韦达定理之推导

a x^{2} + b x + c = 0

$ax^2+bx+c = 0$

$\\ \\$

a (x^{2} + \frac{b}{a} x + \frac{c}{a}) = 0

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

$\\ \\$

x^{2} + \frac{b}{a} x = - \frac{c}{a}

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

$\\ \\$

x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2 = -\frac{c}{a}+(\frac{b}{2a})^2$

$\\ \\$

(x + \frac{b}{2 a})^{2} = \frac{b^{2} - 4 a c}{4 a^{2}}

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

$\\ \\$

x_{1} = \frac{\sqrt{b^{2} - 4 a c} - b}{2 a}

$x_{1} = \frac{\sqrt[]{b^2-4ac}-b}{2a}$

$\\ \\$

x_{2} = \frac{- \sqrt{b^{2} - 4 a c} - b}{2 a}

$x_{2} = \frac{-\sqrt[]{b^2-4ac}-b}{2a}$

$\\ \\$

设 b^{2} - 4 a c = Δ

$设b^2-4ac=\Delta$

$\\ \\$

∵ x_{1} = \frac{\sqrt{Δ} - b}{2 a}

$\because x_{1}=\frac{ \sqrt[]{\Delta}-b}{2a}$

$\\ \\$

∵ x_{2} = \frac{- \sqrt{Δ} - b}{2 a}

$\because x_{2}=\frac{ -\sqrt[]{\Delta}-b}{2a}$

$\\ \\$

∴ x_{1} + x_{2} = - \frac{b}{a}

$\therefore x_{1}+x_{2}=-\frac{b}{a}$

$\\ \\$

∴ x_{1} \cdot x_{2} = \frac{c}{a}

$\therefore x_{1} \cdot x_{2}=\frac{c}{a}$

posted @ 2022-10-05 10:28 Preparing 阅读(351) 评论(0) 编辑收藏举报

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