韦达定理
ren_gao_zu
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学习·文化课
一元二次方程 ax^2+bx+c=0(a≠0)
先求根，假设有2个不同实根x_{1},x_{2}
x^2-4x+3=0\to x_{1}=1,x_{2}=3\to x^2-4x+3=(x-1)(x-3)
2x^2-5x+2=0\to x_{1}=\frac{1}{2},x_{2}=2\to2x^2-5x+2=2(x-\frac{1}{2})(x-2)
x^2-4x-1\to x_{1}=2-\sqrt{5},x_{2}=2+\sqrt{5}\to x^2-4x=(x-2+\sqrt{5})(x-2-\sqrt{5})
再在R上因式分解：ax^2+bx+c=a(x-x_{1})(x-x_{2})\Longleftrightarrow ax^2+bx+c=ax^2-a(x_{1}+x_{2})x+ax_{1}x_{2}
故\begin{cases}b=-a(x_{1}+x_{2})\\c=ax_{1}x_{2}\end{cases}\Rightarrow \begin{cases}x_{1}+x_{2}=-\frac{b}{a}\\x_{1}x_{2}=\frac{c}{a}\end{cases}(韦达定理)
一般的问题中，考虑的是实根，需要先验证△
推论：|x_{1}-x_{2}|=\sqrt{x_{1}^2-2x_{1}x_{2}+x_{2}^2}=\sqrt{(x_{1}+x_{2})^2-4x_{1}x_{2}}=\sqrt{(-\frac{b}{a})^2-4\frac{c}{a}}=\sqrt{\frac{b^2-4ac}{a^2}}=\frac{\sqrt{\Delta}}{|a|}