Rationalizing Denominators: Understanding the Process

  • Thread starter Thread starter majormuss
  • Start date Start date
AI Thread Summary
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.
majormuss
Messages
124
Reaction score
4

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

Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
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..
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top