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# 65288;x+y）^2<X^2. Which of the following is

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Manager
Joined: 04 Dec 2005
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（x+y）^2<X^2. Which of the following is possible?

Ⅰ. xy<0

Ⅱ. y（y+2x）<0

Ⅲ. y<X
Senior Manager
Joined: 19 Feb 2005
Posts: 486
Location: Milan Italy
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I'll give it a try
let's first cancel out x^2

y^2+2xy<0

y(y+2x)<0 -> Great! St 2 is TRUE and possible, isn't it?

what follows afterwards is not very reliable

Case I: y<0 and y+2x>0 -> y>-2x -> -y[this is positive]<[the sign flips]2x[if 2x is greater than a positive number, than x is positive]

Great! St 1 is possible because y is negative and x is positive , but this also means that St 3 is possible because a negative number is always smaller than a positive one.

Don't think it is necessary to go ahead with Case II, since the question asks only for possibility, not necessity

Hope this helps and the other guys correct me as long as I'm wrong
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VP
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Re: PS - Inequivality [#permalink]  04 Jun 2006, 09:45
withme wrote:
（x+y）^2<X^2. Which of the following is possible?

Ⅰ. xy<0
Ⅱ. y（y+2x）<0
Ⅲ. y<X

given that
（x+y）^2<X^2
x^2+2xy+y^2<X^2
2xy+y^2<0 ===== st 1 is also true cuz when y^2 is +ve, 2xy must be -ve. so xy<0.
y(2x+y) <0 ===== st 2 is true..

but we do not know which is -ve and +ve. either could be +ve or -ve. so III can be true (however is possible). in that case we can say I, II, III.
Re: PS - Inequivality   [#permalink] 04 Jun 2006, 09:45
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