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

`sqrt2x^2-3x-2sqrt2=0`

#### Solution

We have been given that,

`sqrt2x^2-3x-2sqrt2=0`

Now divide throughout by `sqrt2`. We get,

`x^2-3/sqrt2x-2=0`

Now take the constant term to the RHS and we get

`x^2-3/sqrt2x=2`

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

`x^2-3/sqrt2x+(3/(2sqrt2))^2=(3/(2sqrt2))^2+2`

`x^2+(3/(2sqrt2))^2-2(3/(2sqrt2))x=25/8`

`(x-3/(2sqrt2))^2=25/8`

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-3/(2sqrt2)=+-5/(2sqrt2)`

`x=3/(2sqrt2)+-5/(2sqrt2)`

Now, we have the values of ‘*x*’ as

`x=3/(2sqrt2)+5/(2sqrt2)`

`x=8/(2sqrt2)=4/(sqrt2)=2sqrt2`

Also we have,

`x=3/(2sqrt2)-5/(2sqrt2)`

`x=-1/sqrt2`

Therefore the roots of the equation are `2sqrt2` and `-1/sqrt2`.