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

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Since <math> x \spadesuit y = (x+y)(x-y) </math>:  
 
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>
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<math> 3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) =  3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{\textbf{(A)}-72}</math>
  
 
== See Also ==
 
== See Also ==
*[[2006 AMC 10B Problems]]
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{{AMC10 box|year=2006|ab=B|num-b=1|num-a=3}}
  
*[[2006 AMC 10B Problems/Problem 1|Previous Problem]]
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[[Category:Introductory Algebra Problems]]
 
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{{MAA Notice}}
*[[2006 AMC 10B Problems/Problem 3|Next Problem]]
 

Latest revision as of 10:33, 19 December 2021

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

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

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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