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If x does not equal to 0 and x = root(4xy - 4y^2), then in

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If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 29 Oct 2010, 23:42
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If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

A. 2y
B. y
C. y/2
D. 4y^2/(1-4y)
E. -2y
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 30 Oct 2010, 02:03
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tonebeeze wrote:
If x does not = 0 and \(x = \sqrt{4xy-4y^2}\) , then in terms of y, x =

a. 2y
b. y
c. \(y/2\)
d. \(-4^2/1-4y\)
e. -2y


\(x = \sqrt{4xy-4y^2}\)

Square both sides

\(x^2=4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\)
\(\Rightarrow x-2y=0\)
\(x=2y\)

Answer : (A)
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 19 Feb 2011, 04:11
is it possible to solve this one by plugging in numbers? any help will be appreciated.

thanks.
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 19 Feb 2011, 06:49
Since the question and options both involve variables, plugging the numbers is probably going to make it more complicated. Better to solve it by squaring both sides and solving for x
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 19 Feb 2011, 07:02
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well, i kinda figured it out, and i think it will be good to know both systems to solve.
in order to make it more simpler i will plug in Y=1

so i will have

x^2=4x-4
x^2-4x+4=0
(x-2)^2=0
x=2.

if u look on the answers - and plug y=1, the only one that will give u x=2 is A.

therefore u have the answer very easy and quickly with almost no algebra.

what do u think?

thanks.
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 19 Feb 2011, 07:42
1
Well - even after plugging you still need to square both sides and solve for x and then you need to plug it in four choices ( which is up to five extra steps - it just so happens that in this case, only one choice is suitable). Therefore, as a general rule, I would prefer to not plug in if both question and choices have variables.
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If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 21 Nov 2013, 02:06
1
If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

A. 2y
B. y
C. y/2
D. 4y^2/(1-4y)
E. -2y
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 21 Nov 2013, 02:15
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 21 Nov 2013, 02:16
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 11 Jul 2015, 13:12
tonebeeze wrote:
If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

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


This is not a 700+ question!
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If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 11 Dec 2015, 23:08
Bunuel wrote:
registerincog wrote:
If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

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


Square the expression: \(x^2 = 4xy - 4y^2\) --> \(x^2-4xy+4y^2=(x-2y)^2=0\) --> \(x=2y\).

Answer: A.



Can you explain how you were able to get to (x-2y)^2? What is the strategy to know how to determine that quickly?
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 12 Dec 2015, 08:31
gmatczar wrote:
Bunuel wrote:
registerincog wrote:
If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

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


Square the expression: \(x^2 = 4xy - 4y^2\) --> \(x^2-4xy+4y^2=(x-2y)^2=0\) --> \(x=2y\).

Answer: A.



Can you explain how you were able to get to (x-2y)^2? What is the strategy to know how to determine that quickly?


This is basic algebra.

\(a^2 - 2ab + b^2 = (a - b)^2\).
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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New post 25 Aug 2017, 11:05
tonebeeze wrote:
If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

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



We can square both sides of the given equation and obtain:

x^2 = 4xy - 4y^2

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

(x - 2y)(x - 2y) = 0

(x - 2y)^2 = 0

x - 2y = 0

x = 2y

Answer: A
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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in  [#permalink]

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Re: If x does not equal to 0 and x = root(4xy - 4y^2), then in   [#permalink] 28 Aug 2018, 15:04
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