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inqeualities ds

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inqeualities ds

by ern5231 » Tue Aug 18, 2009 4:39 pm
Is x^2+x>-y^2?
(1) x^2+y^2=1
(2) x>0
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Re: inqeualities ds

by yezz » Tue Aug 18, 2009 5:30 pm
ern5231 wrote:Is x^2+x>-y^2?
(1) x^2+y^2=1
(2) x>0
the question is wrong

x^2+x>-y^2

-(x^2+x)<y^2 is sure no need for either statments.

plz revise and re post
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by varunkh70 » Tue Aug 18, 2009 9:55 pm
How did you conclude that?
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by anand0408 » Wed Aug 19, 2009 7:35 am
ya how did you conclude that?

What if x = -0.1...then x^2 = 0.01

therefore: x^2 + x = 0.01-0.1 = -0.09

If: y=0.4, y^2 = 0.16..therefore: -(y^2) = -0.16

Therefore: x^2+x > -(y^2)


Anyway IMO:

Answer is B
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by rish » Wed Aug 19, 2009 1:17 pm
IMO B

from (1), -1<=x<=1

is x^2 + x > -y^2
1 - y^2 + x > -y^2 if -1 < x <= 1
1 - y^2 + x = -y^2 if x = -1

INSUF

from (2), x > 0

x^2 + x > 0 > -y^2

SUFF
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by PussInBoots » Wed Aug 19, 2009 4:50 pm
x^2+x>-y^2 <=> x^2 + y^2 + x > 0?
(1) x^2+y^2=1
INSUFF, with x = -1, y = 0, the equation is exactly 0
(2) x>0
SUFF

(B)
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