# 韦达定理

## 内容

${\displaystyle P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}$ 是一个一元 n 次（或）係數多項式，首項系數 ${\displaystyle a_{n}\neq 0}$，令 P 的 n 個根為 ${\displaystyle x_{1},x_{2},\dots ,x_{n}}$，则根 ${\displaystyle \{x_{i}\}}$和係數 ${\displaystyle \{a_{j}\}}$之間滿足關係式

${\displaystyle {\begin{cases}x_{1}+x_{2}+\dots +x_{n-1}+x_{n}=-{\dfrac {a_{n-1}}{a_{n}}}\\(x_{1}x_{2}+x_{1}x_{3}+\cdots +x_{1}x_{n})+(x_{2}x_{3}+x_{2}x_{4}+\cdots +x_{2}x_{n})+\cdots +x_{n-1}x_{n}={\dfrac {a_{n-2}}{a_{n}}}\\{}\quad \vdots \\x_{1}x_{2}\dots x_{n}=(-1)^{n}{\dfrac {a_{0}}{a_{n}}}\end{cases}}}$

${\displaystyle \sum _{1\leq i_{1}

## 证明

${\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}=a_{n}(x-x_{1})(x-x_{2})\cdots (x-x_{n})}$

${\displaystyle {\begin{cases}a_{n-1}=-a_{n}(x_{1}+x_{2}+\dots +x_{n-1}+x_{n})\\a_{n-2}=a_{n}\left((x_{1}x_{2}+x_{1}x_{3}+\cdots +x_{1}x_{n})+(x_{2}x_{3}+x_{2}x_{4}+\cdots +x_{2}x_{n})+\cdots +x_{n-1}x_{n}\right)\\{}\quad \vdots \\a_{0}=(-1)^{n}a_{n}x_{1}x_{2}\dots x_{n}\end{cases}}}$

## 特例

### n=2

${\displaystyle x_{1},x_{2}}$ 是一元二次多項式 ${\displaystyle ax^{2}+bx+c}$ 的两根，則由${\displaystyle ax^{2}+bx+c=a(x-x_{1})(x-x_{2})=ax^{2}-a(x_{1}+x_{2})x+ax_{1}x_{2}}$

${\displaystyle x_{1}+x_{2}=-{\frac {b}{a}},\quad x_{1}x_{2}={\frac {c}{a}}}$

${\displaystyle x_{1}+x_{2}={\frac {-b+{\sqrt {b^{2}-4ac}}+\left(-b\right)-{\sqrt {b^{2}-4ac}}}{2a}}=-{\frac {b}{a}}}$
${\displaystyle x_{1}x_{2}={\frac {\left(-b+{\sqrt {b^{2}-4ac}}\right)\left(-b-{\sqrt {b^{2}-4ac}}\right)}{\left(2a\right)^{2}}}={\frac {c}{a}}}$

### n=3

${\displaystyle x_{1},x_{2},x_{3}}$ 是一元三次多項式 ${\displaystyle ax^{3}+bx^{2}+cx+d}$ 的三根，則

${\displaystyle x_{1}+x_{2}+x_{3}=-{\frac {b}{a}},\quad x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}={\frac {c}{a}},\quad x_{1}x_{2}x_{3}=-{\frac {d}{a}}}$

## 歷史

...[Girard 是] 理解關於各次方項係數的和與積公式的一般性學說的第一人。他是找到關於將任意方程式的根的次方加總的規則的第一人。

## 參考資料

• Djukić, Dušan; et al, The IMO compendium: a collection of problems suggested for the International Mathematical Olympiads, 1959–2004, Springer, New York, NY, 2006, ISBN 0-387-24299-6