Question

Solve the following quadratic equation for x. $4x^2-4a^{2}x+(a^4-b^4)=0$

Answer 1

We have
$4x^2-4a^{2}x+(a^4-b^4)=0$
$\Rightarrow (4x^2-4a^{2}x+a^4)-b^4=0$
$\Rightarrow \left[\right(2x{)}^2-2\left(2x\right)\left(a\right)+\left(a^2{)}^2\right]-b^4=0$
$\Rightarrow (2x-a^2)2-(b^2{)}^2=0$ $[\because (a-b{)}^2=a^2+b^2-2ab]$
$\Rightarrow (2x-a^2+b^2)(2x-a^2-b^2)=0$ $[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$
$\Rightarrow 2x=a^2-b^{2}or2x=a^2+b^2$
$\Rightarrow x=\frac{a^2-b^2}{2}or\frac{a^2+b^2}{2}$