9.034. $5 x-20 \leq x^{2} \leq 8 x$.

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9.034. $5 x-20 \leq x^{2} \leq 8 x$.
## Solution.

Let's rewrite this inequality as a system of inequalities

$$
\left\{\begin{array}{l}
x^{2}-5 x+20 \geq 0 \\
x^{2}-8 x \leq 0
\end{array}\right.
$$

Since $x^{2}-5 x+20>0$ for $x \in R$, the given inequality is equivalent to the inequality $x^{2}-8 x \leq 0, x(x-8) \leq 0$. Using the interval method, we find $x \in[0 ; 8]$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-529.jpg?height=148&width=496&top_left_y=1566&top_left_x=430)

Answer: $\quad x \in[0 ; 8]$