Question

Solve the following quadratic equations by factorization method :
$4x^2-4a^{2}x+a^4-b^4=0$

Answer 1

$4x^2-4a^{2}x+a^4-b^4=0$

$⟹4x^2-2\left[2\right(a^2+b^2)+2(a^2-b^2\left)\right]x+(a^2-b^2)(a^2+b^2)=0$

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

$⟹2x[2x-(a^2+b^2\left)\right]=\left(a^{-}{b}^2\right)[2x-(a^2+b^2\left)\right]=0$

$⟹[2x-(a^2-b^2\left)\right][2x-(a^2+b^2\left)\right]=0$

$\therefore 2x=a^2-b^2$

$⟹x=\frac{a^2-b^2}{2}$

or, $2x=a^2+b^2$

$⟹x=\frac{a^2+b^2}{2}$