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

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<math> \mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72 </math>
 
<math> \mathrm{(A) \ } -72\qquad \mathrm{(B) \ } -27\qquad \mathrm{(C) \ } -24\qquad \mathrm{(D) \ } 24\qquad \mathrm{(E) \ } 72 </math>
  
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== Solution ==
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== Solution 1==
 
Since <math> x \spadesuit y = (x+y)(x-y) </math>:  
 
Since <math> x \spadesuit y = (x+y)(x-y) </math>:  
  
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<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>
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== Solution 2 ==
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From [[difference of squares]], we have <math> x \spadesuit y = (x+y)(x-y) = x^2 - y^2</math>
  
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So:
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<math>3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit(4^2 - 5^2) = 3 \spadesuit (16 - 25) = 3 \spadesuit (-9)= (3^2) - (-9)^2 =  9 - 81 = \boxed{\textbf{(A)}-72}</math>
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 +
~anabel.disher
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2006|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2006|ab=B|num-b=1|num-a=3}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 16:00, 10 April 2025

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 1

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

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

Solution 2

From difference of squares, we have $x \spadesuit y = (x+y)(x-y) = x^2 - y^2$

So:

$3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit(4^2 - 5^2) = 3 \spadesuit (16 - 25) = 3 \spadesuit (-9)= (3^2) - (-9)^2 = 9 - 81 = \boxed{\textbf{(A)}-72}$

~anabel.disher

See Also

2006 AMC 10B (Problems • Answer Key • Resources)
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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