2002 AMC 12A Problem 13

Attempt Problem 13 of the 2002 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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13.

Two different positive numbers

a

and

b

each differ from their reciprocals by

1.

What is

a + b?

1

2

\sqrt{5}

\sqrt{6}

3

A number

x

with

\left|x - \dfrac1x\right| = 1

satisfies

x^2 - x - 1 = 0

x^2 + x - 1 = 0

The two positive solutions are reciprocals of each other; add them using the quadratic formula

A positive number

x

differs from its reciprocal by

1

when

x - \dfrac1x = 1

x - \dfrac1x = -1,

i.e.

x^2 - x - 1 = 0

x^2 + x - 1 = 0.

The positive roots are

\dfrac{1+\sqrt5}{2}

and

\dfrac{-1+\sqrt5}{2},

which are reciprocals of each other. Their sum is

a + b = \sqrt5.

Thus, the correct answer is C.

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