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

Solution:

We have to find the roots of given quadratic equation by the method of completing the square. We have,

$2 x^{2}-7 x+3=0$

We should make the coefficient of $x^{2}$ unity. So,

$x^{2}-\frac{7}{2} x+\frac{3}{2}=0$

Now shift the constant to the right hand side,

$x^{2}-\frac{7}{2} x=-\frac{3}{2}$

Now add square of half of coefficient ofon both the sides,

$x^{2}-2\left(\frac{7}{4}\right) x+\left(\frac{7}{4}\right)^{2}=-\frac{3}{2}+\left(\frac{7}{4}\right)^{2}$

We can now write it in the form of perfect square as,

$\left(x-\frac{7}{4}\right)^{2}=-\frac{3}{2}+\frac{49}{16}$

$=\frac{25}{16}$

Taking square root on both sides,

$\left(x-\frac{7}{4}\right)=\sqrt{\frac{25}{16}}$

So the required solution of,

$x=\frac{7}{4} \pm \frac{5}{4}$

$=3, \frac{1}{2}$