\section*{Exercise 6 - 021116}

Determine all real numbers \(x\) that satisfy the inequality

\[
\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}
\]

Verify the result!
}

First, one can already derive the following condition from the roots: \(-1 \leq x \leq 3\). Next, the points need to be calculated where the inequality changes its truth value, i.e., where \(\sqrt{3-x}-\) \(\sqrt{x+1}=\frac{1}{2}\) holds.

\[
\begin{array}{r}
\sqrt{3-x}-\sqrt{x+1}=\frac{1}{2} \\
\sqrt{3-x}=\frac{1}{2}+\sqrt{x+1} \\
3-x=\frac{1}{4}+x+1+\sqrt{x+1} \\
\frac{7}{4}-2 x=\sqrt{x+1} \\
\frac{49}{16}+4 x^{2}-7 x=x+1 \\
4 x^{2}-8 x+\frac{33}{16}=0 \\
x^{2}-2 x+\frac{33}{64}=0 \\
x_{1,2}=1 \pm \sqrt{1-\frac{33}{64}} \\
x_{1}=1+\frac{\sqrt{31}}{8} \quad ; \quad x_{2}=1-\frac{\sqrt{31}}{8}
\end{array}
\]

Since squaring was performed, there can be extraneous solutions. Therefore, it must still be substituted.

\(\sqrt{3-x_{1}}-\sqrt{x_{1}+1}=-\frac{1}{2} \rightarrow\) extraneous solution \(\sqrt{3-x_{2}}-\sqrt{x_{2}+1}=\frac{1}{2} \rightarrow\) solution \(x_{2}\) is thus the desired boundary.

The inequality is true for all \(x\) for which: \(-1 \leq x<1-\frac{\sqrt{31}}{8}\).