8.43 For what values of $p$ are both roots of the quadratic trinomial $x^{2}+2(p+1) x+9 p-5$ negative?
8.43 The quadratic trinomial $x^{2}+2(p+1) x+9 p-5$ has roots if $D=4(p+1)^{2}-4(9 p-5) \geq 0$. Let $x_{1}$ and $x_{2}$ be the roots of the trinomial; then by the condition $x_{1}0$, $x_{1}+x_{2}=-2(p+1) 0 , } \\
{ p + 1 > 0 , }
\end{array} \text { or } \left\{\begin{array}{l}
p^{2}-7 p+6 \geq 0 \\
p>\frac{5}{9} \\
p>-1
\end{array}\right.\right.
$$

or $\left\{\begin{array}{l}p \leq 1, p \geq 6, \\ p>\frac{5}{9}, \\ p>-1 .\end{array}\right.$

Using Fig. 8.14, we find the solution to the last system: $\frac{5}{9}<p \leq 1$ or $p \geq 6$.

Answer: for $\frac{5}{9}<p \leq 1$ or $p \geq 6$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-284.jpg?height=169&width=591&top_left_y=1173&top_left_x=362)

Fig. 8.14