5. The parabolas in the figure are obtained by shifting the parabola $f(x)=x^{2}$ and are arranged such that the points of their intersections with the $O X$ axis respectively coincide in pairs, and all vertices lie on the same straight line. There are a total of 2020 parabolas. The length of the segment on the $O X$ axis enclosed between the roots of the first parabola is 1; the length of the segment enclosed between the roots of the second parabola is 2. Find the length of the segment on the $O X$ axis enclosed between the roots of the last parabola.

![](https://cdn.mathpix.com/cropped/2024_05_06_eefe83f285a2477dfcd6g-2.jpg?height=357&width=539&top_left_y=1689&top_left_x=770)
Solution. Note that the arrangement of parabolas can be changed by shifting the entire set to the left or right (the lengths of the segments on the $O X$ axis, enclosed between the roots of the parabolas, do not change with the shift). The intersection point of the line drawn through the vertices of the parabolas will not be needed. The lengths of the segments are determined by the angle of inclination of the drawn line.

Let's recall the expression for the distance between the roots of a parabola. Let $x_{1}$ and $x_{2}$ be the roots of the quadratic polynomial $x^{2}+b x+c$. Then

$$
\left|x_{2}-x_{1}\right|=\frac{-b+\sqrt{D}}{2}-\frac{-b-\sqrt{D}}{2}=\sqrt{D}
$$

Here $D$ is the discriminant of this polynomial.

![](https://cdn.mathpix.com/cropped/2024_05_06_eefe83f285a2477dfcd6g-3.jpg?height=654&width=874&top_left_y=273&top_left_x=594)

Let $x_{0}=-\frac{b}{2}$ be the abscissa of the vertex of the parabola $f(x)=x^{2}+b x+c$. The value of the polynomial at this point is

$$
f\left(x_{0}\right)=x_{0}^{2}+b x_{0}+c=\frac{b^{2}}{4}-b \cdot \frac{b}{2}+c=-\frac{D}{4}=-\frac{\left(x_{2}-x_{1}\right)^{2}}{4}
$$

In our problem, we need to calculate the tangent of the angle of inclination of the drawn line (the angle between the positive direction of the $O X$ axis and the line). Let this angle be $\varphi$. Let $x_{01}$ and $x_{02}$ be the abscissas of the vertices of the first and second parabolas, and $x_{1}, x_{2}, x_{3}$ be their roots in ascending order (see the figure).

$$
\operatorname{tg} \varphi=\frac{f\left(x_{02}\right)-f\left(x_{01}\right)}{x_{02}-x_{01}}
$$

We have

$$
\begin{gathered}
x_{02}-x_{01}=x_{02}-x_{2}+x_{2}-x_{01}=\frac{1}{2}\left(\left(x_{3}-x_{2}\right)+\left(x_{2}-x_{1}\right)\right)=\frac{1}{2}(2+1)=\frac{3}{2} \\
f\left(x_{02}\right)-f\left(x_{01}\right)=-1-\left(-\frac{1}{4}\right)=-\frac{3}{4}
\end{gathered}
$$

We get

$$
\operatorname{tg} \varphi=\frac{-\frac{3}{4}}{\frac{3}{2}}=-\frac{1}{2}
$$

Now let's find how the lengths of the segment on the $O X$ axis between the roots of the parabolas with numbers $n+1$ and $n$ are related, expressing this length through the roots $x_{n}, x_{n+1}, x_{n+2}$, the values at the vertices $f\left(x_{0 n}\right)$ and $f\left(x_{0 n+1}\right)$, the discriminants $D_{n}$ and $D_{n+1}$, and the found tangent. We use the fact that the distance between the abscissas of the vertices is the sum of half the distances between the roots of adjacent parabolas.

$x_{0 n+1}-x_{0 n}=x_{0 n+1}-x_{n+1}+x_{n+1}-x_{0 n}=\frac{x_{n+2}-x_{n+1}}{2}+\frac{x_{n+1}-x_{n}}{2}=\frac{\sqrt{D_{n+1}}}{2}+\frac{\sqrt{D_{n}}}{2}$

$$
\begin{gathered}
\operatorname{tg} \varphi=-\frac{1}{2}=\frac{f\left(x_{0 n+1}\right)-f\left(x_{0 n}\right)}{x_{0 n+1}-x_{0 n}}=\frac{-\frac{D_{n+1}}{4}+\frac{D_{n}}{4}}{\frac{\sqrt{D_{n+1}}}{2}+\frac{\sqrt{D_{n}}}{2}}= \\
=\frac{D_{n}-D_{n+1}}{2\left(\sqrt{D_{n+1}}+\sqrt{D_{n}}\right)}
\end{gathered}
$$

From this equation

$$
\sqrt{D_{n+1}}-\sqrt{D_{n}}=1
$$

Hence

$$
\begin{gathered}
\sqrt{D_{n+1}}=\sqrt{D_{n}}+1 \\
x_{n+2}-x_{n+1}=\left(x_{n+1}-x_{n}\right)+1
\end{gathered}
$$

That is, the length of the segment between the roots of each subsequent parabola increases by 1. From this, we get that the length of the segment on the $O X$ axis, enclosed between the roots of the parabola with number 2020, is 2020.

Answer. 2020.