9.119. For what values of $a$ does the sum $a+\frac{-1+9 a+4 a^{2}}{a^{2}-3 a-10}$ take only positive values
Solution.

Let's write down the inequality

$$
\begin{aligned}
& a+\frac{-1+9 a+4 a^{2}}{a^{2}-3 a-10}>0, \frac{a^{3}-3 a^{2}-10 a-1+9 a+4 a^{2}}{a^{2}-3 a-10}>0 \\
& \frac{a^{3}+a^{2}-a-1}{a^{2}-3 a-10}>0, \frac{a^{2}(a+1)-(a+1)}{(a+2)(a-5)}>0, \frac{(a+1)\left(a^{2}-1\right)}{(a+2)(a-5)}>0 \\
& (a+1)^{2}(a-1)(a+2)(a-5)>0
\end{aligned}
$$

Using the interval method, we find $a \in(-2 ;-1) \bigcup(-1 ; 1) \bigcup(5 ; \infty)$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0654.jpg?height=131&width=634&top_left_y=1226&top_left_x=355)

Answer: $a \in(-2 ;-1) \cup(-1 ; 1) \cup(5 ; \infty)$.