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 $\fbox{A}$ .

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1970 AHSME Problems/Problem 14

Problem

Solution

See also