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If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =

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If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 26 Mar 2011, 06:34
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If x≠0 and \(x = \sqrt{4xy-4y^2}\), then, in terms of y, x =

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

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petrifiedbutstanding

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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 26 Mar 2011, 06:51
2
5
petrifiedbutstanding wrote:
If x≠0 and x = \sqrt{4xy-4y^2} , then, in
terms of y, x =

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

Can someone please help?


\(x = \sqrt{4xy-4y^2}\)
Square both sides;

\(x^2 = 4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\) Note: \((a-b)^2=a^2+b^2-2ab\)
\((x-2y)^2=0\)
\(x-2y=0\)
\(x=2y\)

Ans: "A"
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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 26 Mar 2011, 06:49
1
@petrifiedbutstanding:-

Well it means :
Squaring we get:
x^2 =4y(x-y)
Assume X^2=m >>>>>>>>>>>1
Than ,m=4y(m^1/2-y)
on solving:m-4y(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 = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 26 Mar 2011, 06:53
If i am correct donot hesitate to kudo me .

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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 26 Mar 2011, 06:57
Thnkss Flukee!!:)
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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 26 Mar 2011, 12:27
x^2 = 4xy-4(y^2) => (x-2y)^2 = 0

=> x = 2y

Answer A
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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 26 Mar 2011, 17:20
x = sqrt{4xy-4y^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 = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 14 Jul 2018, 18:11
petrifiedbutstanding wrote:
If x≠0 and \(x = \sqrt{4xy-4y^2}\), then, in terms of y, x =

A. 2y
B. y
C. y/2
D. -(4y^2)/(1-2y)
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 = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 29 Dec 2018, 01:42
petrifiedbutstanding wrote:
If x≠0 and \(x = \sqrt{4xy-4y^2}\), then, in terms of y, x =

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

Project PS Butler : Question #105


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

squaring both sides-

x^2=4xy - 4y^2

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

hence A)
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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 03 Jan 2019, 12:22
1
fluke wrote:
petrifiedbutstanding wrote:
If x≠0 and x = \sqrt{4xy-4y^2} , then, in
terms of y, x =

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

Can someone please help?


\(x = \sqrt{4xy-4y^2}\)
Square both sides;

\(x^2 = 4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\) Note: \((a-b)^2=a^2+b^2-2ab\)
\((x-2y)^2=0\)
\(x-2y=0\)
\(x=2y\)

Ans: "A"



Gladiator59


Hi Gladi,

can you please breakdown this for me

how from here \(x^2+4y^2-4xy=0\)

we got this \((x-2y)^2=0\)


and how from this \((x-2y)^2=0\)

we got this \(x-2y=0\)

thank you :)
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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 03 Jan 2019, 12:26
1
dave13 wrote:
fluke wrote:
petrifiedbutstanding wrote:
If x≠0 and x = \sqrt{4xy-4y^2} , then, in
terms of y, x =

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

Can someone please help?


\(x = \sqrt{4xy-4y^2}\)
Square both sides;

\(x^2 = 4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\) Note: \((a-b)^2=a^2+b^2-2ab\)
\((x-2y)^2=0\)
\(x-2y=0\)
\(x=2y\)

Ans: "A"



Gladiator59


Hi Gladi,

can you please breakdown this for me

how from here \(x^2+4y^2-4xy=0\)

we got this \((x-2y)^2=0\)


and how from this \((x-2y)^2=0\)

we got this \(x-2y=0\)

thank you :)


What number when squared gives 0? Only 0^2 = 0. So, if \((x-2y)^2=0\), then \(x-2y=0\).
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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 03 Jan 2019, 18:10
1
dave13,

Also, (a-b)^2= a^2-2ab+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,
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If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 04 Jan 2019, 07:43
Gladiator59 wrote:
dave13,

Also, (a-b)^2= a^2-2ab+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 \(x^2+4y^2-4xy=0\) that a = x and b = -2y ? :?
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Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 04 Jan 2019, 07:51
1
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 light-bulb 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 \(x^2+4y^2-4xy=0\) that a = x and b = -2y ? :?

Attachments

Algebraic-Identities.png
Algebraic-Identities.png [ 17.5 KiB | Viewed 611 times ]


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If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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New post 16 Feb 2019, 10:44
petrifiedbutstanding wrote:
If x≠0 and \(x = \sqrt{4xy-4y^2}\), then, in terms of y, x =

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

Project PS Butler : Question #105


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Overthinking costs you time

Now if you see \(x = \sqrt{4xy-4y^2}\)

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 = (4xy-4y^2)^1/2, then, in terms of y, x =   [#permalink] 16 Feb 2019, 10:44
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