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

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Manager
Joined: 19 Oct 2010
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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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26 Mar 2011, 07:34
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Question Stats:

68% (01:53) correct 32% (02:24) wrong based on 281 sessions

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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

Project PS Butler : Question #105

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petrifiedbutstanding
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Joined: 20 Dec 2010
Posts: 1571
Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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26 Mar 2011, 07:51
4
6
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

$$x = \sqrt{4xy-4y^2}$$
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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Manager
Joined: 11 Feb 2011
Posts: 108
Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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26 Mar 2011, 07:49
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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26 Mar 2011, 07:53
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 = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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26 Mar 2011, 07:57
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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26 Mar 2011, 13:27
x^2 = 4xy-4(y^2) => (x-2y)^2 = 0

=> x = 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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26 Mar 2011, 18:20
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

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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14 Jul 2018, 19:11
1
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

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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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29 Dec 2018, 02:42
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

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

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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Joined: 09 Mar 2016
Posts: 1230
Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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03 Jan 2019, 13:22
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

$$x = \sqrt{4xy-4y^2}$$
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"

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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Joined: 02 Sep 2009
Posts: 58464
Re: If x≠0 and x = (4xy-4y^2)^1/2, then, in terms of y, x =  [#permalink]

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03 Jan 2019, 13:26
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

$$x = \sqrt{4xy-4y^2}$$
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"

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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03 Jan 2019, 19:10
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,
G

Posted from my mobile device
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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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04 Jan 2019, 08:43
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

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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04 Jan 2019, 08:51
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:

but where do you see here $$x^2+4y^2-4xy=0$$ that a = x and b = -2y ?

Attachments

Algebraic-Identities.png [ 17.5 KiB | Viewed 1821 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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16 Feb 2019, 11:44
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

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, 11:44
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