Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
$\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$
We have been given that,
$\sqrt{2} x^{2}-3 x-2 \sqrt{2}=0$
Now divide throughout by $\sqrt{2}$. We get,
$x^{2}-\frac{3}{\sqrt{2}} x-2=0$
Now take the constant term to the RHS and we get
$x^{2}-\frac{3}{\sqrt{2}} x=2$
Now add square of half of co-efficient of ‘x’ on both the sides. We have,
$x^{2}-\frac{3}{\sqrt{2}} x+\left(\frac{3}{2 \sqrt{2}}\right)^{2}=\left(\frac{3}{2 \sqrt{2}}\right)^{2}+2$
$x^{2}+\left(\frac{3}{2 \sqrt{2}}\right)^{2}-2\left(\frac{3}{2 \sqrt{2}}\right) x=\frac{25}{8}$
$\left(x-\frac{3}{2 \sqrt{2}}\right)^{2}=\frac{25}{8}$
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
$x-\frac{3}{2 \sqrt{2}}=\pm \frac{5}{2 \sqrt{2}}$
$x=\frac{3}{2 \sqrt{2}} \pm \frac{5}{2 \sqrt{2}}$
Now, we have the values of ‘x’ as
$x=\frac{3}{2 \sqrt{2}}+\frac{5}{2 \sqrt{2}}$
$=2 \sqrt{2}$
Also we have,
$x=\frac{3}{2 \sqrt{2}}-\frac{5}{2 \sqrt{2}}$
$=-\frac{1}{\sqrt{2}}$
Therefore the roots of the equation are $2 \sqrt{2}$ and $-\frac{1}{\sqrt{2}}$.
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