Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
$x^{2}-(\sqrt{2}+1) x+\sqrt{2}=0$
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 $\square$.
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