Question

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
$2x^2-7x+3=0$

Answer 1

$\begin{array}{cc}2x^2-7x+3& =0\\ x^2-\frac{7}{2}x+\frac{3}{2}& =0\\ x^2-2\times x\times \frac{7}{4}+{\left(\frac{7}{4}\right)}^2-{\left(\frac{7}{4}\right)}^2+\frac{6}{4}& =0\\ {\left(x-\frac{7}{4}\right)}^2+\frac{24}{16}-\frac{49}{16}& =0\\ {\left(x-\frac{7}{4}\right)}^2& =\frac{25}{16}\dots \left(i\right)\end{array}$

Take the square root of both sides of the equation $\left(i\right),$

$\begin{array}{cc}\sqrt{{\left(x-\frac{7}{4}\right)}^2}& =\sqrt{\frac{25}{16}}\\ x-\frac{7}{4}& =\pm \frac{5}{4}\\ x& =\frac{7\pm 5}{4}\end{array}$

So, the roots will be,

$\begin{array}{cc}x& =\frac{7+5}{4}\\ & =\frac{12}{4}\\ & =3\\ & \\ x& =\frac{7-5}{4}\\ & =\frac{2}{4}\\ & =\frac{1}{2}\end{array}$