二次方程

一元二次方程

判別式

32x2+12x43
43x2+43x13
x2+12

${\displaystyle \Delta =b^{2}-4ac\,}$

方程的根和判別式的關係

${\displaystyle \Delta >0\,}$，該方程有兩個不相等的實數根： ${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

${\displaystyle \Delta =0\,}$，該方程有兩個相等的實數根： ${\displaystyle x_{1,2}={\frac {-b}{2a}}}$

${\displaystyle \Delta <0\,}$，該方程有一對共軛複數

根與係數的關係

${\displaystyle x_{1}\,}$${\displaystyle x_{2}\,}$是一元二次方程 ${\displaystyle ax^{2}+bx+c=0\,}$ 的兩根，那麼

(兩根之和)${\displaystyle x_{1}+x_{2}=-{\frac {b}{a}}}$，(兩根之積)${\displaystyle x_{1}x_{2}={\frac {c}{a}}}$

求根公式的由來

{\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}}}

極值

極值的公式

${\displaystyle y=ax^{2}+bx+c\,}$

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

${\displaystyle {\frac {\mathop {\mbox{d}} y}{\mathop {\mbox{d}} x}}=0}$，可得 ${\displaystyle x\,}$${\displaystyle y\,}$ 中的極值極大值極小值${\displaystyle x_{e}\,}$滿足：

{\displaystyle {\begin{aligned}2ax_{e}+b&=0\\x_{e}&=-{\frac {b}{2a}}\end{aligned}}}

${\displaystyle x_{e}=-{\frac {b}{2a}}}$ 代入 ${\displaystyle y\,}$，可得 ${\displaystyle y\,}$ 的極值 ${\displaystyle y_{e}\,}$

{\displaystyle {\begin{aligned}y_{e}&=a\left(-{\frac {b}{2a}}\right)^{2}+b\left(-{\frac {b}{2a}}\right)+c\\y_{e}&={\frac {b^{2}}{4a}}-{\frac {b^{2}}{2a}}+c\\y_{e}&={\frac {b^{2}-2b^{2}}{4a}}+c\\y_{e}&=-{\frac {b^{2}}{4a}}+c\end{aligned}}}

極值的類型

${\displaystyle f''(x_{e})<0\,}$${\displaystyle x_{e}\,}$${\displaystyle f(x)\,}$極大值點
${\displaystyle f''(x_{e})>0\,}$${\displaystyle x_{e}\,}$${\displaystyle f(x)\,}$極小值點
${\displaystyle f''(x_{e})=0\,}$${\displaystyle x_{e}\,}$${\displaystyle f(x)\,}$拐點)。

${\displaystyle {\frac {\mathop {\mbox{d}} ^{2}y}{\mathop {\mbox{d}} x^{2}}}=2a}$可知：
${\displaystyle a<0\,}$${\displaystyle y\,}$ 的極值${\displaystyle y_{e}\,}$為極大值；
${\displaystyle a>0\,}$${\displaystyle y\,}$ 的極值${\displaystyle y_{e}\,}$為極小值；
${\displaystyle a=0\,}$${\displaystyle y\,}$ 並非二次函數。