# 二次方程

## 一元二次方程

### 求根公式

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

${\displaystyle \Delta >0\,}$，则该方程有两个不相等的实数根： ${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac))}{2a))\,}$

${\displaystyle \Delta =0\,}$，则该方程有两个相等的实数根： ${\displaystyle x_{1,2}=-{\frac {b}{2a))\,}$

${\displaystyle \Delta <0\,}$，则该方程有一对共轭复数根： ${\displaystyle x_{1,2}={\frac {-b\pm i{\sqrt {4ac-b^{2)))){2a))\,}$

### 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)\,}$ 的极大值点，
${\displaystyle f''(x)>0\,}$${\displaystyle x\,}$${\displaystyle f(x)\,}$ 的极小值点；
${\displaystyle {\frac {\mathop {\mbox{d)) ^{2}y}{\mathop {\mbox{d)) x^{2))}=2a}$，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。