The curve $C$ has equation $y=ax+a+\dfrac{a-3}{x-1}$, where $a$ is a non-zero constant.
Prove that if $C$ has two stationary points, then $a<0$ or $a>3$.
My attempt: From the equation of the curve, we obtain
$$ax^2-yx+(y-3)=0$$
I stuck here. I don't know how to use the $2$ stationary points on the equation above. Any hint?
Computing the derivative $y'$ and setting it equal to $0$ gives us
$$a = \frac{a-3}{(x-1)^2}$$
which is equivalent (for $x\neq 1$) to $ax^2 - 2ax + 3 = 0$
Considering the discriminant, we conclude that two solutions to this exist if and only if $4a(a - 3)>0$, so $a<0$ or $a>3$.