Difference between revisions of "1991 AHSME Problems/Problem 27"

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== Problem ==
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If <math>x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20</math> then <math>x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=</math>
 
If <math>x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20</math> then <math>x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=</math>
  
 
(A) <math>5.05</math> (B) <math>20</math> (C) <math>51.005</math> (D) <math>61.25</math> (E) <math>400</math>
 
(A) <math>5.05</math> (B) <math>20</math> (C) <math>51.005</math> (D) <math>61.25</math> (E) <math>400</math>
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== Solution ==
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<math>\fbox{}</math>
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== See also ==
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{{AHSME box|year=1991|num-b=26|num-a=28}} 
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[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:57, 28 September 2014

Problem

If $x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20$ then $x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=$

(A) $5.05$ (B) $20$ (C) $51.005$ (D) $61.25$ (E) $400$

Solution

$\fbox{}$

See also

1991 AHSME (ProblemsAnswer KeyResources)
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

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