2次方程式の解の公式

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

$ D = b^2 - 4ac

$ D = a^2(x_1 - x_2)^2

code:tex

\begin{aligned}

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

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

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

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

&\iff x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

\end{aligned}

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

$ x_1 + x_2 = \frac{-b + \sqrt{D}}{2a} + \frac{-b - \sqrt{D}}{2a} = -\frac{b}{a}

$ x_1 - x_2 = \frac{-b + \sqrt{D}}{2a} - \left(\frac{-b - \sqrt{D}}{2a}\right) = \frac{\sqrt{D}}{a}

$ x_1 \times x_2 = \frac{-b + \sqrt{D}}{2a} \times \frac{-b - \sqrt{D}}{2a} = \frac{4ac}{2a} = \frac{c}{a}