Find the roots of the following quadratic equations

Solution:

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 $-\frac{1}{\sqrt{2}}$.