Find the set of values of $\displaystyle a$ for which the equation, $\displaystyle \frac{2x^2-4ax+a^2+8}{(x-a)^2}$ has 2 distinct roots.

While I know that this question can be solved by using $\displaystyle b^2-4ac>0$ using the equation $\displaystyle 2x^2-4ax+a^2+8$, I do not understand why $\displaystyle (x-a)^2$ can be ignored. Why is it that only the top of the fraction is considered?