5. While studying the polynomial $p(x)=x^{2}+2 x-6$, Enrica discovered two distinct real numbers $\alpha$ and $\beta$ such that $p(\alpha)=\beta$ and $p(\beta)=\alpha$. What is the value of $\alpha+\beta$?
(A) -3
(B) $-2 \sqrt{2}$
(C) 0
(D) 2
(E) 6
5. The answer is $\mathbf{( A )}$. The conditions imposed translate into the system

$$
\left\{\begin{array}{l}
\alpha^{2}+2 \alpha-6=\beta \\
\beta^{2}+2 \beta-6=\alpha
\end{array}\right.
$$

Subtracting the two equations, we get $\alpha^{2}+2 \alpha-\left(\beta^{2}+2 \beta\right)=\beta-\alpha$; writing $\alpha^{2}-\beta^{2}=$ $(\alpha-\beta)(\alpha+\beta)$ and factoring out a factor $\alpha-\beta$, this leads to

$$
(\alpha-\beta)(\alpha+\beta+2+1)=0
$$

Since $\alpha \neq \beta$, it follows that $\alpha+\beta=-3$.