Problem 5. (20 points) A point is randomly thrown onto the segment $[5 ; 7]$ and let $k-$ be the obtained value. Find the probability that the roots of the equation $\left(k^{2}-3 k-4\right) x^{2}+(3 k-7) x+2=0$ satisfy the condition $x_{1} \leq 2 x_{2}$.
# Solution.

By Vieta's theorem

$\left\{\begin{array}{l}x_{1}+x_{2}=\frac{7-3 k}{k^{2}-3 k-4} \\ x_{1} \cdot x_{2}=\frac{2}{k^{2}-3 k-4}\end{array}\right.$

Find the value of $k$ under the condition that $x_{1}=2 x_{2}$, and then use the method of intervals. $\left\{\begin{array}{l}3 x_{2}=\frac{7-3 k}{k^{2}-3 k-4} ; \\ 2 x_{2}^{2}=\frac{2}{k^{2}-3 k-4} .\end{array} \Rightarrow\left\{\begin{array}{l}x_{2}=\frac{7-3 k}{3\left(k^{2}-3 k-4\right)} ; \\ x_{2}^{2}=\frac{1}{k^{2}-3 k-4} .\end{array} \Rightarrow \frac{(7-3 k)^{2}}{9\left(k^{2}-3 k-4\right)^{2}}=\frac{1}{k^{2}-3 k-4} ;\right.\right.$

Since $k^{2}-3 k-10>0$ for $k \in(-\infty ;-1) \cup(4 ;+\infty)$, multiplying both sides of the equation by the square of this expression, we get

$$
\frac{(7-3 k)^{2}}{9}=k^{2}-3 k-4
$$

$9 k^{2}-42 k+49=9 k^{2}-27 k-36,15 k=85, k=\frac{17}{3}$.

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-34.jpg?height=125&width=851&top_left_y=563&top_left_x=320)

This means that the condition $x_{1} \leq 2 x_{2}$ is satisfied for $k \leq \frac{17}{3}$. Then $P=\frac{17 / 3-5}{7-5}=\frac{1}{3}$ Answer. $\frac{1}{3}$.