4. Given the equation $x^{2}+2 p x-q=0$ has no real roots. Then the value of $p+q$ ( ).
(A) is less than $\frac{1}{4}$
(B) is equal to $\frac{1}{4}$
(C) is greater than $\frac{1}{4}$
(D) cannot be determined
4. A.

From the problem, we have $4 p^{2}+4 q<0$. Therefore, $q<-p^{2}$. Then, $p+q<-p^{2}+p=-\left(p-\frac{1}{2}\right)^{2}+\frac{1}{4} \leqslant \frac{1}{4}$. Hence, $p+q<\frac{1}{4}$.