An algebra problem by Munem Shahriar

$\begin{aligned} x^2 - \frac{12}{x^2} & = 1 \\ x^2 - 1 - \frac{12}{x^2} & = 0 \\ \left(x - \frac 4x \right) \left(x + \frac 3x \right) & = 0 \end{aligned}$

$\begin{cases} x - \dfrac 4x = 0 & \implies x = \dfrac 4x & \implies x^2 = 4 & \implies x = \boxed{\pm 2} \\ x + \dfrac 3x = 0 & \implies x = - \dfrac 3x & \implies x^2 = -3 & \implies \small \color{#D61F06} \text{No real solution.} \end{cases}$