TITLE: If $A$ and $B$ are positive constants, show that $\frac{A}{x-1} + \frac{B}{x-2}$ has a solution on $(1,2)$
QUESTION [19 upvotes]: I have problem which I couldn't figure out how to solve;
If $A$ and $B$ are positive constants, show that $$0=\frac{A}{x-1} + \frac{B}{x-2}$$ has a solution on the open interval  $(1,2)$.
If you support your answers with rigorous proof, I appreciate that.
What I thought was taking interval roughly close to the end points from the inside i.e $[1.1,1.9]$, but then it wouldn't be rigorous solution to this problem. After that, I totally stuck since I couldn't determine to closed interval, which prevented me from using any useful theorem.
Note: The problem is taken from G.Simmons Calculus with Analytic Geometry 2nd.

REPLY [7 votes]: In the interval $(1,2)$ you may freely multiply by $(x-1)(x-2)$ and
$$A(x-2)+B(x-1)=0$$ or
$$(A+B)x=2A+B,$$
$$x=\frac{2A+B}{A+B}=1+\frac A{A+B}.$$
Clearly,
$$0<\frac A{A+B}<1,$$ which substantiates the claim.

REPLY [2 votes]: To add on DonAntonio's Answer:
$$f(x) = \frac{A}{x-1} + \frac{B}{x-2} = \frac{(A+B)x - 2A - B}{(x-1)(x-2)}.$$
Now, $1 < \frac{2A+B}{A+B} < 2$, since $A,B > 0$. Moreover, one calculates
$f(\frac{2A+B}{A+B}) = 0$.