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

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==Problem 7==
 
==Problem 7==
For real numbers <math>a</math> and <math>b</math>, define <math>a&#036;</math>b = (a - b)^2<math>. What is </math>(x - y)^2&#036;<math>(y - x)^2</math>?
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For real numbers <math>a</math> and <math>b</math>, define <math>a\textdollar b = (a - b)^2</math>. What is <math>(x - y)^2\textdollar(y - x)^2</math>?
  
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy</math>
 
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy</math>

Revision as of 10:34, 2 February 2015

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

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

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

$[0]^2$

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

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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