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If x not eq to 0 and x=sqrt(4xy-4y^2) , then, in terms of y,

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Director
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If x not eq to 0 and x=sqrt(4xy-4y^2) , then, in terms of y, [#permalink] New post 17 Jan 2006, 04:15
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If x not eq to 0 and x=sqrt(4xy-4y^2) , then, in terms of y, x=
(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/(1-2y)
(E) -2y
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Last edited by Yurik79 on 17 Jan 2006, 07:17, edited 1 time in total.
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 [#permalink] New post 17 Jan 2006, 06:05
Yurik,

are you sure that the stem and the answers are correct?

Plugged in the answers, but none of them is true.
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Re: One more Algebra [#permalink] New post 17 Jan 2006, 06:31
Yurik79 wrote:
If x not eq to 0 and x=sqrt(4xy-y^2) , then, in terms of y, x=
(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/(1-2y)
(E) -2y


I guess there's some typo , it should be sqrt(2xy-y^2)
Anyway, we can solve to get the answer, even though the result doesn't appear among answer choices.
x^2 = [sqrt(4xy-y^2)] ^2 = 4xy- y^2
---> x^2-4xy+ y^2 = 0
We can consider this a quaratic equation of variable x and constant y so we can find x in term of y
Discriminant of this quadratic equation = 16y^2 - 4y^2 = 12 y^2
----> x= [4y + or - sqrt(12y^2)] / 2
= 2y + or - y*sqrt3= y ( 2-sqrt3) or y ( 2+sqrt3)
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 [#permalink] New post 17 Jan 2006, 07:17
allabout wrote:
Yurik,

are you sure that the stem and the answers are correct?

Plugged in the answers, but none of them is true.

Sorry my fault fixed the typo
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Re: One more Algebra [#permalink] New post 17 Jan 2006, 07:22
Yurik79 wrote:
If x not eq to 0 and x=sqrt(4xy-4y^2) , then, in terms of y, x=
(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/(1-2y)
(E) -2y


x=sqrt(4xy-4y^2) ---> x^2= 4xy-4y^2 ---> x^2 - 4xy+4y^2 = (x-2y)^2 =0 ----> x = 2y
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Re: One more Algebra [#permalink] New post 17 Jan 2006, 07:27
laxieqv wrote:
Yurik79 wrote:
If x not eq to 0 and x=sqrt(4xy-4y^2) , then, in terms of y, x=
(A) 2y
(B) y
(C) y/2
(D) (-4y^2)/(1-2y)
(E) -2y


x=sqrt(4xy-4y^2) ---> x^2= 4xy-4y^2 ---> x^2 - 4xy+4y^2 = (x-2y)^2 =0 ----> x = 2y

Laxie, can you see the red hat on my avatar?I am taking it off !!!Good explanation!
The OA is A
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 [#permalink] New post 17 Jan 2006, 07:35
x = 2y

is my answere too.
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 [#permalink] New post 17 Jan 2006, 14:41
A

Square both sides and we get

x^2 = 4xy-4y^2
i.e x^2 - 4xy + 4y^2 = 0
i.e (x -2y)^2 = 0

so x = 2y
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 [#permalink] New post 17 Jan 2006, 16:02
Squaring,

x^2 = 4yx - 4y^2 => x^2 - 4xy - y^2 = (x-2y)^2 = 0.
=> x = 2y.
  [#permalink] 17 Jan 2006, 16:02
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If x not eq to 0 and x=sqrt(4xy-4y^2) , then, in terms of y,

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