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

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Plug in a perfect square-4
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==Problem==
4 * sqrt(4) = 4 * 2 = 8
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For [[real number]]s <math>a</math> and <math>b</math>, define <math>a * b=(a-b)^2</math>. What is <math>(x-y)^2*(y-x)^2</math>?
cbrt(8)
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2 * 2 * 2 = 8
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<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
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that would be the sqrt
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==Solution==
x ^ (1/2) means sqrt(x)
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Since <math>(-a)^2 = a^2</math>, it follows that <math>(x-y)^2 = (y-x)^2</math>, and
so the answer is x^(1/2) (D)
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<cmath>(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.</cmath>
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==See also==
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{{AMC10 box|year=2008|ab=B|num-b=4|num-a=6}}
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[[Category:Articles with dollar signs]]
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 11:26, 4 July 2013

Problem

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

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

Solution

Since $(-a)^2 = a^2$, it follows that $(x-y)^2 = (y-x)^2$, and \[(x-y)^2 * (y-x)^2 = [(x-y)^2 - (y-x)^2]^2 = [(x-y)^2 - (x-y)^2]^2 = 0\ \mathrm{(A)}.\]

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
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

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