\left[\begin{array}{l}\text { Integral and Area } \\ \text { [Quadratic Equations. Vieta's Theorem] }\end{array}\right]

Author: Sergiev I.N.

The numbers $p$ and $q$ are such that the parabolas $y=-2 x^{2}$ and $y=x^{2}+p x+q$ intersect at two points, enclosing a certain figure.

Find the equation of the vertical line that divides the area of this figure in half.
If $x_{1}<x_{2}$ are the abscissas of the points of intersection of the given parabolas, then for each number $x_{0}$ from the interval $\left(x_{1}, x_{2}\right)$, the area of the part of the original figure located to the left of the vertical line $x=x_{0}$ is

$S\left(x_{0}\right)=\int_{x_{1}}^{x_{0}}\left(-2 x^{2}-\left(x^{2}+p x+q\right)\right) d x=\int_{x_{1}}^{x_{0}}\left(-3 x^{2}-p x-q\right) d x$, which coincides with the area of the corresponding part of the new figure bounded by the x-axis and the parabola $y=-3 x^{2}-p x-q$. This parabola intersects the x-axis at points $x_{1}, x_{2}$, and therefore is symmetric with respect to the line $x=1 / 2\left(x_{1}+x_{2}\right)$. Consequently, the area of the new figure (as well as the original one) will be divided in half when $x_{0}=\frac{x_{1}+x_{2}}{2}=\frac{p}{2 \cdot(-3)}=-\frac{p}{6}$.

## Answer

$x=-p / 6$.