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If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =
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24 Mar 2018, 05:11
24 Mar 2018, 05:11
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Question Stats:
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If \(x^2 + y^2 = 16 - 2xy\), then \((x + y)^4 =\)
A. 4
B. 32
C. 48
D. 64
E. 256
A. 4
B. 32
C. 48
D. 64
E. 256
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Joined: 15 Jan 2018
Posts: 147
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Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 =
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25 Mar 2018, 19:45
25 Mar 2018, 19:45
1
Expert Reply
Key to this one is noticing the hidden special equation. x^2 + y^2 isn't one, but it's close. If you shuffle the equation around a bit you get:
x^2 + 2xy + y^2 = 16
which can be reformatted to:
(x + y)^2 = 16
Since (x + y)^4 is (x + y)^2 squared, we can say that:
(x + y)^4 = 16^2
which is 256, giving us answer choice E.
x^2 + 2xy + y^2 = 16
which can be reformatted to:
(x + y)^2 = 16
Since (x + y)^4 is (x + y)^2 squared, we can say that:
(x + y)^4 = 16^2
which is 256, giving us answer choice E.
gmatclubot
Re: If x^2 + y^2 = 16 - 2xy, then (x + y)^4 = [#permalink]
25 Mar 2018, 19:45
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