Difference between revisions of "1969 AHSME Problems/Problem 5"
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Rockmanex3 (talk | contribs) (Solution to Problem 5) |
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== Solution == | == Solution == | ||
| − | <math>\ | + | Write an equation from the given information. |
| + | <cmath>N - \frac{4}{N} = R</cmath> | ||
| + | <cmath>N^2 - 4 = RN</cmath> | ||
| + | <cmath>N^2 - RN - 4 = 0</cmath> | ||
| + | By [[Vieta's Formulas]], the sum of all possible values of <math>N</math> for a given <math>R</math> is <math>R</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 02:18, 7 June 2018
Problem
If a number 
, diminished by four times its reciprocal, equals a given real constant 
, then, for this given , the sum of all such possible values of 
is
Solution
Write an equation from the given information.
![\[N - \frac{4}{N} = R\]](http://latex.artofproblemsolving.com/c/b/5/cb5033ecb13da31354292f779672838e5eaa4850.png)
![\[N^2 - 4 = RN\]](http://latex.artofproblemsolving.com/1/4/3/1435badee44fa7f4077646f2b90a169ef1e38be9.png)
![\[N^2 - RN - 4 = 0\]](http://latex.artofproblemsolving.com/f/e/5/fe5c2eaf4ffca9eeb028886b192760af3357e343.png)
By Vieta's Formulas, the sum of all possible values of for a given is , so the answer is 
.
See also
| 1969 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
