1989 AHSME Problems/Problem 18

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 $\mathrm{(B)}$

1989 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1989 AHSME Problems/Problem 18

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