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> | ||
| − | == Solution == | + | == Solution 1== |
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 \ | + | <math> 3 \spadesuit (4 \spadesuit 5) = 3 \spadesuit((4+5)(4-5)) = 3 \spadesuit (-9) = (3+(-9))(3-(-9)) = \boxed{\textbf{(A)}-72}</math> |
| + | == Solution 2 == | ||
| + | From [[difference of squares]], we have <math> x \spadesuit y = (x+y)(x-y) = x^2 - y^2</math> | ||
| + | So: | ||
| + | |||
| + | <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> | ||
| + | |||
| + | ~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}} | ||
Latest revision as of 16:00, 10 April 2025
Contents
Problem
For real numbers 
and 
, define 
. What is 
?
Solution 1
Since :
Solution 2
From difference of squares, we have 
So:
~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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
