Difference between revisions of "2006 AMC 10B Problems/Problem 2"

Revision as of 18:50, 13 July 2006

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

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

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 == Problem ==
+For real numbers <math>x</math> and <math>y</math>, define <math> x \spadesuit y = (x+y)(x-y) </math>. What is <math> 3 \spadesuit (4 \spadesuit 5) </math>?
+<math> \mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72 </math>
 == Solution ==
+Since <math> x \spadesuit y = (x+y)(x-y) </math>:
+<math> 3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) =  3 \spadesuit (-9) = (3+(-9))(3-(-9)) = -72 \Rightarrow A</math>
 == See Also ==
 *[[2006 AMC 10B Problems]]