Question

Find the roots of the following quadratic equation by the method of completing the square:
$2x^2+x-4=0$

• A. $\frac{\sqrt{33}-1}{2}$ or $x=\frac{-\sqrt{33}-1}{2}$
• B. $\frac{\sqrt{33}-1}{4}$ or $x=\frac{-\sqrt{33}-1}{4}$
• C. $\frac{\sqrt{35}-1}{4}$ or $x=\frac{-\sqrt{35}-1}{4}$
• D. $\frac{\sqrt{35}-1}{2}$ or $x=\frac{-\sqrt{35}-1}{2}$

Answer 1

The correct option is B $\frac{\sqrt{33}-1}{4}$ or $x=\frac{-\sqrt{33}-1}{4}$

$2x^2+x=4$
On dividing both sides of the equation by 2, we get

$x^2+\frac{x}{2}=2$

On adding the square of one-half of the coefficient of $x$ to both sides [i.e., $(\frac{1}{4}{)}^2$] to both sides of the equation, we get,

$(x{)}^2+\frac{x}{2}+(\frac{1}{4}{)}^2=2+(\frac{1}{4}{)}^2$

$\Rightarrow (x{)}^2+2\times x\times \frac{1}{4}+(\frac{1}{4}{)}^2=2+(\frac{1}{4}{)}^2$

$\Rightarrow (x+\frac{1}{4}{)}^2=\frac{33}{16}$ [Using $(a+b{)}^2=a^2+b^2+2ab$]

$\Rightarrow x+\frac{1}{4}=\pm \frac{\sqrt{33}}{4}$

$\Rightarrow x$ = $\pm \frac{\sqrt{33}}{4}-\frac{1}{4}$

$\Rightarrow x$ = $\frac{\pm \sqrt{33}-1}{4}$

$\Rightarrow x$ = $\frac{\sqrt{33}-1}{4}$ or $x=\frac{-\sqrt{33}-1}{4}$

Therefore, the roots of the given quadratic equation are $x=\frac{\sqrt{33}-1}{4},\frac{-\sqrt{33}-1}{4}$