4. Given $A=\left\{x \mid x^{2}+(p+2) x+1=0, x \in \mathbf{R}\right\}$, and $A \cap \mathbf{R}^{+}=\varnothing$, then the range of real number $p$ is . $\qquad$
4. $p>-2$ From $A \cap \mathbf{R}^{+}=\varnothing$ we know $A=\varnothing$ or the equation $x^{2}+(p+2) x+1=0$ has only non-positive roots. If $A=\varnothing$ then $\Delta=$ $(p+2)^{2}-4<0$, which gives $-4<p<0$. If the equation $x^{2}+(p+2) x+1=0$ has only non-positive roots, then
$$
\left\{\begin{array}{l}
\Delta \geqslant 0 \\
-(p+2) \leqslant 0
\end{array} \quad ; \quad \begin{array}{l}
p \geqslant 0 \\
p \geqslant-2
\end{array} \Rightarrow p \geqslant 0 \text {, hence }(-2,0) \cup[0,+\infty) .\right.
$$