+{\sqrt {w^{2}-4v}} \over 2},}$
${\displaystyle x_{4}={-w-{\sqrt {w^{2}-4v}} \over 2}.}$

## 其它方法

### 求根公式

${\displaystyle {x_{1}=-{\frac {b}{4a}}+{\frac {1}{2}}{\sqrt {\left({\frac {b}{2a}}\right)^{2}-{\frac {2c}{3a}}+{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}-{\frac {1}{2}}{\sqrt {{\frac {b^{2}}{2a^{2}}}-{\frac {4c}{3a}}-{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}-{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2}}a}}+{\frac {-b^{3}+4abc-8a^{2}d}{4a^{3}{\sqrt {\left({\dfrac {b}{2a}}\right)^{2}-{\dfrac {2c}{3a}}+{\dfrac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\dfrac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}}}}}}}$

${\displaystyle {x_{2}=-{\frac {b}{4a}}+{\frac {1}{2}}{\sqrt {\left({\frac {b}{2a}}\right)^{2}-{\frac {2c}{3a}}+{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}+{\frac {1}{2}}{\sqrt {{\frac {b^{2}}{2a^{2}}}-{\frac {4c}{3a}}-{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}-{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}-{\frac {b^{3}-4abc+8a^{2}d}{4a^{3}{\sqrt {\left({\dfrac {b}{2a}}\right)^{2}-{\dfrac {2c}{3a}}+{\dfrac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\dfrac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}}}}}}}$

${\displaystyle {x_{3}=-{\frac {b}{4a}}-{\frac {1}{2}}{\sqrt {\left({\frac {b}{2a}}\right)^{2}-{\frac {2c}{3a}}+{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}-{\frac {1}{2}}{\sqrt {{\frac {b^{2}}{2a^{2}}}-{\frac {4c}{3a}}-{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}-{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}+{\frac {b^{3}-4abc+8a^{2}d}{4a^{3}{\sqrt {\left({\dfrac {b}{2a}}\right)^{2}-{\dfrac {2c}{3a}}+{\dfrac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\dfrac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}}}}}}}$

${\displaystyle {x_{4}=-{\frac {b}{4a}}-{\frac {1}{2}}{\sqrt {\left({\frac {b}{2a}}\right)^{2}-{\frac {2c}{3a}}+{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}+{\frac {1}{2}}{\sqrt {{\frac {b^{2}}{2a^{2}}}-{\frac {4c}{3a}}-{\frac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}-{\frac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}+{\frac {b^{3}-4abc+8a^{2}d}{4a^{3}{\sqrt {\left({\dfrac {b}{2a}}\right)^{2}-{\dfrac {2c}{3a}}+{\dfrac {\sqrt[{3}]{2(c^{2}-3bd+12ae)}}{3a{\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}}}+{\dfrac {\sqrt[{3}]{2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace+{\sqrt {(2c^{3}-9bcd+27ad^{2}+27b^{2}e-72ace)^{2}-4(c^{2}-3bd+12ae)^{3}}}}}{3{\sqrt[{3}]{2a}}}}}}}}}}}}$

${\displaystyle {\Delta =256a^{3}e^{3}-192a^{2}bde^{2}-128a^{2}c^{2}e^{2}+144a^{2}cd^{2}e-27a^{2}d^{4}+144ab^{2}ce^{2}-6ab^{2}d^{2}e-80abc^{2}de+18abcd^{3}+16ac^{4}e-4ac^{3}d^{2}-27b^{4}e^{2}+18b^{3}cde-4b^{3}d^{3}-4b^{2}c^{3}e+b^{2}c^{2}d^{2}}}$ [來源請求]

PlanetMath指出，这四个形式直接使用，即使是在计算机上也过于复杂。[2]这四个解的推导过程的最后几步有较为简单的中间形式可以采用。得到这些解需要用到三次方程的求根公式。[1]

## 文獻

1. The Quartic Formula Derivation. [2021-07-14]. （原始内容存档于2021-07-14）.
Galois-theoretic derivation of the quartic formula. planetmath.org. [2021-07-14]. （原始内容存档于2021-01-18）.
2. quartic formula. planetmath.org. [2021-07-14]. （原始内容存档于2021-04-11）.