Solve each of the following quadratic equations:

We write, $4 b x=2(a+b) x-2(a-b) x$ as $4 x^{2} \times\left[-\left(a^{2}-b^{2}\right)\right]=-4\left(a^{2}-b^{2}\right) x^{2}=2(a+b) x \times[-2(a-b) x]$

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

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

$\Rightarrow 2 x[2 x+(a+b)]-(a-b)[2 x+(a+b)]=0$

$\Rightarrow[2 x+(a+b)][2 x-(a-b)]=0$

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

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

Hence, $-\frac{a+b}{2}$ and $\frac{a-b}{2}$ are the roots of the given equation.