Problem

Find all real values of x for which
\sqrt{3-x}-\sqrt{x+1} \gt \frac{1}{2}.

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

\sqrt{3-x}-\sqrt{x+1} = \frac{1}{2}.

Squaring gives

(3-x) - 2\sqrt{ (3-x)(x+1) } + (x+1) = \frac{1}{4}.

After a simplification,

15 = 8\sqrt{ (3-x)(x+1) }.

Squaring once more,

225 = 64(3-x)(x+1),

or, equivalently,

64x^2 - 128x + 33 = 0.

The quadratic formula yields two solutions

x=\frac{64 \pm 8\sqrt{31}}{64} = \frac{8 \pm \sqrt{31}}{8},

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

f(\frac{8 - \sqrt{31}}{8}) = \frac{1}{2} and f(\frac{8 + \sqrt{31}}{8}) = -\frac{1}{2},

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