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Is x^4-x^3+x^2-x>0? 1) x>1 2) |x^3|>|x|

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Is x^4-x^3+x^2-x>0? 1) x>1 2) |x^3|>|x|  [#permalink]

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New post 09 Mar 2017, 22:45
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Is \(x^4-x^3+x^2-x>0\)?

1) \(x>1\)

2) \(|x^3|>|x|\)

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Re: Is x^4-x^3+x^2-x>0? 1) x>1 2) |x^3|>|x|  [#permalink]

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New post 10 Mar 2017, 00:11
ziyuen wrote:
Is \(x^4-x^3+x^2-x>0\)?

1) \(x>1\)

2) \(|x^3|>|x|\)



I chose D,

1- I tried substituting values and found out the bigger x gets, the greater positive number ends up being the result.
2- This rules out x being 0 or being a fraction and I figure this may just be the only way for x not being > 0. However I could not put my finger on exactly how to prove that. Can someone please explain this further on how 2 is helping us in this respect.
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Re: Is x^4-x^3+x^2-x>0? 1) x>1 2) |x^3|>|x|  [#permalink]

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New post 10 Mar 2017, 00:39
\(x^4-x^3+x^2-x>0\)
or \(x(x+1)(x-1)^2 >0\) therefore x <-1 or x>0

St 1: x>1. it will be positive as for x>0 it is positive and the condition holds true.
St 2: \(|x^3|>|x|\)
condition holds true for x>1 and x<-1. holds true as the values satisfying this equation are the subset for the above question statement

Option D
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Re: Is x^4-x^3+x^2-x>0? 1) x>1 2) |x^3|>|x|  [#permalink]

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New post 11 Jun 2017, 14:21
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hazelnut wrote:
Is \(x^4-x^3+x^2-x>0\)?

1) \(x>1\)

2) \(|x^3|>|x|\)


I solved this as follows:
Statement can be simplified to \(x^4-x^3+x^2-x = (x^3+x)(x-1)\)
\((x^3+x)(x-1)>0\) goes to proof that \((x^3+x)\) and \((x-1)\) have the same sign (both pos or both neg), AND \(x\) is not 0 or 1.

1) \(x>1\), implies that \(x-1>0\) and \(x#0\), and \(x^3+x>1>0\) =>Sufficient.

2) \(|x^3|>|x|\) implies that \(x#0, x#1\) and \(x#-1\), and since \(|x|\) is positive, we can divide both sides by \(|x|\) without changing the inequation, yielding
\(\frac{|x^3|}{|x|}>1\) or \(|x^2|>1\). This means \(x>1\) or \(x<-1\).
If \(x>1\), we are back to statement one => Sufficient
If \(x<-1\), \(x-1<0\) and \(x^3+x>1<0\) =>Sufficient.
Answer D.
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Re: Is x^4-x^3+x^2-x>0? 1) x>1 2) |x^3|>|x|  [#permalink]

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Re: Is x^4-x^3+x^2-x>0? 1) x>1 2) |x^3|>|x|   [#permalink] 16 Aug 2018, 14:06
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