1. Determine all real numbers $s$ for which the equation

$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=0
$$

has four distinct real roots, with the product of two of them being equal to -2.
1. Assume that the number $s$ satisfies the problem statement, and denote the roots $x_{1}, x_{2}, x_{3}, x_{4}$ of the given equation such that

$$
x_{1} x_{2}=-2 \text {. }
$$

From the factorization into root factors

$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=4\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)
$$

after expanding and comparing coefficients of the same powers of $x$, we obtain Viète's formulas

$$
\begin{aligned}
x_{1}+x_{2}+x_{3}+x_{4} & =5, \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4} & =\frac{s}{4}, \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4} & =-\frac{11}{2}, \\
x_{1} x_{2} x_{3} x_{4} & =-\frac{1}{2} .
\end{aligned}
$$

From equations (0) and (4) it immediately follows that

$$
x_{3} x_{4}=\frac{1}{4}
$$

From equation (3) rearranged as

$$
\left(x_{1}+x_{2}\right) x_{3} x_{4}+\left(x_{3}+x_{4}\right) x_{1} x_{2}=-\frac{11}{2}
$$

after substituting the values of $x_{1} x_{2}$ and $x_{3} x_{4}$, we get the equation

$$
\frac{1}{4}\left(x_{1}+x_{2}\right)-2\left(x_{3}+x_{4}\right)=-\frac{11}{2},
$$

which, together with equation (1), forms a system of two linear equations for the unknown sums $x_{1}+x_{2}$ and $x_{3}+x_{4}$. A simple calculation shows that the solution to this system is the pair of values

$$
x_{1}+x_{2}=2 \quad \text { and } \quad x_{3}+x_{4}=3 .
$$

Substituting all the known values into equation (2) rearranged as

$$
x_{1} x_{2}+\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)+x_{3} x_{4}=\frac{s}{4},
$$

we find that necessarily $s=17$.

Now we need to perform the verification: from the equations

$$
x_{1}+x_{2}=2 \quad \text { and } \quad x_{1} x_{2}=-2
$$

it follows that the numbers $x_{1,2}$ are the roots of the quadratic equation

$$
x^{2}-2 x-2=0, \quad \text { thus } \quad x_{1,2}=1 \pm \sqrt{3}
$$

from the equations

$$
x_{3}+x_{4}=3 \quad \text { and } \quad x_{3} x_{4}=\frac{1}{4}
$$

it follows that the numbers $x_{3,4}$ are the roots of the quadratic equation

$$
x^{2}-3 x+\frac{1}{4}=0, \quad \text { thus } \quad x_{3,4}=\frac{3}{2} \pm \sqrt{2}
$$

We see that $x_{1,2,3,4}$ are indeed four distinct real numbers that satisfy the system of equations (1)-(4) for the value $s=17$, so they are the roots of the equation from the problem statement. Note that the problem statement did not require calculating these roots. It would not be enough to just verify that each of the quadratic equations (5) and (6) has two distinct real roots (which happens precisely when their discriminants are positive numbers), in addition, it would also be necessary to show that these two equations do not have a common root.

The sought number $s$ is unique and has the value $s=17$.

Another solution. Let $x_{1,2}$ be the roots of the given equation for which $x_{1} x_{2}=-2$ holds. The polynomial on the left side of the equation is divisible by the polynomial $\left(x-x_{1}\right)\left(x-x_{2}\right)$, i.e., by a polynomial of the form $x^{2}+p x-2$ (where $p=-x_{1}-x_{2}$), so the factorization

$$
4 x^{4}-20 x^{3}+s x^{2}+22 x-2=\left(x^{2}+p x-2\right)\left(4 x^{2}+q x+r\right) \text {. }
$$

Expanding and comparing coefficients of the same powers of $x$ gives the system

$$
-20=4 p+q, s=-8+p q+r, 22=-2 q+p r,-2=-2 r
$$

From the fourth equation we have $r=1$, substituting into the third gives $22=-2 q+p$, which together with the first equation gives $p=-2$ and $q=-12$. From the remaining (second) equation we then determine the value $s=17$. We know that for this value the polynomial from the given equation has the factorization

$$
4 x^{4}-20 x^{3}+17 x^{2}+22 x-2=\left(x^{2}-2 x-2\right)\left(4 x^{2}-12 x+1\right),
$$

it remains to perform the verification (as in the first method).

For a complete solution, award 6 points, of which 4 points for finding the value $s=17$, 1 point for finding both trinomials with roots $x_{1,2}$ and $x_{3,4}$ respectively, and 1 point for the verification. If the solver only correctly writes down the system of Viète's formulas (and possibly also determines the value of the product $x_{3} x_{4}$ from it), award 2 points.