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If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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26 Mar 2011, 06:34
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If x≠0 and \(x = \sqrt{4xy4y^2}\), then, in terms of y, x = A. 2y B. y C. y/2 D. (4y^2)/(12y) E. 2y Project PS Butler : Question #105 Subscribe to get Daily Email  Click Here  Subscribe via RSS  RSS
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petrifiedbutstanding




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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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26 Mar 2011, 06:51
petrifiedbutstanding wrote: If x≠0 and x = \sqrt{4xy4y^2} , then, in terms of y, x =
A. 2y B. y C. y/2 D. (4y^2)/(12y) E. 2y
Can someone please help? \(x = \sqrt{4xy4y^2}\) Square both sides; \(x^2 = 4xy4y^2\) \(x^2+4y^24xy=0\) \((x2y)^2=0\) Note: \((ab)^2=a^2+b^22ab\) \((x2y)^2=0\) \(x2y=0\) \(x=2y\) Ans: "A"
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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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26 Mar 2011, 06:49
@petrifiedbutstanding: Well it means : Squaring we get: x^2 =4y(xy) Assume X^2=m >>>>>>>>>>>1 Than ,m=4y(m^1/2y) on solving:m4y(m)^1/2+4y^2=0 >> quadratic in y s solve for y. Y comes out to be =(m^1/2)/2 Hence,put m =x^2 from 1. So x=2y The answer shd be 2y. i.e A. Pls confirm it .
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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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26 Mar 2011, 06:53
If i am correct donot hesitate to kudo me . Your support is required to bang GMAT.
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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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26 Mar 2011, 06:57
Thnkss Flukee!!:)
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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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26 Mar 2011, 12:27
x^2 = 4xy4(y^2) => (x2y)^2 = 0
=> x = 2y
Answer A



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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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26 Mar 2011, 17:20
x = sqrt{4xy4y^2} Squaring both side : => x^2 = 4xy  4y^2 Substituting answer choices, e.g., choice A : => (2y)^2 = 4*2y*y  4y^2 => 4y^2  8y^2 =  4y^2 So Answer  A Or, x^2 + 4y^2  4xy = 0 => (x 2y)^2 = 0 => x = 2y Note  I thought simplifying further would be faster and more accurate than plugging the answer choices in this case, but situations may vary. Also, the OA seems to be wrong.
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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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14 Jul 2018, 18:11
petrifiedbutstanding wrote: If x≠0 and \(x = \sqrt{4xy4y^2}\), then, in terms of y, x =
A. 2y B. y C. y/2 D. (4y^2)/(12y) E. 2y We first square both sides of the equation and then simplify. We have: x^2 = 4xy  4y^2 4y^2  4xy + x^2 = 0 (2y  x)^2 = 0 2y  x = 0 2y = x Answer: A
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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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29 Dec 2018, 01:42
petrifiedbutstanding wrote: If x≠0 and \(x = \sqrt{4xy4y^2}\), then, in terms of y, x = A. 2y B. y C. y/2 D. (4y^2)/(12y) E. 2y Project PS Butler : Question #105 Subscribe to get Daily Email  Click Here  Subscribe via RSS  RSS \(x = \sqrt{4xy4y^2}\) squaring both sides x^2=4xy  4y^2 => 4y^2  4xy + x^2 = 0 => (2yx)^2 = 0 => 2y  x = 0 => x=2y hence A)



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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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03 Jan 2019, 12:22
fluke wrote: petrifiedbutstanding wrote: If x≠0 and x = \sqrt{4xy4y^2} , then, in terms of y, x =
A. 2y B. y C. y/2 D. (4y^2)/(12y) E. 2y
Can someone please help? \(x = \sqrt{4xy4y^2}\) Square both sides; \(x^2 = 4xy4y^2\) \(x^2+4y^24xy=0\) \((x2y)^2=0\) Note: \((ab)^2=a^2+b^22ab\) \((x2y)^2=0\) \(x2y=0\) \(x=2y\) Ans: "A" Gladiator59 Hi Gladi, can you please breakdown this for me how from here \(x^2+4y^24xy=0\) we got this \((x2y)^2=0\) and how from this \((x2y)^2=0\) we got this \(x2y=0\) thank you



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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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03 Jan 2019, 12:26



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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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03 Jan 2019, 18:10
dave13, Also, (ab)^2= a^22ab+b^2 is an identity which is true for any a,b Put a = x and b = 2y ... And you'll get the first part of your question. Also, as Bunuel sir mentioned.. x^2 = 0 .. means X=0 Hope this answers.. let me know. Regards, G Posted from my mobile device
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If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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04 Jan 2019, 07:43
Gladiator59 wrote: dave13, Also, (ab)^2= a^22ab+b^2 is an identity which is true for any a,b Put a = x and b = 2y ... And you'll get the first part of your question. Also, as Bunuel sir mentioned.. x^2 = 0 .. means X=0 Hope this answers.. let me know. Regards, G Posted from my mobile device gladiator thanks Gladi but where do you see here \(x^2+4y^24xy=0\) that a = x and b = 2y ?



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Re: If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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04 Jan 2019, 07:51
This is pattern recognition. When you have two perfect squares being added ( here \(x^2\) and \((2y)^2 = 4y^2\) are definitely perfect squares) you need to immediately search for the product terms 2ab ... which is present here. \(2*(x)*(2y) = 4xy\) This should flash like a lightbulb and will come only from drilling the identities in. Here is attached a list of identities which you might want to try & drill in. At least the binomial ones dave13 wrote: gladiator thanks Gladi but where do you see here \(x^2+4y^24xy=0\) that a = x and b = 2y ?
Attachments
AlgebraicIdentities.png [ 17.5 KiB  Viewed 611 times ]
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If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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16 Feb 2019, 10:44
petrifiedbutstanding wrote: If x≠0 and \(x = \sqrt{4xy4y^2}\), then, in terms of y, x = A. 2y B. y C. y/2 D. (4y^2)/(12y) E. 2y Project PS Butler : Question #105 Subscribe to get Daily Email  Click Here  Subscribe via RSS  RSSOverthinking costs you time Now if you see \(x = \sqrt{4xy4y^2}\) squaring both sides \(x^2 = 4xy {2y}^2\) \(x^2 + {2y}^2  4xy = 0\) \((x  2y)^2\) = 0 x = 2y A
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If x≠0 and x = (4xy4y^2)^1/2, then, in terms of y, x =
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