1. Let $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ be the real roots of the quadratic equation $x^{2}+p x+q=0$. What conditions must $p$ and $q$ satisfy for $\alpha+\beta=60^{\circ}+k \cdot 180^{\circ}$ to hold?
Solution. Let $\alpha+\beta=60^{\circ}+k \cdot 180^{\circ}$; then $\operatorname{tg}(\alpha+\beta)=\sqrt{3}$, i.e.

$$
\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=\sqrt{3}
$$

By applying Vieta's relations for the solutions of the quadratic equation, we get $\operatorname{tg} \alpha+\operatorname{tg} \beta=-p$ and $\operatorname{tg} \alpha \cdot \operatorname{tg} \beta=q$, so, by substitution in (1) we obtain

$$
p=(q-1) \sqrt{3}
$$

In addition to this condition, the discriminant of the quadratic equation must be non-negative, i.e., $p^{2}-4 q \geq 0$. If we substitute $p$ from (2) into this condition, we get

$$
3 q^{2}-10 q+3 \geq 0
$$

The solution to this inequality is

$$
q \leq \frac{1}{3} \quad \text { or } \quad q \geq 3
$$

Thus, $p$ and $q$ must satisfy conditions (2) and (3).

Conversely, if $p$ and $q$ satisfy conditions (2) and (3), then the discriminant of the quadratic equation will be positive, and there will exist real solutions which, of course, are equal to $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ for suitably chosen $\alpha$ and $\beta$. However, according to Vieta's relations, and considering (2), it is obtained that $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ satisfy equation (1), from which $\operatorname{tg}(\alpha+\beta)=\sqrt{3}$, i.e., $\alpha+\beta=60^{\circ}+k \cdot 180^{\circ}$.