求助一个奇怪的数学题

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。

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