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by ketkoag » Sat Mar 28, 2009 3:23 am
If x = 0 and x = (8xy-16y^2)^.5
, then, in terms of y, x =

A. – 4y
B. y/4
C. y
D. 4y
E. 4y^2

OA: D
Please tell me why x=0 is given in the question.
I think i never used it while g\solving the problem for the correct answer.
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by DanaJ » Sat Mar 28, 2009 3:38 am
It's OK to start by ignoring the fact that x = 0. Use the equation you've got:

x = (8xy-16y^2)^.5 - raise to the second power to get that:
x^2 = 8xy - 16y^2 - take everything to the left side to get:
x^2 - 8xy + 16y^2 = 0 - this is a perfect square, so you get that:
(x - 4y)^2 = 0.
x - 4y = 0
x = 4y.

Maybe that x = 0 is just there to confuse you or smth.... I don't get it either.
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by 2ndShot » Sat Mar 28, 2009 6:08 am
In fact, it does not even make sense with x=0. In that case y=0 and all answer choices are correct. Probably it was x !=0 (not equal)?
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