Difference between revisions of "2008 AMC 12B Problems/Problem 7"

(Solution)
(Solution 2)
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<math>0 \Rightarrow \textbf{(A)}</math>
 
<math>0 \Rightarrow \textbf{(A)}</math>
  
== Solution 2 ==
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== Solution 2==
WLOG, let <math>x</math> and <math>y</math> both be 0. Thus,<math>0^2</math>
+
WLOG, let <math>x</math> and <math>y</math> both be <math>0</math>. Thus, <math>(0-0)^2</math> <math>=</math>  <math>0 \Rightarrow \textbf{(A)}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:25, 29 April 2020

Problem 7

For real numbers $a$ and $b$, define $a\textdollar b = (a - b)^2$. What is $(x - y)^2\textdollar(y - x)^2$?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy$

Solution 1

$\left[ (x-y)^2 - (y-x)^2 \right]^2$

$\left[ (x-y)^2 - (x-y)^2 \right]^2$

$[0]^2$

$0 \Rightarrow \textbf{(A)}$

Solution 2

WLOG, let $x$ and $y$ both be $0$. Thus, $(0-0)^2$ $=$ $0 \Rightarrow \textbf{(A)}$

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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 12 Problems and Solutions

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