刚做数学题发现了一个奇怪的方程 $x^2+5x+25=0$

$\begin{aligned}x^2&=-5x-25\\&=-5(x+5)\end{aligned}$

$x^3=-5(x^2+5x)$

$\begin{aligned}x^3&=-5(x^2+5x)\\&=-5\times (-25)\\&=125\end{aligned}$

$\therefore x=5$ 但是左式$=$75$\ne$右式=0

套了公式也一样： $\boxed{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\quad (\Delta<0)}$

$\Downarrow$

$\boxed{x=\dfrac{-b\pm\sqrt{-(b^2-4ac)}i}{2a}}$

$\therefore x=\dfrac{-5\pm\sqrt{-(5^2-4\times 1\times 25)}i}{2}$

$x=\dfrac{-5\pm\sqrt{75}i}{2}$

$x=\dfrac{-5\pm5\sqrt{3}i}{2}$

$\therefore x^3=-\dfrac{125}{8}\pm\dfrac{375\sqrt{3}i}{8}+\dfrac{1125}{8}\mp\dfrac{375\sqrt{3}i}{8}$

$\therefore x^3=125$

$\therefore x=5$

MnZn求解。