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2006 AMC 10B Problems/Problem 2

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Problem

For real numbers x and y, define x \spadesuit y = (x+y)(x-y). What is 3 \spadesuit (4 \spadesuit 5)?

\mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72

Solution

Since x \spadesuit y = (x+y)(x-y):

3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = -72 \Rightarrow A

See Also

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Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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