Problem 5. (20 points) A point is randomly thrown onto the segment $[8 ; 13]$ and let $k-$ be the obtained value. Find the probability that the roots of the equation $\left(k^{2}-2 k-35\right) x^{2}+(3 k-9) x+2=0$ satisfy the condition $x_{1} \leq 2 x_{2}$.
# Solution.

By Vieta's theorem

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{9-3 k}{k^{2}-2 k-35} \\
x_{1} \cdot x_{2}=\frac{2}{k^{2}-2 k-35}
\end{array}\right.
$$

Find the value of $k$ under the condition that $x_{1}=2 x_{2}$, and then use the method of intervals. $\left\{\begin{array}{l}3 x_{2}=\frac{9-3 k}{k^{2}-2 k-35} ; \\ 2 x_{2}^{2}=\frac{2}{k^{2}-2 k-35} .\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=\frac{9-3 k}{3\left(k^{2}-2 k-35\right)} ; \\ x_{2}^{2}=\frac{1}{k^{2}-2 k-35} .\end{array} \Rightarrow \frac{(9-3 k)^{2}}{9\left(k^{2}-2 k-35\right)^{2}}=\frac{1}{k^{2}-2 k-35} ;\right.\right.$

Since $k^{2}-2 k-35>0$ for $k \in(-\infty ;-5) \cup(7 ;+\infty)$, multiplying both sides of the equation by the square of this expression, we get

$\frac{(9-3 k)^{2}}{9}=k^{2}-2 k-35$,

$9 k^{2}-54 k+81=9 k^{2}-18 k-315,30 k=396, k=11$.

Plot the obtained value of $k$ on the number line, and check which part of the line satisfies the condition $x_{1}-2 x_{2} \leq 0$.

![](https://cdn.mathpix.com/cropped/2024_05_06_424f80418dd801e5b28ag-17.jpg?height=126&width=851&top_left_y=231&top_left_x=320)

This means that the condition $x_{1} \leq 2 x_{2}$ is satisfied for $k \leq 11$. Then $P=\frac{11-8}{13-8}=\frac{3}{5}=0.6$.

Answer. 0.6.