2000 AMC 12 Problem 11

Attempt Problem 11 of the 2000 AMC 12 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 2000 AMC 12 solutions, or check the answer key.

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11.

Two non-zero real numbers,

a

and

b,

satisfy

ab = a - b.

Find a possible value of

\frac{a}{b} + \frac{b}{a} - ab.

-2

-\dfrac{1}{2}

\dfrac{1}{3}

\dfrac{1}{2}

2

Combine over a common denominator:

\dfrac{a}{b} + \dfrac{b}{a} - ab = \dfrac{a^2 + b^2 - (ab)^2}{ab}

Replace

ab

with

a - b

in the numerator and expand

Combining over a common denominator,

\frac{a}{b} + \frac{b}{a} - ab = \frac{a^2 + b^2 - (ab)^2}{ab}.

Replacing

ab

with

a - b

in the numerator,

\begin{gathered} a^2 + b^2 - (a - b)^2 \\ = a^2 + b^2 \\ {}- (a^2 - 2ab + b^2) \\ = 2ab. \end{gathered}

Therefore the expression equals

\dfrac{2ab}{ab} = 2.

Thus, the correct answer is E.

Problem 11 in other years