
1. Find 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 and the product of two of these roots is -2 .


Solution. Assume that $s$ is a number as above, and denote the four roots of the equation by $x_{1}, x_{2}, x_{3}$ and $x_{4}$ in such a way that

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

From the factorization

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

we obtain, upon multiplying out the brackets and comparing the coefficients at like powers of $x$ on both sides, the familiar Vièta's relations

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

From the equalities (0) and (4) it follows immediately that

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

Rewriting (3) as

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

and substituting the known values for $x_{1} x_{2}$ and $x_{3} x_{4}$ we obtain

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

which together with the equation (1) forms a system of two linear equations for the unknown sums $x_{1}+x_{2}$ and $x_{3}+x_{4}$. An easy calculation shows that its solution is given by

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

Inserting all this into the equality (2) rewritten in the form

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

we find that necessarily $s=17$.

Conversely, from the equalities

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

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

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

and from the equalities

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

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

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

We see that $x_{1,2,3,4}$ are indeed four mutually different real numbers which satisfy the system (1)-(4) for the value $s=17$, hence are the roots of the original equation from the statement of the problem.

There is thus only one such number $s$, namely $s=17$.
