Question

Find the roots of the following equation $4x^2+4bx-(a^2-b^2)=0$ by the method of completing the square.

Answer 1

we have,

$4x^2+4bx-(a^2-b^2)=0$

Taking $4$ common, we get:

$x^2+bx-\left(\frac{a^2-b^2}{4}\right)=0$

Rewriting $bx$ as $2\times \frac{b}{2}x$,

$x^2+2\left(\frac{b}{2}\right)x=\frac{a^2-b^2}{4}$

Adding ${\left(\frac{b}{2}\right)}^2$ on both sides,

$x^2+2\left(\frac{b}{2}\right)x+{\left(\frac{b}{2}\right)}^2=\frac{a^2-b^2}{4}+{\left(\frac{b}{2}\right)}^2$

Using the identity $(a+b{)}^2=a^2+2ab+b^2$, the equation can be written as:

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

Now, taking square root on both sides,

$x+\frac{b}{2}=\pm \frac{a}{2}$

$\Rightarrow x=\frac{-b}{2}\pm \frac{a}{2}\Rightarrow x=\frac{-b-a}{2},\frac{-b+a}{2}$.

Hence, the roots are $\left(\frac{-a-b}{2}\right)$ and $\left(\frac{a-b}{2}\right)$.