Difference between revisions of "1970 AHSME Problems/Problem 14"
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== Solution == | == Solution == | ||
| − | <math>\fbox{A}</math> | + | From the quadratic equation, the two roots of the equation are <math>\frac{-p + \sqrt{p^2 - 4q}}{2}</math> and <math>\frac{-p - \sqrt{p^2 - 4q}}{2}</math>. The positive difference between these roots is <math>\sqrt{p^2 - 4q}</math>. If <math>\sqrt{p^2 - 4q} = 1</math>, then <math>p^2 - 4q = 1</math>. This leads to <math>p^2 = 1 + 4q</math>, and <math>p = \sqrt{1 + 4q}</math>, which is answer <math>\fbox{A}</math>. |
== See also == | == See also == | ||
Revision as of 21:18, 13 July 2019
Problem
Consider 
, where 
and 
are positive numbers. If the roots of this equation differ by 1, then equals
Solution
From the quadratic equation, the two roots of the equation are 
and 
. The positive difference between these roots is 
. If 
, then 
. This leads to 
, and 
, which is answer 
.
See also
| 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. 
