Rationalizing Denominators: Understanding the Process

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The discussion revolves around the confusion regarding the rationalization of a denominator involving roots. The key point is that to rationalize the denominator of a term like 1/√[n]{x^m}, one must multiply by √[n]{x^{n-m}} to achieve a perfect power inside the root. In this case, using x^2 instead of x^3 was necessary to ensure that the product results in x^5, which simplifies to x when taking the fifth root. The focus is on ensuring that the expression under the root becomes a complete power corresponding to the root being used. The clarification provided helped resolve the initial confusion.
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Homework Statement



I am confused with this equation I found online... It seems wrong to me, I need help.
My question is,why did the person who worked taht equation use a x^2 to rationalize the equation when the actual equation was an x^3??
http://tutorial.math.lamar.edu/Classes/Alg/Radicals_files/eq0081MP.gif

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The Attempt at a Solution

 
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Because the fifth root of x^2*x^3=x^5 is x. How would you do it??
 
Dick said:
Because the fifth root of x^2*x^3=x^5 is x. How would you do it??

no my question is... why was x^2 used instead of x^3 to rationalize the denominator? my thinking is if x^2 is the denominator then why use x^3 to rationalize it?
 
Because x^2*x^3= x^5 as Dick said. The crucial point is that it is the fifth root that is to be rationalized. You have to multiply what ever power is necessary to get a fifth power inside the fifth root.

In general to rationalize the denominator of 1/\sqrt[n]{x^m} you need to multiply numerator and denominator by \sqrt[n]{x^{n-m}}. That way, in the denominator you will have \sqrt[n]{x^nx^{n-m}}= \sqrt[n]{x^n}= x.
 
HallsofIvy said:
Because x^2*x^3= x^5 as Dick said. The crucial point is that it is the fifth root that is to be rationalized. You have to multiply what ever power is necessary to get a fifth power inside the fifth root.

In general to rationalize the denominator of 1/\sqrt[n]{x^m} you need to multiply numerator and denominator by \sqrt[n]{x^{n-m}}. That way, in the denominator you will have \sqrt[n]{x^nx^{n-m}}= \sqrt[n]{x^n}= x.

I get it now thanks..
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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