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Algebra question

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Intern
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Joined: 01 Jan 2011
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Algebra question [#permalink] New post 11 Aug 2012, 18:39
What is the best way to tackle a question of the following type?

Find the range of values of X if X^3< X^(1/3) and X^2> X^(1/3). I am trying to solve this using algebra.
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Re: Algebra question [#permalink] New post 11 Aug 2012, 23:05
sayak636 wrote:
What is the best way to tackle a question of the following type?

Find the range of values of X if X^3< X^(1/3) and X^2> X^(1/3). I am trying to solve this using algebra.


I prefer to get rid of the fractional exponents, therefore I use substitutions.
For the first inequality:

Denote x^{\frac{1}{3}}=y, so x=y^3. Then we have to find the values of y for which y^9<y or y(y^8-1)<0.
You have to check for values of y such that y<-1, \, -1<y<0, \, 0<y<1 and y>1.
It follows that y<-1 or 0<y<1. Translating back to x, we deduce that x<-1 or 0<x<1.

For the second inequality:

Denote x^{\frac{1}{3}}=y, again x=y^3. Then we have to find the values of y for which y^6>y or y(y^5-1)>0.
Now you have to test for values of y such that y<0, \, 0<y<1 and y>1.
You find that y<0 or y>1. In terms of x, this means x<0 or x>1.

Look for most posts about inequalities on the site. Here is just one useful one:
solving-inequalities-134671.html
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Re: Algebra question [#permalink] New post 12 Aug 2012, 05:35
Thank you. That makes good sense. I guess there is no direct way to solve equations such as (y^5 -1) or (y^8 -1).
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Re: Algebra question [#permalink] New post 12 Aug 2012, 05:58
sayak636 wrote:
Thank you. That makes good sense. I guess there is no direct way to solve equations such as (y^5 -1) or (y^8 -1).


You are welcome.

The equations can be solved quite easily:

y^5-1=0 gives y^5=1 or y=1.

y^8-1=0 gives y^8=1, from which y=-1 or y=1.

The above values (solutions) define the intervals on the number line in which we should test the corresponding inequalities.
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Re: Algebra question   [#permalink] 12 Aug 2012, 05:58
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