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# Algebra question

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Intern
Joined: 01 Jan 2011
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Location: Kansas, USA
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11 Aug 2012, 19:39
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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12 Aug 2012, 00:05
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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Intern
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12 Aug 2012, 06:35
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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12 Aug 2012, 06:58
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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Love GMAT Quant questions and running.

Re: Algebra question   [#permalink] 12 Aug 2012, 06:58
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