###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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# Rationalizing the Denominator with Higher Roots - Problem 1

Carl Horowitz
###### Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

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Rationalizing a denominator of a root of Higher power. So what we have here is we're trying to rationalize a denominator, we're dealing with a 4th root. So there's going to be two ways of actually dealing with this problem. But the main thing you need to remember is, when you're dealing with a higher root, we're going to need these many components in order to pull it out of a square root.

So, we could just leave it as the 4th root of 9 and think that I need 4 9s in order to pull this out. I have 1 here, so therefore I need 3 more. So I can multiply by the 4th root of 9Â³ over the 4th root of 9Â³. Distribute this through, we now have 2 the 4th of 9Â³ and we chose 9Â³ because in order to make something come out of a 4th root, we need 4 of them. So I have 1 9 here, 3 9s here, so that makes 9 to the 4th and giving us the 4th root of 9 to the 4th which is just going to be 9. So this is one way of doing it.

The other way of doing it actually is a little bit easier in my book because we're dealing with smaller numbers. So when I rewrite the problem, we have 2 over the 4th root of 9. What we need to do in this case is think about how we can simplify 9. 9 is the same thing as 3Â², so this then becomes 2 over the 4th root of 3Â².

We're still dealing with the 4th root, so we still need 4 over component in order to pull it out of the root. But now I already have 2, so I only need 2 more. So if I multiply by the 4th root of 3Â², I now have 4 3's in the denominator which I can cancel out with that 4th root to give me 3 in the denominator. So what I have here then is 2 times the 4th root of 3Â², 3Â² is just 9. And then 4th root of 3Â² times the 4th root of 3Â², 4th root of 3 to the 4th which is going to be 3.

So two different ways of simplifying the exact same expression. I prefer this one just because these numbers turned out to be a little bit nicer. I don't know what 9 to the third is off the top of my head, but either one is perfectly fine. Just making sure that you match up the root with the number of terms that you actually have inside of that root.