# Difference between revisions of "1970 AHSME Problems/Problem 14"

## Problem

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

$\text{(A) } \sqrt{4q+1}\quad \text{(B) } q-1\quad \text{(C) } -\sqrt{4q+1}\quad \text{(D) } q+1\quad \text{(E) } \sqrt{4q-1}$

## Solution

From the quadratic equation, the two roots of the equation are $\frac{-p + \sqrt{p^2 - 4q}}{2}$ and $\frac{-p - \sqrt{p^2 - 4q}}{2}$. The positive difference between these roots is $\sqrt{p^2 - 4q}$. If $\sqrt{p^2 - 4q} = 1$, then $p^2 - 4q = 1$. This leads to $p^2 = 1 + 4q$, and $p = \sqrt{1 + 4q}$, which is answer $\fbox{A}$.