, 2012

Answer: $\sqrt{5}$.
By condition, the trinomial $x^2 + p x + q$ has the roots, so its discriminant $D = p^2 - 4q \ge 0$. Since $p^2 = 4q + D$ and $p$ and $q$ are integer, we have $D \ne 2$ and $D \ne 3$ because the square of the integer number is congruent neither to 2 nor to 3 modulo 4.
Moreover, the trinomial $x^2 + p x + q$ has the roots
$$\alpha_1 = 2^{-1}(-p - \sqrt{D}) \quad \text{и} \quad \alpha_2 = 2^{-1}(-p + \sqrt{D}). \quad (*)$$
If the trinomial is irrational, but $\alpha_1$ and $\alpha_2$ are irrational numbers if and only if $D$ is different from the square of some integer number. Thus, in particular, $D \ne 0, D \ne 1$ and $D \ne 4$. Therefore, $D \ge 5$.
Since the roots $\alpha_1$ and $\alpha_2$ of the irrational trinomial are different from 0, there are only two possibilities:

1) $\alpha_1 \cdot \alpha_2 > 0$

2) $\alpha_1 \cdot \alpha_2 < 0$

For case 1) we have $q = \alpha_1 \cdot \alpha_2 > 0$ (the Vieta theorem). So $q \ge 1$. From (*) it follows that $|\alpha_1| + |\alpha_2| = |p|$. Since $D \ge 5$, we have $p^2 = 4q + D \ge 4 \cdot 1 + 5 = 9$, i.e. $|p| \ge 3$. Thus, $|\alpha_1| + |\alpha_2| \ge 3$ for case 1).

For case 2) from (*) it follows that $|\alpha_1| + |\alpha_2| = \sqrt{D} \ge \sqrt{5}$.
It suffices to show that the estimate $\sqrt{5}$ is admissible.

We find all irrational trinomials with the discriminants $D = 5$. By Vieta's theorem, $\alpha_1 \cdot \alpha_2 < 0$ if and only if $q = \alpha_1 \cdot \alpha_2 < 0$, i.e. case 2) holds if and only if $q \le -1$. If $q = -1$, then $D = 5$ only if $p^2 = 1$, but if $q \le -2$, then $D = p^2 - 4q \ge -4q \ge -4 \cdot (-2) = 8$. Therefore, there exist exactly two irrational trinomials with $D = 5$ and $\alpha_1 \cdot \alpha_2 < 0$: $x^2 - x - 1$ and $x^2 + x - 1$.

Therefore, the smallest value of the sum of the modules of the roots of the irrational trinomial is equal to $\sqrt{5}$.