Question

Solve the following quadratic equation for x:

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

Answer 1

The given equation is $4x^2+4bx-(a^2-b^2)=0$ ....(i)

Comparing equation (i) with quadratic equation $A{x}^2+Bx+C=0$, we get

$A=4,B=4b,C=-(a^2-b^2)$

By quadratic formula

$x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

$x=\frac{-4b\pm \sqrt{16b^2+4\times 4\times (a^2-b^2)}}{2\times 4}$

$x=\frac{-4b\pm \sqrt{16b^2+16a^2-16b^2}}{8}$

$x=\frac{-4b\pm 4a}{8}$

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

Therefore, $x=\frac{-b-a}{2}\Rightarrow -\left(\frac{a+b}{2}\right)$

or $x=\frac{-b+a}{2}\Rightarrow \frac{a-b}{2}$

Hence, $x=-\left(\frac{a+b}{2}\right)$ and $x=\frac{a-b}{2}$.