Find the roots of the following quadratic equations

We have been given that,

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

Now divide throughout by 4. We get,

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

Now take the constant term to the RHS and we get

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

Now add square of half of co-efficient of ‘x’ on both the sides. We have,

$x^{2}+2\left(\frac{\sqrt{3}}{2}\right) x+\left(\frac{\sqrt{3}}{2}\right)^{2}=\left(\frac{\sqrt{3}}{2}\right)^{2}-\frac{3}{4}$

$x^{2}+2\left(\frac{\sqrt{3}}{2}\right) x+\left(\frac{\sqrt{3}}{2}\right)^{2}=0$

$\left(x+\frac{\sqrt{3}}{2}\right)^{2}=0$

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

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

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

Now, we have the values of ‘x’ as

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

Also we have,

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

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