Difference between revisions of "1989 AHSME Problems/Problem 18"
| Line 14: | Line 14: | ||
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> | ||
| + | {{MAA Notice}} | ||
Revision as of 12:48, 5 July 2013
Problem
The set of all real numbers for which
![\[x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}\]](http://latex.artofproblemsolving.com/1/4/f/14fbbf1d306af9fed15519af602e8092a4b49982.png)
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 
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
