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 $\frac{-p\pm\sqrt{p^2-4q}}{2}$ . 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 $\boxed{\text{(A)}}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1970 AHSME Problems/Problem 14

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