Question

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

Answer 1

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

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

$\Rightarrow 2(x^2-2\cdot \frac{7}{4}x+\frac{49}{16})-\frac{49}{8}+3=0$

$\Rightarrow 2{(x-\frac{7}{4})}^2-\frac{25}{8}=0$

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

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

$\Rightarrow x=3$ or $\frac{1}{4}$

$\therefore$ The roots are $3$ and $\frac{1}{2}$.