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27 Jun 2008, 04:40
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Is x^3+x^2+x<0?
(1) 9<x^2<16
(2) x^3+x<0
(1) 9<x^2<16
(2) x^3+x<0
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27 Jun 2008, 05:25
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ritula
I think its B.
x(x^2+x+1) < 0
if x<0 then this satisfies the equation.
1) x can be 3.1,3.5 or -3.1,-3.5 so insufficient
2) x(x^2 + 1) <0 so x<0 since x^2 + 1 is always positive...sufficient
27 Jun 2008, 05:36
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Answer is (B)
We have \(x^3+x^2+x = x*(x^2+x+1)\) and \(x^2+x+1>0\) for any x
Therefore the sign of \(x^3+x^2+x\) is simply the sign of x
(1) is insufficient since we can have either -4<x<-3 or 3<x<4
(2) is sufficient since \(x^3+x<0\) is equivalent to \(x*(x^2+1)<0\)
As \(x^2+1>0\)for any x,we can conclude \(x<0\)
We have \(x^3+x^2+x = x*(x^2+x+1)\) and \(x^2+x+1>0\) for any x
Therefore the sign of \(x^3+x^2+x\) is simply the sign of x
(1) is insufficient since we can have either -4<x<-3 or 3<x<4
(2) is sufficient since \(x^3+x<0\) is equivalent to \(x*(x^2+1)<0\)
As \(x^2+1>0\)for any x,we can conclude \(x<0\)
