Question

Solve the equation $2x^2-5x+3=0$ by the method of completing square.

Answer 1

We have,
$2x^2-5x+3=0$

$\Rightarrow x^2-\frac{5}{2}x+\frac{3}{2}=0$ (Dividing throughough by $2$)

$\Rightarrow x^2-\frac{5}{2}x=-\frac{3}{2}$ (Shifting the constant term on RHS)

$\Rightarrow x^2-2\left(\frac{5}{4}\right)x+{\left(\frac{5}{4}\right)}^2={\left(\frac{5}{4}\right)}^2-\frac{3}{2}$ (Adding ${(\frac{1}{2}Coeff.ofx)}^2$ onboth sides)

$\Rightarrow {(x-\frac{5}{4})}^2=\frac{25}{16}-\frac{3}{2}\Rightarrow {(x-\frac{5}{4})}^2=\frac{1}{16}\Rightarrow x-\frac{5}{4}=\pm \frac{1}{4}\Rightarrow x=\frac{5}{4}\pm \frac{1}{4}$

$\Rightarrow x=\frac{5}{4}+\frac{1}{4}=\frac{6}{4}orx=\frac{5}{4}-\frac{1}{4}=\frac{4}{4}\Rightarrow x=\frac{3}{2}orx=1$

Here the roots of the equation are $\frac{3}{2}$ and $1$