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If x<>0 and x=^(1/2) , then, in terms of y, x= (A) 2y

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If x<>0 and x=^(1/2) , then, in terms of y, x= (A) 2y [#permalink] New post 08 Sep 2003, 11:58
If x<>0 and x=[ 4xy - 4(y^2) ]^(1/2) , then, in terms of y, x=


(A) 2y
(B) y
(C) y/2
(D) (-4(y^2))/(1-4y)
(E) -2y



Any ideas on how to solve this WITHOUT backsolving?
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Re: Square root Problem [#permalink] New post 08 Sep 2003, 15:48
MartinMag wrote:
If x<>0 and x=[ 4xy - 4(y^2) ]^(1/2) , then, in terms of y, x=


(A) 2y
(B) y
(C) y/2
(D) (-4(y^2))/(1-4y)
(E) -2y



Any ideas on how to solve this WITHOUT backsolving?



Step 1..

x^2 = 4xy - 4y^2

Step 2

x^2 - 4xy + 4y^2 = 0

Step 3

(x - 2y) ^2 = 0

Step 4

x -2y = 0

therefore x =2y

thanks
praetorian
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 [#permalink] New post 08 Sep 2003, 18:54
Step 2

x^2 - 4xy + 4y^2 = 0

Step 3

(x - 2y) ^2 = 0


NICE MOVE!!! Thanks
  [#permalink] 08 Sep 2003, 18:54
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If x<>0 and x=^(1/2) , then, in terms of y, x= (A) 2y

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