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 $\mathrm{(B)}$

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