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

Revision as of 17:25, 29 April 2020

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

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

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

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

$[0]^2$

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

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

2008 AMC 12B (Problems • Answer Key • Resources)
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

@@ Line 14: / Line 14: @@
 <math>0 \Rightarrow \textbf{(A)}</math>
-== Solution 2 ==
+== 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==
 {{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}}
 {{MAA Notice}}