1. Determine all values of real parameters $p$ and $q$ for which each of the equations

$$
x(x-p)=3+q, \quad x(x+p)=3-q
$$

has two distinct real roots, and the arithmetic mean of these roots is one of the roots of the other equation.
1. From Vieta's formulas for the roots of a quadratic equation (which, incidentally, follow from the factorization of the given quadratic trinomial into a product of root factors), we easily find that the sum of the roots of the first equation is $p$, so their arithmetic mean is $\frac{1}{2} p$. This number must be a root of the second equation, hence

$$
\frac{p}{2} \cdot \frac{3 p}{2}=3-q \text {. }
$$

Similarly, the sum of the roots of the second equation is $-p$, their arithmetic mean is $-\frac{1}{2} p$, and therefore

$$
-\frac{p}{2} \cdot\left(-\frac{3 p}{2}\right)=3+q .
$$

By comparing both equations (1) and (2), we have $3-q=3+q$ or $q=0$, and from (1) we get $p=2$ or $p=-2$.

Both solutions lead to the same pair of equations $x(x-2)=3, x(x+2)=3$. The roots of the first are the numbers -1 and 3, their arithmetic mean is 1. The roots of the second equation are the numbers 1 and -3, their arithmetic mean is -1.

For a complete solution, award 6 points. Award 2 points for determining that the arithmetic means of the roots are $\frac{1}{2} p$ and $-\frac{1}{2} p$, 1 point for the system of equations (1), (2), 2 points for solving the system, and 1 point for verifying that each of the equations $x(x-2)=3, x(x+2)=3$ has two distinct real roots.