Problem 9.7. The sides of the square $A B C D$ are parallel to the coordinate axes, with $A B$ lying on the y-axis, and the square is positioned as shown in the figure. The parabola defined by the equation

$$
y=\frac{1}{5} x^{2}+a x+b
$$

passes through points $B$ and $C$. Additionally, the vertex of this parabola (point $E$) lies on the segment $A D$. Find the sum of the roots of the quadratic trinomial whose graph is the parabola.

![](https://cdn.mathpix.com/cropped/2024_05_06_790dd471148872cd0846g-05.jpg?height=511&width=517&top_left_y=1416&top_left_x=468)
Answer: 20.

Solution. Note that the parabola is symmetric with respect to the vertical axis passing through its vertex, point $E$. Since points $B$ and $C$ are on the same horizontal line, they are symmetric with respect to this axis. This means that this axis passes through the midpoint of $B C$, and therefore, through the midpoint of $A D$.

Now that we have established that $E$ is the midpoint of $A D$, let's find the length of the side of the square $A B C D$. By completing the square in the quadratic polynomial, we get

$$
y=\frac{1}{5}\left(x-x_{E}\right)^{2}+y_{E}
$$

where $\left(x_{E}, y_{E}\right)$ are the coordinates of the vertex of the parabola (we will similarly denote the coordinates of other marked points). Substituting $x=x_{B}, y=y_{B}$ (the equality will hold since $B$ lies on the parabola) and moving $y_{E}$ to the left side, we get:

$$
y_{B}-y_{E}=\frac{1}{5}\left(x_{B}-x_{E}\right)^{2}
$$

If we denote the unknown side length of the square by $L$, then $y_{B}-y_{E}=L$ and $x_{B}-x_{E}=L / 2$. We obtain the relation $L=\frac{1}{5}(L / 2)^{2}$, from which it is not difficult to extract $L=20$.

It remains to recall that the points on the graph corresponding to the roots of the polynomial are also symmetric with respect to the axis of the parabola, which means that the semi-sum of the roots is equal to the abscissa of the vertex $x_{E}=L / 2$. Therefore, the sum of the roots is $L=20$.