5. Given the real-coefficient quadratic equation $x^{2}+p x+q=0$ with real solutions $\alpha, \beta$ satisfying $|\alpha|+|\beta| \leqslant 1$. Then the range of $m=$ $p^{2}+4 q$ is ( ).
(A) $[-1,1]$
(B) $[-1,2]$
(C) $[0,1]$
(D) $[0,2]$
5.B.

By Vieta's formulas, we have $\alpha+\beta=-p, \alpha \beta=q$.
Then $m=(\alpha+\beta)^{2}+4 \alpha \beta=2(\alpha+\beta)^{2}-(\alpha-\beta)^{2}$.
$$
\begin{array}{l}
\text { Also }|\alpha+\beta| \leqslant|\alpha|+|\beta| \leqslant 1, \\
|\alpha-\beta| \leqslant|\alpha|+|\beta| \leqslant 1,
\end{array}
$$

Therefore, $m \geqslant-(\alpha-\beta)^{2} \geqslant-1$, with equality if and only if $\alpha=\frac{1}{2}, \beta=$ $-\frac{1}{2}$ or $\alpha=-\frac{1}{2}, \beta=\frac{1}{2}$; $m \leqslant 2(\alpha+\beta)^{2} \leqslant 2$, with equality if and only if $\alpha=\beta= \pm \frac{1}{2}$.
Thus, $-1 \leqslant m \leqslant 2$.