# 1957 AHSME Problems/Problem 22

## Problem

If $\sqrt{x - 1} - \sqrt{x + 1} + 1 = 0$, then $4x$ equals:

$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 4\sqrt{-1}\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\frac{1}{4}\qquad \textbf{(E)}\ \text{no real value}$

## Solution

By repeatedly rearranging the equation and squaring both sides, we can solve for $x$: \begin{align*} \sqrt{x-1}-\sqrt{x+1}+1 &= 0 \\ \sqrt{x-1}+1 &= \sqrt{x+1} \\ x-1+2\sqrt{x-1}+1 &= x+1 \\ 2\sqrt{x-1} &= 1 \\ \sqrt{x-1} &= \frac{1}{2} \\ x-1 &= \frac{1}{4} \\ x &= \frac{5}{4} \end{align*} After checking for extraneous solutions, we see that $x=\tfrac{5}{4}$ does indeed solve the equation. Thus, $4x=5$, and so our answer is $\boxed{\textbf{(A) }5}$.

## See Also

 1957 AHSC (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

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