# 二次方程

（重定向自二次方程式

## 一元二次方程

### 求根公式

${\displaystyle y={\frac {3}{2}}x^{2}+{\frac {1}{2}}x-{\frac {4}{3}}\,}$
${\displaystyle y=-{\frac {4}{3}}x^{2}+{\frac {4}{3}}x+{\frac {1}{3}}\,}$
${\displaystyle y=x^{2}+{\frac {1}{2}}\,}$

${\displaystyle \Delta >0\,}$ ，則該方程有两個不相等的實数根： ${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\,}$

${\displaystyle \Delta =0\,}$ ，則該方程有两個相等的實数根： ${\displaystyle x_{1,2}=-{\frac {b}{2a}}\,}$

${\displaystyle \Delta <0\,}$ ，則該方程有一對共軛複數根： ${\displaystyle x_{1,2}={\frac {-b\pm i{\sqrt {4ac-b^{2}}}}{2a}}\,}$

### 根与系数的关系

${\displaystyle x_{1}\,}$ ${\displaystyle x_{2}\,}$ 是一元二次方程 ${\displaystyle ax^{2}+bx+c=0\,}$ ${\displaystyle a\neq 0\,}$  ）的两根，则

### 求根公式的由来

{\displaystyle {\begin{aligned}ax^{2}+bx+c&=0\\x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}&=0\\x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}-\left({\frac {b}{2a}}\right)^{2}+{\frac {c}{a}}&=0\\\left(x+{\frac {b}{2a}}\right)^{2}-{\frac {b^{2}}{4a^{2}}}+{\frac {c}{a}}&=0\\\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}\\\left(x+{\frac {b}{2a}}\right)^{2}&={\frac {b^{2}-4ac}{4a^{2}}}\\x+{\frac {b}{2a}}&={\frac {\pm {\sqrt {b^{2}-4ac}}}{2a}}\\x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\end{aligned}}}

{\displaystyle {\begin{aligned}ax^{2}+bx+c&=0\\ax^{2}+bx+\left({\frac {b}{2{\sqrt {a}}}}\right)^{2}&=\left({\frac {b}{2{\sqrt {a}}}}\right)^{2}-c\\\left(x{\sqrt {a}}+{\frac {b}{2{\sqrt {a}}}}\right)^{2}&=\left({\frac {b}{2{\sqrt {a}}}}\right)^{2}-c\\x{\sqrt {a}}+{\frac {b}{2{\sqrt {a}}}}&=\pm {\sqrt {\left({\frac {b}{2{\sqrt {a}}}}\right)^{2}-c}}\\x{\sqrt {a}}+{\frac {b}{2{\sqrt {a}}}}&=\pm {\sqrt {{\frac {b^{2}}{4a}}-c}}\\x+{\frac {b}{2a}}&=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}}\\x+{\frac {b}{2a}}&=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {4ac}{4a^{2}}}}}\\x&=-{\frac {b}{2a}}\pm {\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}\\x&={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}\end{aligned}}}

### 对应函数的极值

${\displaystyle y=ax^{2}+bx+c\,}$ ${\displaystyle a\neq 0\,}$ ），
${\displaystyle x\,}$ 求导，得

${\displaystyle {\frac {\mathop {\mbox{d}} y}{\mathop {\mbox{d}} x}}=2ax+b}$

${\displaystyle {\frac {\mathop {\mbox{d}} y}{\mathop {\mbox{d}} x}}=0}$ ，得

{\displaystyle {\begin{aligned}x&=-{\frac {b}{2a}}\end{aligned}}}

{\displaystyle {\begin{aligned}y&=-{\frac {b^{2}-4ac}{4a}}\end{aligned}}}

${\displaystyle f''(x)<0\,}$ ${\displaystyle x\,}$ ${\displaystyle f(x)\,}$ 极大值点
${\displaystyle f''(x)>0\,}$ ${\displaystyle x\,}$ ${\displaystyle f(x)\,}$ 极小值点
${\displaystyle {\frac {\mathop {\mbox{d}} ^{2}y}{\mathop {\mbox{d}} x^{2}}}=2a}$ ，可知：
${\displaystyle a<0\,}$ 时（抛物线开口向下），${\displaystyle x=-{\frac {b}{2a}}}$ ${\displaystyle y\,}$ 的极大值点；
${\displaystyle a>0\,}$ 时（抛物线开口向上），${\displaystyle x=-{\frac {b}{2a}}}$ ${\displaystyle y\,}$ 的极小值点。

## 参考

1. ^ 一般二次方程的讨论. [2012-12-29]. （原始内容存档于2019-07-24）.页面存档备份，存于互联网档案馆