Question

Solve the given quadratic equation by factorization method
$4x^2-4a^{2}x+(a^4-b^4)=0$

Answer 1

We have
$4x^2-4a^{2}x+(a^4-b^4)=0$

Here. Constant term $=a^4-b^4=(a^2-b^2)(a^2+b^2)$

and coefficient of the middle term $=-4a^2$

Also. coeffcient of the middle term $=-4a^2=-\{2(a^2+b^2)+2(a^2-b^2)\}$

$\therefore 4x^2-4a^{2}x+(a^4-b^4)=0$

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

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

$\Rightarrow \{2x-(a^2+b^2)\}\{2x-(a^2-b^2)\}=0$

$\Rightarrow 2x-(a^2+b^2)=0or2x-(a^2-b^2)=0$

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