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

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

See Also

2006 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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