Difference between revisions of "1991 AHSME Problems/Problem 27"
m (→Solution) |
(Added a solution with explanation) |
||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
| − | + | We have <math>\frac{1}{x-\sqrt{x^2-1}} = \frac{x+\sqrt{x^2-1}}{x^2-(x^2-1)} = x+\sqrt{x^2-1}.</math> Hence the given equation becomes <math>2(x+\sqrt{x^2-1}) = 20 \implies x+\sqrt{x^2-1} = 10.</math> Thus <math>\frac{1}{x-\sqrt{x^2-1}} = 10</math>, so <math>x-\sqrt{x^2-1} = \frac{1}{10}</math>. Adding this to <math>x+\sqrt{x^2-1} = 10</math> gives <math>2x = 10 + \frac{1}{10} \implies x = \frac{101}{20}.</math> Now the quantity we need to find is <math>x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}} = x^2 + \sqrt{x^4-1} + \frac{x^2 - \sqrt{x^4-1}}{x^4 - (x^4 - 1)} = 2x^2 = 2(\frac{101}{20})^2 = 51.005</math>, which is <math>\boxed{C}.</math> | |
| − | |||
| − | |||
| − | |||
| − | |||
== See also == | == See also == | ||
{{AHSME box|year=1991|num-b=26|num-a=28}} | {{AHSME box|year=1991|num-b=26|num-a=28}} | ||
Revision as of 06:02, 24 February 2018
Problem
If 
then 
(A) 
(B) 
(C) 
(D) 
(E) 
Solution
We have 
Hence the given equation becomes 
Thus 
, so 
. Adding this to 
gives 
Now the quantity we need to find is 
, which is 
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 26 |
Followed by Problem 28 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
