When $(a+b)^2=(x-3a)(x-b)e^x$ has only one positive solution, find the relationship between a and b. Here, a and b are constants and satisfy $a>b>0$.
Hint: consider the graph of $y=A\cdot e^{-x}$, where $A$ is constant.
https://i.stack.imgur.com/kQQwb.jpg
So far, I rearranged it so that
$$(a+b)^2e^{-x}=(x-3a)(x-b)$$
Since
$$A\cdot e^{-x}>0 ,$$
$$(x-3a)(x-b)>0.$$
Not sure what's next.