Solve each of the following quadratic equations:

We write, $a x=2 a x-a x$ as $2 x^{2} \times\left(-a^{2}\right)=-2 a^{2} x^{2}=2 a x \times(-a x)$

$\therefore 2 x^{2}+a x-a^{2}=0$

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

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

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

$\Rightarrow x+a=0$ or $2 x-a=0$

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

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