Question

By completing square method find the roots of the following quadratic equation ${2n}^2-7n+3=0$

Answer 1

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

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

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

$\Rightarrow n^2-2\times \frac{7}{4}n+\frac{49}{16}-\frac{49}{16}+\frac{3}{2}=0$

$\Rightarrow {(n-\frac{7}{4})}^2+\frac{-49+24}{16}=0$

$\Rightarrow {(n-\frac{7}{4})}^2-\frac{25}{16}=0$

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

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

$\Rightarrow n=\frac{7\pm 5}{4}$

$\Rightarrow n=\frac{7+5}{4},\frac{7-5}{4}$

$\therefore n=3,\frac{1}{2}$