Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AHSME Problems/Problem 18"
m (→‎See also)
 
(2 intermediate revisions by 2 users not shown)
Line 13: Line 13:
 
== Solution ==
 
== 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>. 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>. <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}}
 
{{MAA Notice}}

Latest revision as of 16:53, 30 September 2024

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 (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AHSME_Problems/Problem_18&oldid=228674"