Task B-3.4. Determine the real number $p$ such that $\operatorname{tg} \alpha \text{ and } \operatorname{tg} \beta$ are the roots of the quadratic equation $\quad x^{2}+p x-\sqrt{3}=0$, if $\quad \alpha+\beta=\frac{\pi}{3}$.
## Solution.

Since $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ are real roots, the discriminant of the given equation must be $\geq 0$.

$D=p^{2}+4 \sqrt{3}$, which is positive for all real numbers $p$.

$$
x^{2}+p x-\sqrt{3}=0, \quad x_{1}=\operatorname{tg} \alpha, \quad x_{2}=\operatorname{tg} \beta
$$

Since $x_{1}+x_{2}=-p$, we have

$$
\operatorname{tg} \alpha+\operatorname{tg} \beta=-p
$$

Since $x_{1} \cdot x_{2}=-\sqrt{3}$, we have

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

If $\alpha+\beta=\frac{\pi}{3}$, then $\operatorname{tg}(\alpha+\beta)=\operatorname{tg} \frac{\pi}{3}$, that is,

$$
\begin{gathered}
\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}=\sqrt{3} \\
\frac{-p}{1+\sqrt{3}}=\sqrt{3} \quad \text { so } \quad p=-3-\sqrt{3}
\end{gathered}
$$