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

Revision as of 01:34, 25 April 2008

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

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

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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: @@
-Plug in a perfect square-4
+==Problem==
-* sqrt(4) = 4 * 2 = 8
+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>?
-cbrt(8)
-* 2 * 2 = 8
+<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>
-so you start with a 4 and end up with a 2
-that would be the sqrt
+==Solution==
-x ^ (1/2) means sqrt(x)
+{{solution}}
-so the answer is x^(1/2) (D)
+==See also==
+{{AMC10 box|year=2008|ab=B|num-b=4|num-a=6}}