The given equation is $a^{2} b^{2} x^{2}-\left(4 b^{4}-3 a^{4}\right) x-12 a^{2} b^{2}=0$.

Comparing it with $A x^{2}+B x+C=0$, we get $A=a^{2} b^{2}, B=-\left(4 b^{4}-3 a^{4}\right)$ and $c=-12 a^{2} b^{2}$

$\therefore$ Discriminant,

$\begin{array}{l}
B^{2}-4 A C=\left[-\left(4 b^{4}-3 a^{4}\right)\right]^{2}-4 \times a^{2} b^{2} \times\left(-12 a^{2} b^{2}\right)=16 b^{8}-24 a^{4} b^{4}+9 a^{8}+48 a^{4} b^{4} \\
=16 b^{8}+24 a^{4} b^{4}+9 a^{8}=\left(4 b^{4}+3 a^{4}\right)^{2}>0
\end{array}$

So, the given equation has real roots

Now, $\sqrt{D}=\sqrt{\left(4 b^{4}+3 a^{4}\right)^{2}}=4 b^{4}+3 a^{4}$

$\begin{array}{l}
\therefore \alpha=\frac{-B+\sqrt{D}}{2 A}=\frac{-\left[-\left(4 b^{4}-3 a^{4}\right)\right]+\left(4 b^{4}+3 a^{4}\right)}{2 \times a^{2} b^{2}}=\frac{8 b^{4}}{2 a^{2} b^{2}}=\frac{4 b^{2}}{a^{2}} \\
\beta=\frac{-B-\sqrt{D}}{2 A}=\frac{-\left[-\left(4 b^{4}-3 a^{4}\right)\right]-\left(4 b^{4}+3 a^{4}\right)}{2 \times a^{2} b^{2}}=\frac{-6 a^{4}}{2 a^{2} b^{2}}=-\frac{3 a^{2}}{b^{2}}
\end{array}$

Hence, $\frac{4 b^{2}}{a^{2}}$ and $-\frac{3 a^{2}}{b^{2}}$ are the roots of the given equation.