2006 AMC 10B Problems/Problem 2

Problem

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

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

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

$3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{\textbf{(A)}-72}$

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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