Consider $x^2+px+q=0$, where $p$ and $q$ are positive numbers. If the roots of this equation differ by 1, then $p$ equals
$\text{(A) } \sqrt{4q+1}\quad \text{(B) } q-1\quad \text{(C) } -\sqrt{4q+1}\quad \text{(D) } q+1\quad \text{(E) } \sqrt{4q-1}$
From the quadratic equation, the two roots of the equation are $\frac{-p\pm\sqrt{p^2-4q}}{2}$. The positive difference between these roots is $\sqrt{p^2 - 4q}$. Setting $\sqrt{p^2-4q}=1$ and isolating $p$ gives $\sqrt{4q+1}$, or choice $\boxed{\text{(A)}}$.