Algebra

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by Brent@GMATPrepNow » Mon Oct 15, 2012 8:03 am
neeti2711 wrote:If x≠0 and x=√ (4xy-4y^2), then in terms of y, x=

(A) 2y
(B) y
(C) y/2
(D) 4y^2/(1-2y)
(E) -2y
x=√(4xy-4y^2)
Square both sides to get: x^2 = 4xy-4y^2
Set equal to zero: 4y^2 - 4xy + x^2 = 0
Factor: (2y - x)(2y - x) = 0
This tells us that 2y - x = 0
In other words, 2y = x

Answer = A

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by GMATGuruNY » Mon Oct 15, 2012 12:30 pm
neeti2711 wrote:If x≠0 and x=√ (4xy-4y^2), then in terms of y, x=

(A) 2y
(B) y
(C) y/2
(D) 4y^2/(1-2y)
(E) -2y
An alternate approach is to plug in a value for y and solve for x.
Let y=1.
Plugging y=1 into x=√(4xy-4y²), we get:

x = √(4x*1 - 4*1²)

x = √(4x - 4)

x² = 4x - 4

x² - 4x + 4 = 0

(x-2)² = 0

x=2. This is our target.

Now we plug y=1 into the answers to see which yield our target of 2.
Only A works:
2y = 2*1 = 2.

The correct answer is A.
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