Question

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

`x^2-4sqrt2x+6=0`

Answer 1

We have been given that,

`x^2-4sqrt2x+6=0`

Now we take the constant term to the right hand side and we get

`x^2-4sqrt2x=-6`

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

`x^2 - 4sqrt2x+(2sqrt2)^2=(2sqrt2)^2-6`

`x^2+(2sqrt2)^2-2(2sqrt2)x=2`

`(x-2sqrt2)^2=2`

Since right hand side is a positive number, the roots of the equation exist.

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

`x-2sqrt2=+-sqrt2`

`x=2sqrt2+-sqrt2`

Now, we have the values of ‘x’ as

`x=2sqrt2+sqrt2=3sqrt2`

Also we have,

`x=2sqrt2-sqrt2=sqrt2`

Therefore the roots of the equation are `3sqrt2` and `sqrt2`.