Question

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(ii) $2x^2+x-4=0$

Answer 1

(ii) $2x^2+x-4=0$
We know;
D = $b^2–4ac$
= $1^2–4\times 2\times (-4)$
= 1+ 32
= 33
Since, $D>0$, hence roots are possible for this equation.
$\Rightarrow 2x^2+x=4$
On dividing both sides of the equation, we get;
$\Rightarrow x^2+\frac{x}{2}=2$
On adding $(\frac{1}{4}{)}^2$ to both sides of the equation, we get
$\Rightarrow (x{)}^2+2\times x\times \frac{1}{4}+(\frac{1}{4}{)}^2=2+(\frac{1}{4}{)}^2\Rightarrow (x+\frac{1}{4}{)}^2=\frac{33}{16}\Rightarrow x+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}$
$\Rightarrow x$ = $\pm \frac{\sqrt{33}}{4}-\frac{1}{4}$
$\Rightarrow x$ = $\pm \frac{\sqrt{33}-1}{4}$
$\Rightarrow x$= $\frac{\sqrt{33}-1}{4}$ or $x=\frac{-\sqrt{33}-1}{4}$