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Quadratic where roots differ by 1

Source: 1970 AHSME Problem 14

March 24, 2014
quadraticsAMC

Problem Statement

Consider x2+px+q=0x^2+px+q=0 where pp and qq are positive numbers. If the roots of this equation differ by 11, then pp equals
<spanclass=′latex−bold′>(A)</span>4q+1<spanclass=′latex−bold′>(B)</span>q−1<spanclass=′latex−bold′>(C)</span>−4q+1<spanclass=′latex−bold′>(D)</span>q+1<spanclass=′latex−bold′>(E)</span>4q−1<span class='latex-bold'>(A) </span>\sqrt{4q+1}\qquad<span class='latex-bold'>(B) </span>q-1\qquad<span class='latex-bold'>(C) </span>-\sqrt{4q+1}\qquad<span class='latex-bold'>(D) </span>q+1\qquad <span class='latex-bold'>(E) </span>\sqrt{4q-1}
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