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

By Vieta's theorem

$\left\{\begin{array}{l}x_{1}+x_{2}=\frac{5-3 k}{k^{2}-2 k-3} \\ x_{1} \cdot x_{2}=\frac{2}{k^{2}-2 k-3}\end{array}\right.$

Find the value of $k$ under the condition that $x_{1}=2 x_{2}$, and then use the method of intervals. $\left\{\begin{array}{l}3 x_{2}=\frac{5-3 k}{k^{2}-2 k-3} ; \\ 2 x_{2}^{2}=\frac{2}{k^{2}-2 k-3} .\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=\frac{5-3 k}{3\left(k^{2}-2 k-3\right)} ; \\ x_{2}^{2}=\frac{1}{k^{2}-2 k-3} .\end{array} \Rightarrow \frac{(5-3 k)^{2}}{9\left(k^{2}-2 k-3\right)^{2}}=\frac{1}{k^{2}-2 k-3} ;\right.\right.$

Since $k^{2}-2 k-3>0$ for $k \in(-\infty ;-1) \cup(3 ;+\infty)$, multiplying both sides of the equation by the square of this expression, we get

$\frac{(5-3 k)^{2}}{9}=k^{2}-2 k-3$

$9 k^{2}-30 k+25=9 k^{2}-18 k-27,12 k=52, k=\frac{13}{3}$.

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_dd63a07d88d2ea60fea0g-23.jpg?height=123&width=831&top_left_y=675&top_left_x=338)

This means that the condition $x_{1} \leq 2 x_{2}$ is satisfied for $k \leq \frac{13}{3}$. Then $P=\frac{\frac{13}{3}-3}{7-3}=\frac{1}{3}$

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