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Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0

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earnit
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Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   Updated on: 10 Aug 2022, 01:43
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Is x > y?

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

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

Originally posted by earnit on 03 Oct 2014, 15:38.
Last edited by Bunuel on 10 Aug 2022, 01:43, edited 2 times in total.
Edited the question.
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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   03 Oct 2014, 15:48
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Is x > y?


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

Re-arrange: \(x + y = -2\sqrt{xy}\);

Square: \(x^2 + 2xy + y^2 = 4xy\);

Re-arrange: \(x^2 - 2xy + y^2 = 0\);

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

\(x=y\).

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.
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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   25 Feb 2016, 20:15
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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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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] 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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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   26 Aug 2018, 03:32
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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? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] 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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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   19 Sep 2018, 05:10
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T1101 wrote:

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!

____________________
Consider x = y = -1.
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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   19 Dec 2019, 13:50
Statement 1:
x+2 √ (xy)+ y=0
or, (√x)^2+2. √x.√y.+(√y)^2=0
or, (√x+√y)^2=0
or, (√x+√y)=0
or, √x=-√y
Since any value inside the square root CAN NOT be negative, x and y are both ZERO or BOTH positive.
If x and y are both positive, the second segment of the equation will be negative and first portion will be positive, which is impossible in equation.
So, x and y can NOT be both positive. Thus, x and y are both ZERO.
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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   19 Dec 2019, 14:03
Statement 2: x^2 - y^2 = 0
or, (x+y) (x-y)=0, implies that
(1) only (x+y) is zero, or
(2) only(x-y) is zero or
(3) Both (x+y) and (x-y) is zero.

for example, if x+y=0
x and y can be (1, -1 : YES to main question) or (-1 and 1, NO to main question)
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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   19 Dec 2019, 14:06
Number picking can also be a easy solution for statement 2.
Statement 2: x^2 - y^2 = 0
if x=1, y=1 (No to main question) =======> (1)^2-(1)^2=0
if x=1, y=-1 (YES to main question) =======> (1)^2-(-1)^2=0

Insufficient
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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink] New post   11 Jan 2024, 02:08
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Re: Is x > y? (1) x + 2(xy)^(1/2) + y = 0 (2) x^2 - y^2 = 0 [#permalink]
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