On the previous page, all the fractions containing radicals (or radicals containing fractions) had denominators that cancelled off or else simplified to whole numbers. What if we get an expression where the denominator insists on staying messy?
The numerator contains a perfect square, so I can simplify this:
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This looks very similar to the previous exercise, but this is the "wrong" answer. Why? Because the denominator contains a radical. The denominator must contain no radicals, or else it's "wrong".
(Why "wrong", in quotes? Because this issue may matter to your instructor right now, but it probably won't matter to other instructors in later classes. It's like when you were in elementary school and improper fractions were "wrong" and you had to convert everything to mixed numbers instead. But now that you're in algebra, improper fractions are fine, even preferred. Similarly, once you get to calculus or beyond, they won't be so uptight about where the radicals are.)
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To get the "right" answer, I must "rationalize" the denominator. That is, I must find some way to convert the fraction into a form where the denominator has only "rational" (fractional or whole number) values. But what can I do with that radical-three? I can't take the 3 out, because I don't have a pair of threes inside the radical.
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Thinking back to those elementary-school fractions, you couldn't add the fractions unless they had the same denominators. To create these "common" denominators, you would multiply, top and bottom, by whatever the denominator needed. Anything divided by itself is just 1, and multiplying by 1 doesn't change the value of whatever you're multiplying by that 1. But multiplying that "whatever" by a strategic form of 1 could make the necessary computations possible, such as when adding fifths and sevenths:
For the two-fifths fraction, the denominator needed a factor of 7, so I multiplied by , which is just 1. For the three-sevenths fraction, the denominator needed a factor of 5, so I multiplied by , which is just 1. We can use this same technique to rationalize radical denominators.
I could take a 3 out of the denominator of my radical fraction if I had two factors of 3 inside the radical. I can create this pair of 3's by multiplying my fraction, top and bottom, by another copy of root-three. If I multiply top and bottom by root-three, then I will have multiplied the fraction by a strategic form of 1. I won't have changed the value, but simplification will now be possible:
This last form, "five, root-three, divided by three", is the "right" answer they're looking for.
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Nothing simplifies, as the fraction stands, and nothing can be pulled from radicals. So all I really have to do here is "rationalize" the denominator.
I need to get rid of the root-three in the denominator; I can do this by multiplying, top and bottom, by root-three. When I'm finished with that, I'll need to check to see if anything simplifies at that point.
Don't stop once you've rationalized the denominator. As the above demonstrates, you should always check to see if, after the rationalization, there is now something that can be simplified.
This expression is in the "wrong" form, due to the radical in the denominator. But if I try to multiply through by root-two, I won't get anything useful:
This doesn't help:
It didn't get rid of the radical underneath.
Multiplying through by another copy of the whole denominator won't help, either:
This doesn't help, either:
This is worse than what I'd started with!
How can I fix this? There's a trick:
Look what happens when I multiply the denominator they gave me by the same numbers as are in that denominator, but with the opposite sign in the middle; that is, when I multiply the denominator by its conjugate:
This multiplication made the radical terms cancel out, which is exactly what I want. This "same numbers but the opposite sign in the middle" thing is the "conjugate" of the original expression. By using the conjugate, I can do the necessary rationalization.
By the way, do not try to reach inside the numerator and rip out the 6 for "cancellation". The only thing that factors out of the numerator is a 3, but that won't cancel with the 2 in the denominator.
You can only cancel common factors in fractions, not parts of expressions. In this case, there are no common factors. Nothing cancels.
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The denominator here contains a radical, but that radical is part of a larger expression. To get rid of it, I'll multiply by the conjugate in order to "simplify" this expression.
The multiplication of the denominator by its conjugate results in a whole number (okay, a negative, but the point is that there aren't any radicals):
The multiplication of the numerator by the denominator's conjugate looks like this:
Then, plugging in my results from above and then checking for any possible cancellation, the simplified (rationalized) form of the original expression is found as:
It can be helpful to do the multiplications separately, as shown above. Don't try to do too much at once, and make sure to check for any simplifications when you're done with the rationalization.
You can use the Mathway widget below to practice simplifying fractions containing radicals (or radicals containing fractions). Try the entered exercise, or type in your own exercise. Then click the button and select "Simplify" to compare your answer to Mathway's.
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