# Find the roots of the following quadratic equations

Question:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

$\sqrt{3} x^{2}+10 x+7 \sqrt{3}=0$

Solution:

We have been given that,

$\sqrt{3} x^{2}+10 x+7 \sqrt{3}=0$

Now divide throughout by $\sqrt{3}$. We get,

$x^{2}+\frac{10}{\sqrt{3}} x+7=0$

Now take the constant term to the RHS and we get

$x^{2}+\frac{10}{\sqrt{3}} x=-7$

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

$x^{2}+\frac{10}{\sqrt{3}} x+\left(\frac{10}{2 \sqrt{3}}\right)^{2}=\left(\frac{10}{2 \sqrt{3}}\right)^{2}-7$

$x^{2}+\left(\frac{10}{2 \sqrt{2}}\right)^{2}+2\left(\frac{10}{2 \sqrt{2}}\right) x=\frac{16}{12}$

$\left(x+\frac{10}{2 \sqrt{3}}\right)^{2}=\frac{16}{12}$

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

$x+\frac{10}{2 \sqrt{3}}=\pm \frac{4}{2 \sqrt{3}}$

$x=-\frac{10}{2 \sqrt{3}} \pm \frac{4}{2 \sqrt{3}}$

Now, we have the values of ‘x’ as

$x=-\frac{10}{2 \sqrt{3}}+\frac{4}{2 \sqrt{3}}$

$=-\sqrt{3}$

Also we have,

$x=-\frac{10}{2 \sqrt{3}}-\frac{4}{2 \sqrt{3}}$

$=-\frac{7}{\sqrt{3}}$

Therefore the roots of the equation are $-\sqrt{3}$ and $-\frac{7}{\sqrt{3}}$.