#### Question

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

`sqrt3x^2+10x+7sqrt3=0`

#### Solution

We have been given that,

`sqrt3x^2+10x+7sqrt3=0`

Now divide throughout by `sqrt3`. We get,

`x^2+10/sqrt3x+7=0`

Now take the constant term to the RHS and we get

`x^2+10/sqrtx=-7`

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

`x^2+10/sqrt3x+(10/(2sqrt3))^2=(10/(2sqrt3))^2-7`

`x^2+(10/(2sqrt3))^2+2(10/(2sqrt3))x=16/12`

`(x+10/(2sqrt3))^2=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+10/(2sqrt3)=+-4/(2sqrt3)`

`x=-10/(2sqrt3)+-4/(2sqrt3)`

Now, we have the values of ‘*x*’ as

`x=-10/(2sqrt3)+4/(2sqrt3)=-sqrt3`

Also we have,

`x=-10/(2sqrt3)-4/(2sqrt3)=-7/sqrt3`

Therefore the roots of the equation are `-sqrt3` and `-7/sqrt3`.