韦达定理之推导
\[ax^2+bx+c = 0
\]
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\]
\[a(x^2+\frac{b}{a}x+\frac{c}{a}) = 0
\]
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\]
\[x^2+\frac{b}{a}x = -\frac{c}{a}
\]
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\]
\[x^2+\frac{b}{a}x+(\frac{b}{2a})^2 = -\frac{c}{a}+(\frac{b}{2a})^2
\]
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\]
\[(x+\frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2}
\]
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\]
\[x_{1} = \frac{\sqrt[]{b^2-4ac}-b}{2a}
\]
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\]
\[x_{2} = \frac{-\sqrt[]{b^2-4ac}-b}{2a}
\]
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\]
\[设b^2-4ac=\Delta
\]
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\]
\[\because x_{1}=\frac{ \sqrt[]{\Delta}-b}{2a}
\]
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\]
\[\because x_{2}=\frac{ -\sqrt[]{\Delta}-b}{2a}
\]
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\]
\[\therefore x_{1}+x_{2}=-\frac{b}{a}
\]
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\]
\[\therefore x_{1} \cdot x_{2}=\frac{c}{a}
\]
