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reagrding inequalities

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reagrding inequalities [#permalink] New post 28 Apr 2013, 03:51
x^3-x<0
x(x^2-1)<0

I am stuck after this step. Can anyone explain what are possible ranges for the value of x?how to proceed further
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Re: reagrding inequalities [#permalink] New post 28 Apr 2013, 04:04
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skamal7 wrote:
x^3-x<0
x(x^2-1)<0

I am stuck after this step. Can anyone explain what are possible ranges for the value of x?how to proceed further



At this point you have to study the sign of x and x^2-1
The first is +ve if x>0
The second in +ve is x^2-1>0 so if x>1 or x<-1
Take a look at the picture. Once you have the intervals you intersect them and find the overall sign.
You want the equation to be <0 so you take the -ve intervals => x<-1 and 0<x<1

Hope it's clear, let me know
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Senior Manager
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Joined: 02 Sep 2012
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Location: United States
Concentration: Entrepreneurship, Finance
GMAT Date: 07-25-2013
GPA: 3.83
WE: Architecture (Computer Hardware)
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Kudos [?]: 70 [0], given: 99

Re: reagrding inequalities [#permalink] New post 28 Apr 2013, 04:36
thanks a lot..can you please tell me if (X^3-x) >0 then ranges are -1<x <0 and x >1
is it correct?
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Senior Manager
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Location: United States
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WE: Architecture (Computer Hardware)
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Re: reagrding inequalities [#permalink] New post 28 Apr 2013, 04:39
thanks a lot..you rock
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Re: reagrding inequalities [#permalink] New post 29 Apr 2013, 19:56
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skamal7 wrote:
x^3-x<0
x(x^2-1)<0

I am stuck after this step. Can anyone explain what are possible ranges for the value of x?how to proceed further


Notice that you can further simplify the inequality

x(x+1)(x - 1) < 0

So your transition points are -1, 0 and 1, in that order on the number line.
The negative regions (since there is a less than sign) will be x < -1 and 0 < x < 1.

For more on how we got these ranges, check this post:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
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Re: reagrding inequalities   [#permalink] 29 Apr 2013, 19:56
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