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Difference between revisions of "2002 AMC 12A Problems/Problem 13"

Problem

Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$. What is $a+b$? $\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }\sqrt 5 \qquad \text{(D) }\sqrt 6 \qquad \text{(E) }3$

Solution

Each of the numbers $a$ and $b$ is a solution to $\left| x - \frac 1x \right| = 1$.

Hence it is either a solution to $x - \frac 1x = 1$, or to $\frac 1x - x = 1$. Then it must be a solution either to $x^2 - x - 1 = 0$, or to $x^2 + x - 1 = 0$.

There are in total four such values of $x$, namely $\frac{\pm 1 \pm \sqrt 5}2$.

Out of these, two are positive: $\frac{-1+\sqrt 5}2$ and $\frac{1+\sqrt 5}2$. We can easily check that both of them indeed have the required property, and their sum is $\frac{-1+\sqrt 5}2 + \frac{1+\sqrt 5}2 = \boxed{(C) \sqrt 5}$.

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