Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

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,

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

$x^{2}+\left(\frac{1}{4}\right)^{2}+2\left(\frac{1}{4}\right) x=\frac{-31}{16}$

$\left(x+\frac{1}{4}\right)^{2}=-\frac{31}{16}$

Since RHS is a negative number, therefore the roots of the equation do not exist as the square of a number cannot be negative.

Therefore the roots of the equation do not exist.