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 == |
Latest 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 | |
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All AHSME Problems and Solutions |
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