Question

Find the roots of the following quadratic equation, by the method of completing the square:
$2x^2-7x+3=0$

Answer 1

$2x^2-7x+3=0$
$\Rightarrow 2x^2-7x=-3$
On dividing both sides of the equation by 2, we get,
$\Rightarrow x^2-\frac{7x}{2}=-\frac{3}{2}$
$\Rightarrow x^2-2\times x\times \frac{7}{4}=-\frac{3}{2}$
On adding $(\frac{7}{4}{)}^2$ to both sides of equation, we get
$\Rightarrow (x{)}^2-2\times x\times \frac{7}{4}+(\frac{7}{4}{)}^2=(\frac{7}{4}{)}^2-\frac{3}{2}$
$\Rightarrow (x-\frac{7}{4}{)}^2=\frac{49}{16}-\frac{3}{2}$
$\Rightarrow (x-\frac{7}{4}{)}^2=\frac{25}{16}$
$\Rightarrow (x-\frac{7}{4})=\pm \frac{5}{4}$
$\Rightarrow x=\frac{7}{4}\pm \frac{5}{4}$
$\Rightarrow x=\frac{7}{4}+\frac{5}{4}orx=\frac{7}{4}-\frac{5}{4}$
$\Rightarrow x=\frac{12}{4}orx=\frac{2}{4}$
$\Rightarrow x=3or\frac{1}{2}$