Solve the following quadratic equations by factorization:

Solution:

We have been given

$4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}=0$

$4 \sqrt{3} x^{2}+8 x-3 x-2 \sqrt{3}=0$

$4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0$

$(\sqrt{3} x+2)(4 x-\sqrt{3})=0$

Therefore,

$\sqrt{3} x+2=0$

$\sqrt{3} x=-2$

$x=\frac{-2}{\sqrt{3}}$

 

or,

$4 x-\sqrt{3}=0$

$4 x=\sqrt{3}$

$x=\frac{\sqrt{3}}{4}$

Hence, $x=\frac{-2}{\sqrt{3}}$ or $x=\frac{\sqrt{3}}{4}$.