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For all real numbers x and y, let x# y = (xy)^2 − x + y^2. What is the

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New post 20 May 2017, 10:35
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A
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Difficulty:

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Question Stats:

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For all real numbers \(x\) and \(y\), let x#y = \((xy)^2 − x + y^2\) . What is the value of y that makes x # y equal to \(–x\) for all values of \(x\) ?

(A) 0
(B) 2
(C) 5
(D) 7
(E) 10

Source: Nova GMAT
Difficulty Level: 500

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Re: For all real numbers x and y, let x# y = (xy)^2 − x + y^2. What is the  [#permalink]

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New post 20 May 2017, 10:59
We have to make x#y = -x
or (x^2)(y^2) -x + y^2 = -x
y^2 (x^2 + 1) = 0

This equation will obviously be true for all values of x if y^2 = 0 or if y=0

Hence A
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Re: For all real numbers x and y, let x# y = (xy)^2 − x + y^2. What is the  [#permalink]

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New post 20 May 2017, 12:05
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The only possibility for the equation,
when x#y = \((xy)^2 − x + y^2\) = -x is when \((xy)^2 + y^2 = 0\).
Only one possibility can give us a 0 for the expression at all possible values of x(y=0)
If y =0, the expression will always yield a zero and hence the value of x#y will always be -x

Therefore, Option A is the correct answer.
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New post 17 Feb 2018, 00:55
for all numbers x and y ,let x club y be defined as x club y=x^2 -2xy +y^2 what is the value of (2 club 4) club 8?
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Re: For all real numbers x and y, let x# y = (xy)^2 − x + y^2. What is the  [#permalink]

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New post 18 Feb 2018, 02:19
(xy)^2-x+y^2=-x
So (y^2)(x^2+1)=0
As x^2+1 cannot be zero.
So y=0
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Re: For all real numbers x and y, let x# y = (xy)^2 − x + y^2. What is the  [#permalink]

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New post 15 Aug 2018, 06:28
Top Contributor
SajjadAhmad wrote:
For all real numbers x and y, let x# y = (xy)^2 − x + y^2 . What is the value of y that makes x # y equal to –x for all values of x ?

(A) 0
(B) 2
(C) 5
(D) 7
(E) 10


APPROACH #1
We want the following equation to hold true: x # y = –x
Replace x # y with its equivalent to get: (xy)² − x + y² = -x
Add x to both sides to get: (xy)² + y² = 0
Simplify (xy)² to get: x²y² + y² = 0
Factor out the y² to get: y²(x² + 1) = 0
So, the equation will hold true when EITHER y² = 0 OR x² + 1 = 0
If y² = 0, then y =0

Answer: A

Cheers,
Brent
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Re: For all real numbers x and y, let x# y = (xy)^2 − x + y^2. What is the  [#permalink]

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New post 15 Aug 2018, 06:29
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Top Contributor
SajjadAhmad wrote:
For all real numbers x and y, let x# y = (xy)^2 − x + y^2 . What is the value of y that makes x # y equal to –x for all values of x ?

(A) 0
(B) 2
(C) 5
(D) 7
(E) 10


APPROACH #2 - Test the answer choices

A) 0
Take x#y = (xy)² − x + y², and replace y with 0
We get: [(x)(0)]² − x + 0² = [0]² − x + 0
= -x
So, when y = 0, x#y = -x
PERFECT!

Answer: A

Cheers,
Brent
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Re: For all real numbers x and y, let x# y = (xy)^2 − x + y^2. What is the   [#permalink] 24 Aug 2019, 14:15
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