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

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== 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>
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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>. <math>2x</math> is only rational if <math>x</math> is rational <math>\mathrm{(B)}</math>
  
 
== See also ==
 
== See also ==

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
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