Prove that

Given:

$3 a^{2} x^{2}+8 a b x+4 b^{2}=0$

On comparing it with $\mathrm{A} x^{2}+B x+C=0$, we get:

$A=3 a^{2}, B=8 a b$ and $C=4 b^{2}$

Discriminant $D$ is given by :

$D=\left(B^{2}-4 A C\right)$

$=(8 a b)^{2}-4 \times 3 a^{2} \times 4 b^{2}$

$=16 a^{2} b^{2}>0$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by:

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-8 a b+\sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}}=\frac{-8 a b+4 a b}{6 a^{2}}=\frac{-4 a b}{6 a^{2}}=\frac{-2 b}{3 a}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-8 a b-\sqrt{16 a^{2} b^{2}}}{2 \times 3 a^{2}}=\frac{-8 a b-4 a b}{6 a^{2}}=\frac{-12 a b}{6 a^{2}}=\frac{-2 b}{a}$

Thus, the roots of the equation are $\frac{-2 b}{3 a}$ and $\frac{-2 b}{a}$.