Find the roots of the following quadratic equations

Solution:

We have been given that,

$2 x^{2}+x-4=0$

Now divide throughout by 2. We get,

$x^{2}+\frac{1}{2} x-2=0$

Now take the constant term to the RHS and we get

$x^{2}+\frac{1}{2} x=2$

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

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

$x^{2}+\left(\frac{1}{4}\right)^{2}+2\left(\frac{1}{4}\right) x=\frac{33}{16}$

$\left(x+\frac{1}{4}\right)^{2}=\frac{33}{16}$

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

$x+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}$

$x=\frac{-1 \pm \sqrt{33}}{4}$

Now, we have the values of ‘x’ as

$x=\frac{-1+\sqrt{33}}{4}$

Also we have,

$x=\frac{-1-\sqrt{33}}{4}$

Therefore the roots of the equation are $\frac{\sqrt{33}-1}{4}$ and $\frac{-1-\sqrt{33}}{4}$.