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

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

Given:

Given quadratic equation is $2x^2 - 7x + 3 = 0$.

To do:

We have to find the roots of the given quadratic equation.

Solution:

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

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

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

$x^2 - 2\times \frac{1}{2}\times \frac{7}{2} x = -\frac{3}{2}$  

$x^2 - 2\times \frac{7}{4} x = -\frac{3}{2}$  

Adding $(\frac{7}{4})^2$ on both sides completes the square. Therefore,

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

$(x-\frac{7}{4})^2=-\frac{3}{2}+\frac{49}{16}$      (Since $(a-b)^2=a^2-2ab+b^2$)

$(x-\frac{7}{4})^2=\frac{49-3\times8}{16}$

$(x-\frac{7}{4})^2=\frac{49-24}{16}$

$(x-\frac{7}{4})^2=\frac{25}{16}$

$x-\frac{7}{4}=\pm \sqrt{\frac{25}{16}}$     (Taking square root on both sides)

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

$x=\frac{7}{4}+\frac{5}{4}$ or $x=\frac{7}{4}-\frac{5}{4}$

$x=\frac{7+5}{4}$ or $x=\frac{7-5}{4}$

$x=\frac{12}{4}$ or $x=\frac{2}{4}$

$x=3$ or $x=\frac{1}{2}$

The values of $x$ are $3$ and $\frac{1}{2}$.