Why is the '+' solution to the quadratic formula always the one that satisfies my constraint Let $A,B \in (0,1)$ be known constants, and $C \in (-\infty, \infty)$ 
be a known constant. Define
\begin{equation}
\xi(x) 
= \log \big( x \big) + \log \big( 1+x-A-B \big) 
- \log \big( A - x \big) 
- \log \big( B - x \big) 
- C
\end{equation}
For a particular application I need to find the root
of $\xi(x)$ that satisfies: 
\begin{equation} {\rm max}\Big(0, A+B-1\Big) \le x \le {\rm
min}\Big(A,B\Big).
\end{equation}
Some simple manipulation shows that the roots of $\xi(x)$ are also the roots of this quadratic equation: 
$$ 
(1-e^{C}) x^{2} + 
\Big(1 - A(1-e^{C}) - B(1-e^{C}) \Big)x 
- ABe^{C}
$$
Which is easily solved with $a=(1-e^{C})$, $b=\Big(1 - A(1-e^{C}) - B(1-e^{C}) \Big)$, and 
$c=- ABe^{C}$. Comparing these solutions with numerical ones using bisection (with the bracket specified by the contraint above), I find that the solution always coincides with the '+' one of the two solutions to the quadratic equation. Any indication on how to show that analytically? 
 A: Assuming $C\neq0$ (since otherwise there is no quadratic equation and your question doesn't arise), we can divide through by $1-\mathrm e^C$, and the quadratic equation becomes
$$x^2+\left(\frac{1}{1-\mathrm e^C}-A-B\right)x-\frac{AB\mathrm e^C}{1-\mathrm e^C}=:x^2+px+q=0\;.$$
Assuming that you labeled the solutions as '$+$' and '$-$' according to the formula for the full quadratic equation, the sign of the square root is reversed if the coefficient we're dividing through by, $1-\mathrm e^C$, is negative, that is if $C>0$. Thus, for the reduced formula with $p$ and $q$, we need to show that for $C<0$ the '$-$' solution violates the bounds and for $C>0$ the '$+$' solution violates the bounds.
Now for $C<0$ we have $1/(1-\mathrm e^C)>1$, and thus $-p<A+B-1$ and $q<0$, and hence
$$-\frac p2-\sqrt{\frac{p^2}4-q}<-\frac p2-\sqrt{\frac{p^2}4}\le-\frac p2-\frac p2=-p<A+B-1\le\max(0,A+B-1)\;,$$
whereas for $C>0$ we have $1/(1-\mathrm e^C)<0$, and hence
$$-\frac p2+\sqrt{\frac{p^2}4-q}\ge-\frac p2\ge\frac{A+B}{2}\ge\min(A,B)\;.$$
Thus, the solution that has a '$-$' in the formula for the full quadratic equation always violates one of the bounds.
