## Task B-2.2.

How many integers $x$ satisfy the inequality $x^{4}-2020 x^{2}+2019<0$ ?
## First Solution.

The given inequality can be written in factored form:

$$
\begin{gathered}
\left(x^{2}-2019\right)\left(x^{2}-1\right)<0 \\
(x-\sqrt{2019})(x+\sqrt{2019})(x-1)(x+1)<0 .
\end{gathered}
$$

Using a sign chart or graphically

![](https://cdn.mathpix.com/cropped/2024_05_30_d3836afd55956f7fff9cg-06.jpg?height=211&width=962&top_left_y=2279&top_left_x=473)
we arrive at the solution of the inequality in the set $\mathbb{R}$

$$
x \in\langle-\sqrt{2019},-1\rangle \cup\langle 1, \sqrt{2019}\rangle
$$

The desired numbers are

$$
x \in \mathbb{Z} \cap(\langle-\sqrt{2019},-1\rangle \cup\langle 1, \sqrt{2019}\rangle)
$$

or

$$
x \in\{-44,-43, \ldots,-3,-2,2,3, \ldots, 44\}
$$

1 point

There are $43 \cdot 2=86$ such integers.