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

We have been given that,

$x^{2}-4 \sqrt{2} x+6=0$

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

$x^{2}-4 \sqrt{2} x=-6$

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

$x^{2}-4 \sqrt{2} x+(2 \sqrt{2})^{2}=(2 \sqrt{2})^{2}-6$

$x^{2}+(2 \sqrt{2})^{2}-2(2 \sqrt{2}) x=2$

$(x-2 \sqrt{2})^{2}=2$

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

$x-2 \sqrt{2}=\pm \sqrt{2}$

$x=2 \sqrt{2} \pm \sqrt{2}$

Now, we have the values of ‘x’ as

$x=2 \sqrt{2}+\sqrt{2}$

$=3 \sqrt{2}$

Also we have,

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

$=\sqrt{2}$

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