Difference between revisions of "2008 AMC 10B Problems/Problem 5"

Revision as of 11:47, 25 April 2008

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

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

Since $(-a)^2 = a^2$ , it follows that $(x-y)^2 = (y-x)^2$ , and

$(x-y)^2$ $(y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.$

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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

@@ Line 1: / Line 1: @@
 ==Problem==
-For [[real number]]s <math>a</math> and <math>b</math>, define <math>a\</math> b=(a-b)^2<math>. What is </math>(x-y)^2\<math> (y-x)^2</math>?
+For [[real number]]s <math>a</math> and <math>b</math>, define <math>a\</math><math> b=(a-b)^2</math>. What is <math>(x-y)^2\</math><math> (y-x)^2</math>?
 <math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ x^2+y^2\qquad\mathrm{(C)}\ 2x^2\qquad\mathrm{(D)}\ 2y^2\qquad\mathrm{(E)}\ 4xy</math>
 ==Solution==
-{{solution}}
+Since <math>(-a)^2 = a^2</math>, it follows that <math>(x-y)^2 = (y-x)^2</math>, and
+<center><math>(x-y)^2\</math><math> (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.</math></center>
 ==See also==
 {{AMC10 box|year=2008|ab=B|num-b=4|num-a=6}}
+[[Category:Articles with dollar signs]]
+[[Category:Introductory Algebra Problems]]