We write, $4 b x=2(a+b) x-2(a-b) x$ as

$\begin{array}{l}
4 x^{2} \times\left[-\left(a^{2}-b^{2}\right)\right]=-4\left(a^{2}-b^{2}\right) x^{2}=2(a+b) x \times[-2(a-b) x] \\
\therefore 4 x^{2}+4 b x-\left(a^{2}-b^{2}\right)=0 \\
\Rightarrow 4 x^{2}+2(a+b) x-2(a-b) x-(a-b)(a+b)=0 \\
\Rightarrow 2 x[2 x+(a+b)]-(a-b)[2 x+(a+b)]=0 \\
\Rightarrow[2 x+(a+b)][2 x-(a-b)]=0 \\
\Rightarrow 2 x+(a+b)=0 \text { or } 2 x-(a-b)=0 \\
\Rightarrow x=-\frac{a+b}{2} \text { or } x=\frac{a-b}{2}
\end{array}$

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