## Task 21/72

Given the equation $x^{4}+x^{3}+x^{2}+(62-k) x+k=0$ with the real solutions $x_{i}(i=1 ; 2 ; 3 ; 4)$. What real values can $k$ take if

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}>5
$$

is to be satisfied?
It is

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=\frac{x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}}{x_{1} x_{2} x_{3} x_{4}}
$$

If a polynomial of degree 4 is represented in the form $x 4+p x^{3}+q x^{2}+r x+s$, then for the roots $x_{i}$ of this polynomial, Vieta's formulas yield

$$
x_{1} x_{2} x_{3} x_{4}=s \quad ; \quad x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}=-r
$$

In our case, $s=k, r=62-k$. Therefore, we have

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=\frac{k-62}{k}>5
$$

We now consider two cases:

a) If $k>0$, solving the inequality leads to a contradiction: $\frac{k-62}{k}>5, 5k<k-62, 4k<-62, k<-\frac{31}{2}$.

(The case $k=0$ is not possible, as then by Vieta's formulas $x_{i}=0$ for at least one $i$, and thus the formation of $\frac{1}{x_{i}}$ would not be possible).

The domain for $k$ is therefore, under the required condition, $-\frac{31}{2}<k<0$.