Question

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

• A. $\frac{1}{3}$, 4
• B. $\frac{1}{3}$, 3
• C. $\frac{1}{2}$, 3
• D. 3 , 2

Answer 1

The correct option is C $\frac{1}{2}$, 3

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

On dividing both sides of the equation by 2, we get,

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

$\Rightarrow x^2-\frac{7x}{2}=-\frac{3}{2}$

On adding the square of one-half of the coefficient of x [i.e.,$(\frac{-7}{4}{)}^2$] to both sides of equation, we get

$\Rightarrow (x{)}^2-2\times x\times \frac{7}{4}+(\frac{-7}{4}{)}^2=(\frac{-7}{4}{)}^2-\frac{3}{2}$

$\Rightarrow (x{)}^2-2\times x\times \frac{7}{4}+(\frac{7}{4}{)}^2=\left(\frac{49}{16}\right)-\frac{3}{2}$

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

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

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

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

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

$\Rightarrow x=\frac{12}{4},x=\frac{2}{4}$

$\Rightarrow x=3,\frac{1}{2}$