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Is x > y?

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Is x > y?  [#permalink]

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New post Updated on: 03 Oct 2014, 15:43
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Is x > y?

(1) \(x + 2\sqrt{xy} + y = 0\)

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

Originally posted by earnit on 03 Oct 2014, 15:38.
Last edited by Bunuel on 03 Oct 2014, 15:43, edited 1 time in total.
Edited the question.
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New post 03 Oct 2014, 15:48
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Re: Is x > y?  [#permalink]

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New post 25 Feb 2016, 20:15
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1
i arrived to the answer choice differently...kind of..

2 - started with this one, since it is easier - clearly insufficient: (x+y)(x-y)=0, so either x=y - which is a definite no, or x=-y. if y=positive, then x<y. if y=negative, then x>y. so 2 outcomes...B and D - out.

now 1.
x+y=-sqrt(4xy)
square everything:
x^2+y^2+2xy = 4xy
x^2+y^2=2xy
x^2+y^2-2xy=0
(x-y)(x-y)=0
we know for sure that x=y. so the answer to the main question is only NO.

A
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Is x > y?  [#permalink]

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New post 25 Aug 2018, 22:51
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Bunuel wrote:
Is x > y?

(1) \(x + 2\sqrt{xy} + y = 0\) --> \((\sqrt{x}+\sqrt{y})^2=0\) --> \(\sqrt{x}+\sqrt{y}=0\). The sum of two non-negative values is 0, only when both are 0, so x=y=0: x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

Answer: A.


Hi Bunuel
From my point of view, in statement 1 we know that xy >= 0, but we can't conclude that x>=0 and y>=0 . Thus, writing \(\sqrt{x}\) and \(\sqrt{y}\) seems not quite right.
Can you please check it?
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New post 26 Aug 2018, 03:32
1
Izzyjolly wrote:
Bunuel wrote:
Is x > y?

(1) \(x + 2\sqrt{xy} + y = 0\) --> \((\sqrt{x}+\sqrt{y})^2=0\) --> \(\sqrt{x}+\sqrt{y}=0\). The sum of two non-negative values is 0, only when both are 0, so x=y=0: x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

Answer: A.


Hi Bunuel
From my point of view, in statement 1 we know that xy >= 0, but we can't conclude that x>=0 and y>=0 . Thus, writing \(\sqrt{x}\) and \(\sqrt{y}\) seems not quite right.
Can you please check it?


Good catch. You are absolutely right. Edited the post. Thank you.
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Re: Is x > y?  [#permalink]

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New post 19 Sep 2018, 05:08
Bunuel wrote:
Izzyjolly wrote:
Bunuel wrote:
Is x > y?

(1) \(x + 2\sqrt{xy} + y = 0\) --> \((\sqrt{x}+\sqrt{y})^2=0\) --> \(\sqrt{x}+\sqrt{y}=0\). The sum of two non-negative values is 0, only when both are 0, so x=y=0: x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

Answer: A.


Hi Bunuel
From my point of view, in statement 1 we know that xy >= 0, but we can't conclude that x>=0 and y>=0 . Thus, writing \(\sqrt{x}\) and \(\sqrt{y}\) seems not quite right.
Can you please check it?


Good catch. You are absolutely right. Edited the post. Thank you.


Hi Bunuel

I do not understand why we can not infer from the red part that x and y have to be 0? Why do we need to square the whole expression in statement 1 and can not use the square roots of x and y? Could you explain it in more detail?

Thank you!
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New post 19 Sep 2018, 05:10
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Re: Is x > y?   [#permalink] 19 Sep 2018, 05:10
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