Question

What are the solutions to the equation over the complex numbers?

Answer 1

$\bf x^2+48=0\implies x^2=-48\implies x=\pm\sqrt{-48}~~\begin{cases}48=2\cdot 2\cdot 2\cdot 2\cdot 3\\\qquad 2^2\cdot 2^2\cdot 3\\\qquad (2^2)^2\cdot 3\\\qquad 4^2\cdot 3\end{cases}\\\\\\x=\pm\sqrt{-1\cdot 4^2\cdot 3}\implies x=\pm\sqrt{-1}\cdot \sqrt{4^2\cdot 3}\implies x=\pm i4\sqrt{3}$