*Problem*

*Solution*

It's a time saver to observe that \sqrt{3-x} and \sqrt{x+1} are both defined only in the interval -1 \le x \le 3.

In that interval, the former is strictly decreasing, the latter strictly increasing so that the function f(x)=\sqrt{3-x}-\sqrt{x+1} is strictly decreasing for -1 \le x \le 3. Furthermore, f(-1)=2 and f(3)=-2.

There is a unique point x_0 between -1 and 3 such that, for all -1\le x \lt x_0, f(x) \gt \frac{1}{2}. Let's find this point from

Squaring gives

After a simplification,

Squaring once more,

or, equivalently,

The quadratic formula yields two solutions

both of which lie in the interval -1 \le x \le 3. Since function f(x) is strictly decreasing, it must be that

so that x_0 = \frac{8 - \sqrt{31}}{8}.