Question

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

Answer 1

(i) $2x^2-7x+3=0$
We know;
$D=b^2–4ac$
= $(-7{)}^2–4\times 2\times 3$
= $49-24$
= $25$
Since $D>0$, roots are possible for this equation.
$\Rightarrow 2x^2-7x=-3$
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
$\Rightarrow (x{)}^2-2\times x\times \frac{7}{4}+(\frac{7}{4}{)}^2=(\frac{7}{4}{)}^2-\frac{3}{2}$
$\Rightarrow (x-\frac{7}{4}{)}^2=\frac{49}{16}-\frac{3}{2}$
$\Rightarrow (x-\frac{7}{4}{)}^2=\frac{25}{16}$
$\Rightarrow (x-\frac{7}{4})=\pm \frac{5}{4}$
$\Rightarrow x=\frac{7}{4}\pm \frac{5}{4}$
$\Rightarrow x=\frac{7}{4}+\frac{5}{4}orx=\frac{7}{4}-\frac{5}{4}$
$\Rightarrow x=\frac{12}{4}orx=\frac{2}{4}$
$\Rightarrow x=3or\frac{1}{2}$