Problem 5. (20 points) A point is randomly thrown onto the segment $[11 ; 18]$ and let $k-$ be the obtained value. Find the probability that the roots of the equation $\left(k^{2}+2 k-99\right) x^{2}+(3 k-7) x+2=0$ satisfy the condition $x_{1} \leq 2 x_{2}$.
# Solution.

By Vieta's theorem

$\left\{\begin{array}{l}x_{1}+x_{2}=\frac{7-3 k}{k^{2}+2 k-99} \\ x_{1} \cdot x_{2}=\frac{2}{k^{2}+2 k-99}\end{array}\right.$

Find the value of $k$ under the condition that $x_{1}=2 x_{2}$, and then use the method of intervals.

$\left\{\begin{array}{l}3 x_{2}=\frac{7-3 k}{k^{2}+2 k-99} ; \\ 2 x_{2}^{2}=\frac{2}{k^{2}+2 k-99} .\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=\frac{7-3 k}{3\left(k^{2}+2 k-99\right)} ; \\ x_{2}^{2}=\frac{1}{k^{2}+2 k-99} .\end{array} \Rightarrow \frac{(7-3 k)^{2}}{9\left(k^{2}+2 k-99\right)^{2}}=\frac{1}{k^{2}+2 k-99} ;\right.\right.$

Since $k^{2}+2 k-99>0$ for $k \in(-\infty ;-11) \cup(9 ;+\infty)$, multiplying both sides of the equation by the square of this expression, we get

$$
\frac{(7-3 k)^{2}}{9}=k^{2}+2 k-99,9 k^{2}-42 k+49=9 k^{2}+18 k-891,60 k=940, k=\frac{47}{3} \text {. }
$$

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-39.jpg?height=128&width=853&top_left_y=2289&top_left_x=316)

This means that the condition $x_{1} \leq 2 x_{2}$ is satisfied for $k \leq \frac{47}{3}$. Then $P \frac{\frac{47}{3}-11}{18-11}=\frac{2}{3}$.

Answer. $\frac{2}{3}$.