Showing that $y = \frac{(x-a)e^x}{(x-b)}$ has stationary points when $a-b<0$ or $a-b>4$

Question

I have the function $$y = \frac{(x-a)e^x}{(x-b)}$$ and I am told that the curve has stationary points under the following conditions -

$$a-b < 0 \quad\text{or}\quad a-b>4$$

I started by differentiating the equation to get $$ y'=\frac{(e^x+xe^x-ae^x)(x-b)-xe^x+ae^x}{(x-b)^2} \tag{1}$$ then simplified to get $$y'=\frac{e^x(x^2-x(a+b)+(a-b+ab))}{(x-b)^2} \tag{2}$$

I then set this equal to zero and attempted to use $b^2-4ac > 0$ on the quadratic. However, I am not really sure if this is the correct way to go about it, as I could not see a solution from this.

Any and all help is appreciated.

Answer 1

Hint: $$ (a+b)^2-4(a-b+ab) = (a-b)^2-4(a-b)$$