Solve this


$\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$


Given :

$\sqrt{3} x^{2}+10 x-8 \sqrt{3}=0$

On comparing it with $a x^{2}+b x+c=0$, we get:

$a=\sqrt{3}, b=10$ and $c=-8 \sqrt{3}$

Discriminant $D$ is given by :

$D=\left(b^{2}-4 a c\right)$

$=(10)^{2}-4 \times \sqrt{3} \times(-8 \sqrt{3})$



Hence, the roots of the equation are real.

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

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-10+\sqrt{196}}{2 \sqrt{3}}=\frac{-10+14}{2 \sqrt{3}}=\frac{4}{2 \sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-10-\sqrt{196}}{2 \sqrt{3}}=\frac{-10-14}{2 \sqrt{3}}=\frac{-24}{2 \sqrt{3}}=\frac{-12}{\sqrt{3}}=\frac{-12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-12 \sqrt{3}}{3}=-4 \sqrt{3}$

Thus, the roots of the equation are $\frac{2 \sqrt{3}}{3}$ and $-4 \sqrt{3}$.


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