Problem 5. (20 points) A point is randomly thrown onto the segment $[6 ; 10]$ and let $k-$ be the obtained value. Find the probability that the roots of the equation $\left(k^{2}-3 k-10\right) x^{2}+(3 k-8) x+2=0$ satisfy the condition $x_{1} \leq 2 x_{2}$.
# Solution.

By Vieta's theorem

$\left\{\begin{array}{l}x_{1}+x_{2}=\frac{8-3 k}{k^{2}-3 k-10} \\ x_{1} \cdot x_{2}=\frac{2}{k^{2}-3 k-10}\end{array}\right.$

Find the value of $k$ under the condition that $x_{1}=2 x_{2}$, and then use the method of intervals. $\left\{\begin{array}{l}3 x_{2}=\frac{8-3 k}{k^{2}-3 k-10} ; \\ 2 x_{2}^{2}=\frac{2}{k^{2}-3 k-10} .\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=\frac{8-3 k}{3\left(k^{2}-3 k-10\right)} ; \\ x_{2}^{2}=\frac{1}{k^{2}-3 k-10} .\end{array} \Rightarrow \frac{(8-3 k)^{2}}{9\left(k^{2}-3 k-10\right)^{2}}=\frac{1}{k^{2}-3 k-10} ;\right.\right.$

Since $k^{2}-3 k-10>0$ for $k \in(-\infty ;-2) \cup(5 ;+\infty)$, multiplying both sides of the equation by the square of this expression, we get

$\frac{(8-3 k)^{2}}{9}=k^{2}-3 k-10$,

$9 k^{2}-48 k+64=9 k^{2}-27 k-90,21 k=\frac{22}{3}, k=\frac{22}{3}$.

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-29.jpg?height=128&width=868&top_left_y=233&top_left_x=317)

This means that the condition $x_{1} \leq 2 x_{2}$ is satisfied for $k \leq \frac{22}{3}$. Then $P=\frac{\frac{22}{3}-6}{10-6}=\frac{1}{3}$.

Answer. $\frac{1}{3}$.