$2 \cdot 61$ Given the equation $x^{2}+p x+q=0$, where $p$ and $q$ are positive numbers, if the difference of its roots is 1, then $p$ equals
(A) $\sqrt{4 q+1}$.
(B) $q-1$.
(C) $-\sqrt{4 q+1}$.
(D) $q+1$.
(E) $\sqrt{4 q-1}$.
(21st American High School Mathematics Examination, 1970)
[Solution] Let the two roots of the equation be $x_{1}, x_{2}$. According to the problem, $\left|x_{1}-x_{2}\right|=1$, i.e., $\left(x_{1}-x_{2}\right)^{2}=1$, or $\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=1$. By Vieta's formulas, we know $p^{2}-4 q=1$.
Given $p>0$, then $p=\sqrt{4 q+1}$.
Therefore, the answer is $(A)$.