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# If (x+y)^2<x^2, which of the following must be true? I.

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Current Student
Joined: 11 May 2008
Posts: 551
If (x+y)^2<x^2, which of the following must be true? I.  [#permalink]

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29 Aug 2008, 04:11
If (x+y)^2<x^2, which of the following must be true?
I. x*y<0
II. y<x
III. y*(y+2x)<0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III

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Manager
Joined: 15 Jul 2008
Posts: 205

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29 Aug 2008, 04:56
arjtryarjtry wrote:
If (x+y)^2<x^2, which of the following must be true?
I. x*y<0
II. y<x
III. y*(y+2x)<0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III- this one for me

expand the LHS and cancel the x^2 terms on both sides. 2xy + y^2 < 0 here y^2 is +ve. so xy <0.
also y(2x+y) <0
1 and 3.
Director
Joined: 01 Jan 2008
Posts: 601

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29 Aug 2008, 05:32
i second E. Similar logic: 2xy+y^2 < 0 -> 2xy < - y^2 -> xy < 0 and y*(2x+y) < 0
Director
Joined: 01 Aug 2008
Posts: 627

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29 Aug 2008, 08:18
I am getting even option II also right which is y < x...

(x+y)^2 < x^2

Case 1):
x+y < x -----> y < 0
Case 2):
x+y < -x ------> 2x+ y < 0
Case 3):
-(x+y) < x = 2x>y ------>y<x

Am I wrong with my case 3?
SVP
Joined: 07 Nov 2007
Posts: 1728
Location: New York

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29 Aug 2008, 08:25
ugimba wrote:
I am getting even option II also right which is y < x...

(x+y)^2 < x^2

Case 1):
x+y < x -----> y < 0
Case 2):
x+y < -x ------> 2x+ y < 0
Case 3):
-(x+y) < x = 2x>y ------>y<x

Am I wrong with my case 3?

do it this way.

(x+y)^2 < x^2
--> x^2+y^2+2xy <x^2
--> y^2+2xy<0
--> y(y+2x)<0
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Director
Joined: 01 Aug 2008
Posts: 627

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29 Aug 2008, 08:30
x2suresh thanks for your response. I agree choice E will be the correct answer using your way of solving the queation.

If I use , sqrt ign both sides, I am getting option II which is (y<x)
Shouldn't I do a sqroot to solve the equations?
SVP
Joined: 07 Nov 2007
Posts: 1728
Location: New York

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Updated on: 29 Aug 2008, 10:26
ugimba wrote:
I am getting even option II also right which is y < x...

(x+y)^2 < x^2

Case 1):
x+y < x -----> y < 0
Case 2):
x+y < -x ------> 2x+ y < 0
Case 3):
-(x+y) < x = 2x>y ------>y<x

Am I wrong with my case 3?

Sorry I thought you are talking about III.

(x+y)^2 < x^2 --> means |x+y| <|x|

here you will get the following four solutions
x+y>0 and x>0 --> x+y <x --> y<0
x+y>0 and x<0 --> x+y < -x --> x+x+y<0 --> x<0
x+y<0 and x>0 --> -( x+y) <x --> y<2x
we can say --> y<x (This is only when X>0 AND X+Y <0)
this is one possible soltuions what if X<0.. it leads to other solutions..

Question stem says which of the following "MUST BE TRUE" -- this is ruled out.
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Originally posted by x2suresh on 29 Aug 2008, 09:23.
Last edited by x2suresh on 29 Aug 2008, 10:26, edited 1 time in total.
Director
Joined: 01 Aug 2008
Posts: 627

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29 Aug 2008, 10:19
Thanks x2suresh for the explanation.

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This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: which is ...? &nbs [#permalink] 29 Aug 2008, 10:19
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