Find the roots of the following quadratic equations

Solution:

We have been given that,

$x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0$

Now take the constant term to the RHS and we get

$x^{2}-(\sqrt{2}+1) x=-\sqrt{2}$

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

$x^{2}-(\sqrt{2}+1) x+\left(\frac{\sqrt{2}+1}{2}\right)^{2}=\left(\frac{\sqrt{2}+1}{2}\right)^{2}-\sqrt{2}$

$x^{2}+\left(\frac{\sqrt{2}+1}{2}\right)^{2}-2\left(\frac{\sqrt{2}+1}{2}\right) x=\frac{3-2 \sqrt{2}}{4}$

$\left(x-\frac{\sqrt{2}+1}{2}\right)^{2}=\frac{(\sqrt{2}-1)^{2}}{2^{2}}$

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

$x-\frac{\sqrt{2}+1}{2}=\pm\left(\frac{\sqrt{2}-1}{2}\right)$

$x=\frac{\sqrt{2}+1}{2} \pm \frac{\sqrt{2}-1}{2}$

Now, we have the values of ‘x’ as

$x=\frac{\sqrt{2}+1}{2}+\frac{\sqrt{2}-1}{2}$

$=\sqrt{2}$

Also we have,

$x=\frac{\sqrt{2}+1}{2}-\frac{\sqrt{2}-1}{2}$

$=1$

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