SOLUTION: Im not sure how to solve problems by rationalizing the denominator

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Question 143431This question is from textbook Algebra
: Im not sure how to solve problems by rationalizing the denominator This question is from textbook Algebra

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The process involves multiplying your fraction expression by 1 in a form such that when the denominators are multiplied, the denominator of the result becomes a rational number -- that is to say you have rid yourself of the radical. The process is also known as "Get that pesky radical OUT of my denominator"

Example 1:

1%2Fsqrt%282%29.

The way to make the denominator rational is to multiply it by sqrt%282%29, but we aren't allowed to change the value of the fraction. Fortunately, a%2A1=a no matter what a is and a%2Fa=1 no matter what a is (as long as it isn't zero), so multiplying 1%2Fsqrt%282%29 by sqrt%282%29%2Fsqrt%282%29 (which is just another way to write 1) is allowed.

%281%2Fsqrt%282%29%29%28sqrt%282%29%2Fsqrt%282%29%29=sqrt%282%29%2F2

Example 2:

1%2Froot%283%2C2%29

This one is a little tricker. Here we need to multiply the denominator by %28root%283%2C2%29%29%5E2. Again, we have to multiply by 1:



Example 3:

%281%2Bsqrt%285%29%29%2F%282-sqrt%282%29%29

This one is a little trickier still. We need to take advantage of the 'difference of two squares' factorization, that is: a%5E2-b%5E2=%28a-b%29%28a%2Bb%29

If we multiply the denominator by what is called its conjugate, 2+%2B+sqrt%282%29 (notice the sign change), then the result will be the difference of the first term squared and the second term squared with no annoying center term containing a radical. Again, and as always, we have to multiply by 1.



Ewww! That's a mess, but at least the denominator is rational. You might want to rephrase the alternate definition I gave earlier for the process to: "Get that pesky radical OUT of my denominator and I don't care how big a mess you make in the numerator"

Write back if you have any questions.