2 If $\alpha, \beta$ are two distinct real roots of the equation $x^{2}+p x+8=0$, and $|\alpha|>|\beta|$, then among the following four conclusions, the one that does not necessarily hold is ( ).
(A) $|\alpha|>2$ and $|\beta|>2$
(B) $|\alpha|+|\beta|>4 \sqrt{2}$
(C) $|\alpha|>\frac{5}{2}$ or $|\beta|>\frac{5}{2}$
(D) $|\alpha|>2 \sqrt{2}$ and $|\beta|<2 \sqrt{2}$
2. (A).

From the given $\Delta=p^{2}-32>0$, i.e., $p>4 \sqrt{2}$ or $p < -4 \sqrt{2}$. Noting that $|\alpha|+|\beta| \geqslant|\alpha+\beta|$, then $|\alpha|+|\beta|>4 \sqrt{2}$. Therefore, (B) holds, and from (B) it is easy to see that (C) holds. This is because at least one of $|\alpha|$ and $|\beta|$ is greater than $2 \sqrt{2}>\frac{5}{2}$. Furthermore, by Vieta's formulas, $|\alpha \beta|=8$. Given $|\alpha|>|\beta|$, thus $|\alpha|>2 \sqrt{2},|\beta|<2 \sqrt{2}$, so (D) holds. Therefore, (A) does not hold.

On the other hand, consider the example $x^{2}-9 x+8=0$, yielding $\alpha=8$ and $\beta=1$. Thus, it is clear that (A) does not hold.