Problem 7. For what values of the parameter $a$ does the equation $x^{3}-11 x^{2}+a x-8=0$ have three distinct real roots that form a geometric progression?
Answer: only 22.

Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-11 x^{2}+a x-8=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:

$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=11 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=8
\end{array}\right.
$$

Since $x_{1}, x_{2}, x_{3}$ form a geometric progression (let's assume in this order), there exist $b$ and $q$ such that $x_{1}=b, x_{2}=b q, x_{3}=b q^{2}$. Then from the equality $x_{1} x_{2} x_{3}=8$, we have $b^{3} q^{3}=8$, from which $x_{2}=b q=2, x_{1}=2 / q, x_{3}=2 q$.

Then $2\left(\frac{1}{q}+1+q\right)=11(*)$, after transformations $2 q^{2}-9 q+2=0$. The discriminant of this expression is $D=9^{2}-4 \cdot 2 \cdot 2>0$, so such $q$, and with it $x_{1}, x_{2}, x_{3}$, will exist. Then

$$
a=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=x_{1} x_{2} x_{3}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=8 \cdot\left(\frac{q}{2}+\frac{1}{2}+\frac{1}{2 q}\right)=4\left(q+1+\frac{1}{q}\right)=2 \cdot 11=22
$$

(In the penultimate transition, we used the equality $(*)$).

Comment. Naturally, $q, x_{1}, x_{2}, x_{3}$ can be calculated explicitly:

$$
q_{1,2}=\frac{9 \pm \sqrt{65}}{4}
$$

Choose $q$ with the “+” (if you choose with the “-”, then $x_{1}$ and $x_{3}$ will swap places, which will not affect the answer); then

$$
x_{1}=\frac{2 \cdot 4}{9+\sqrt{65}}=\frac{9-\sqrt{65}}{2}, x_{2}=2, x_{3}=\frac{9+\sqrt{65}}{2}
$$

$a$ could have been calculated by substituting the obtained numbers into the expression $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$.