Difference between revisions of "1989 AHSME Problems/Problem 18"
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Rationalizing the denominator of <math>\frac{1}{x+\sqrt{x^2+1}}</math>, it simplifies: <math>\frac{1}{x+\sqrt{x^2+1}}</math> = <math>\frac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}</math> = <math>\frac{x-\sqrt{x^2+1}}{-1}</math> = <math>-(x-\sqrt{x^2+1})</math>. Substituting this into the original equation, we get <math>x + \sqrt{x^2+1} - (-(x-\sqrt{x^2+1})) = x+\sqrt{x^2+1} + x - \sqrt{x^2+1} = 2x</math>. 2x is only rational if x is rational <math>\mathrm{(B)}</math> | Rationalizing the denominator of <math>\frac{1}{x+\sqrt{x^2+1}}</math>, it simplifies: <math>\frac{1}{x+\sqrt{x^2+1}}</math> = <math>\frac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}</math> = <math>\frac{x-\sqrt{x^2+1}}{-1}</math> = <math>-(x-\sqrt{x^2+1})</math>. Substituting this into the original equation, we get <math>x + \sqrt{x^2+1} - (-(x-\sqrt{x^2+1})) = x+\sqrt{x^2+1} + x - \sqrt{x^2+1} = 2x</math>. 2x is only rational if x is rational <math>\mathrm{(B)}</math> | ||
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+ | == See also == | ||
+ | {{AHSME box|year=1989|num-b=17|num-a=19}} | ||
+ | |||
+ | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 06:56, 22 October 2014
Problem
The set of all real numbers for which is a rational number is the set of all
(A) integers (B) rational (C) real
(D) for which is rational
(E) for which is rational
Solution
Rationalizing the denominator of , it simplifies: = = = . Substituting this into the original equation, we get . 2x is only rational if x is rational
See also
1989 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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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