6.132. For what value of $p$ is the ratio of the roots of the equation $x^{2}+p x-16=0$ equal to $-4$?
## Solution.

By Vieta's theorem and the condition, we have the system

$\left\{\begin{array}{l}x_{1}+x_{2}=-p, \\ x_{1} \cdot x_{2}=-16, \\ \frac{x_{2}}{x_{1}}=-4\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}+x_{2}=-p, \\ x_{1} \cdot x_{2}=-16, \\ x_{2}=-4 x_{1}\end{array} \Rightarrow\left\{\begin{array}{l}x_{1}-4 x_{1}=-p, \\ x_{1}\left(-4 x_{1}\right)=-16\end{array} \Leftrightarrow\right.\right.\right.$

$\Leftrightarrow\left\{\begin{array}{l}-3 x_{1}=-p, \\ -4 x_{1}^{2}=-16\end{array} \Leftrightarrow\left\{\begin{array}{l}x_{1}=\frac{p}{3}, \\ x_{1}^{2}=4 .\end{array}\right.\right.$

Thus, $\frac{p^{2}}{9}=4$, from which $p^{2}=36$, or $p_{1,2}= \pm 6$.

Answer: $p_{1,2}= \pm 6$.