3B. Determine the values of the parameter $m$ such that for the solutions $x_{1}$ and $x_{2}$ of the equation $x^{2}+(m+3) x+m+21=0$, the following is satisfied: $\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{1}}<1$.
Solution. If we transform the inequality and use Vieta's formulas, we get

$$
\begin{aligned}
& \frac{x_{1}^{2}+x_{2}^{2}}{x_{2} x_{1}}-1 0 } \\
{ m + 2 1 0
\end{aligned} \quad \Rightarrow m \in(-\infty,-21) \cup(-9,-6)\right.\right.
$$