Difference between revisions of "1989 AHSME Problems/Problem 18"

(Created page with "The set of all real numbers for which <cmath>x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}</cmath> is a rational number is the set of all (A) integers <math>x</math> (B) rational <m...")
 
m (Fixed some LaTeX)
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
 +
== Problem ==
 +
 
The set of all real numbers for which  
 
The set of all real numbers for which  
 
<cmath>x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}</cmath>
 
<cmath>x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}</cmath>
Line 8: Line 10:
  
 
(E) <math>x</math> for which <math>x+\sqrt{x^2+1}</math> is rational
 
(E) <math>x</math> for which <math>x+\sqrt{x^2+1}</math> is rational
 +
 +
== Solution ==
 +
 +
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>. <math>2x</math> is only rational if <math>x</math> is rational <math>\mathrm{(B)}</math>
 +
 +
== See also ==
 +
{{AHSME box|year=1989|num-b=17|num-a=19}} 
 +
 +
[[Category: Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:54, 25 February 2018

Problem

The set of all real numbers for which \[x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}\] is a rational number is the set of all

(A) integers $x$ (B) rational $x$ (C) real $x$

(D) $x$ for which $\sqrt{x^2+1}$ is rational

(E) $x$ for which $x+\sqrt{x^2+1}$ is rational

Solution

Rationalizing the denominator of $\frac{1}{x+\sqrt{x^2+1}}$, it simplifies: $\frac{1}{x+\sqrt{x^2+1}}$ = $\frac{x-\sqrt{x^2+1}}{x^2-(x^2+1)}$ = $\frac{x-\sqrt{x^2+1}}{-1}$ = $-(x-\sqrt{x^2+1})$. Substituting this into the original equation, we get $x + \sqrt{x^2+1} - (-(x-\sqrt{x^2+1})) = x+\sqrt{x^2+1} + x - \sqrt{x^2+1} = 2x$. $2x$ is only rational if $x$ is rational $\mathrm{(B)}$

See also

1989 AHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png