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Is x=y?

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Is x=y? [#permalink]

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New post 11 Aug 2017, 01:02
2
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A
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Difficulty:

  55% (hard)

Question Stats:

56% (00:44) correct 44% (00:54) wrong based on 116 sessions

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Is \(x=y?\)

1) \(x^4+y^4=2x^2y^2\)
2) \(x^4+y^4=0\)

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Re: Is x=y? [#permalink]

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New post 11 Aug 2017, 01:36
2
MathRevolution wrote:
Is \(x=y?\)

1) \(x^4+y^4=2x^2y^2\)
2) \(x^4+y^4=0\)



(1) \(x^4+y^4 -2x^2y^2 = 0 \implies (x^2-y^2)^2=0 \implies x^2-y^2=0 \implies (x+y)(x-y)=0\) insufficient.

(2) \(x^4+y^4=0 \implies x=y=0\). Sufficient.

The answer is B
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Re: Is x=y? [#permalink]

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New post 11 Aug 2017, 13:35
Shouldn't we add a condition that x and y are integers.
For statement B, for example, x=1/2 and y=4th root of 15/2 can also be solutions.
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Re: Is x=y? [#permalink]

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New post 12 Aug 2017, 03:24
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Devbek wrote:
Shouldn't we add a condition that x and y are integers.
For statement B, for example, x=1/2 and y=4th root of 15/2 can also be solutions.


Confused on your point.May I know more details about it?


Statement (2)
    IMO, If each number has even exponent and the sum of all numbers equals to Zero, each number must be equal to Zero.Hence X=Y=0


So Statement (2) is Sufficient and Correct Answer is B
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Re: Is x=y? [#permalink]

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New post 12 Aug 2017, 13:46
AbdurRakib wrote:
Devbek wrote:
Shouldn't we add a condition that x and y are integers.
For statement B, for example, x=1/2 and y=4th root of 15/2 can also be solutions.


Confused on your point.May I know more details about it?


Statement (2)
    IMO, If each number has even exponent and the sum of all numbers equals to Zero, each number must be equal to Zero.Hence X=Y=0


So Statement (2) is Sufficient and Correct Answer is B


OMG. My bad. It is one of those days...
I don't know why but I was sure that x^4+y^4=1 not 0.

Thanks for the answer.
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Re: Is x=y? [#permalink]

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New post 13 Aug 2017, 18:18
==> In the original condition, there are 2 variables (x,y) and in order to match the number of variables to the number of equations, there must be 2 equations. Since there is 1 for con 1) and 1 for con 2), C is most likely to be the answer. By solving con 1) and con 2), for con 1), from \(x^4+y^4-2x^2y^2=0, (x^2-y^2)^2=0\), you get \(x^2=y^2\), and from x=±y, yes and no coexists, hence it is not sufficient. For con 2), you only get x=y=0, hence yes, it is sufficient.

The answer is B.
Answer: B
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Re: Is x=y? [#permalink]

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New post 13 Aug 2017, 20:54
I didn't think B is correct because x could be negative, and y could be positive.
can someone correct my logic please?
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Re: Is x=y? [#permalink]

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New post 25 Aug 2017, 05:52
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pclawong wrote:
I didn't think B is correct because x could be negative, and y could be positive.
can someone correct my logic please?
Thank you


It can't be.

\(x^4\)>=0
\(y^4\) >=0

Therefore in order for \(x^4\) + \(y^4\) =0, x = y= 0
Re: Is x=y?   [#permalink] 25 Aug 2017, 05:52
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