Question

Solve the given quadratic equation by completing the square, $4x^2+x-5=0$

Answer 1

The given quadratic equation is $4x^2+x-5=0$.

Add and subtract $\frac{1}{16}$ in the equation $4x^2+x-5=0$ to make a perfect square as shown below:

$4x^2+x-5=0\Rightarrow 4x^2+x+\frac{1}{16}-\frac{1}{16}-5=0\Rightarrow (2x{)}^2+(2\times 2x\times \frac{1}{4})+{\left(\frac{1}{4}\right)}^2+(-\frac{1}{16}-5)=0\Rightarrow {(2x+\frac{1}{4})}^2-\frac{81}{16}=0(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow {(2x+\frac{1}{4})}^2=\frac{81}{16}\Rightarrow 2x+\frac{1}{4}=\pm \sqrt{\frac{81}{16}}\Rightarrow 2x+\frac{1}{4}=\pm \frac{9}{4}\Rightarrow 2x+\frac{1}{4}=-\frac{9}{4},2x+\frac{1}{4}=\frac{9}{4}\Rightarrow 2x=-\frac{9}{4}-\frac{1}{4},2x=\frac{9}{4}-\frac{1}{4}\Rightarrow 2x=-\frac{10}{4},2x=\frac{8}{4}\Rightarrow 2x=-\frac{5}{2},2x=2\Rightarrow x=-\frac{5}{2\times 2},x=\frac{2}{2}\Rightarrow x=-\frac{5}{4},x=1$

Hence, $x=-\frac{5}{4}$ or $x=1$.