Question

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

Answer 1

it is in the form of

$a{x}^2+bx+c$

the solution is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

here $a=4,b=-4a^2,c=(a^4-b^4)$

$\Rightarrow x=\frac{-4a^2\pm \sqrt{{(-4a^2)}^2-4\left(4\right)(a^4-b^4)}}{2\left(4\right)}\Rightarrow x=\frac{-4a^2\pm \sqrt{+16b^4}}{8}\Rightarrow x=\frac{-4a^2\pm 4b^2}{8}$

therefore

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