# 二次方程

(由一元二次方程跳轉過嚟)

## 解二次方程

### 解析解

${\displaystyle ax^{2}+bx+c=0}$

${\displaystyle x^{2}+{\frac {b}{a}}x+{\frac {c}{a}}=0}$

${\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}-{\frac {b^{2}}{4a^{2}}}+{\frac {c}{a}}=0}$

${\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}}$

${\displaystyle x+{\frac {b}{2a}}=\pm {\sqrt {\frac {b^{2}-4ac}{4a^{2}}}}}$

${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$

### 巴比倫人嘅解法

「I have subtracted from the area the side of my square: 14.30. Take 1, the coefficient. Divide 1 into two parts: 30. Multiply 30 and 30:15. You add to 14.30, and 14.30.15 has the root 29.30. You add to 29.30 the 30 which you have multiplied by itself: 30, and this is the side of the square.[1]

## 判別式

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

${\displaystyle \Delta >0}$ ，有兩個唔同嘅實根；

${\displaystyle \Delta =0}$ ，有兩個一樣嘅實根，叫重根；

${\displaystyle \Delta <0}$ ，因為負數開方虛數，所以冇實根，而係有一對共軛嘅複數根。

## 參考

1. Jean-Pierre Tignol, Galois' Theory of Algebraic Equations. QA211 .T5413 2001