Solve the following quadratic equations by factorization:

Solution:

We have been given

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

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

$2 x(2 x+a+b)-(a-b)(2 x+a+b)=0$

 

$(2 x-(a-b))(2 x+a+b)=0$

Therefore,

$2 x-(a-b)=0$

$2 x=a-b$

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

or,

$2 x+a+b=0$

$2 x=-(a+b)$

$x=\frac{-(a+b)}{2}$

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