How to Rationalize a Denominator

Sometimes in algebra it is desirable to find an equivalent expression for a radical expression that doesn't have any radicals in the denominator. This process is called rationalizing the denominator. You can also do the same thing with the numerator, but the denominator is more common. I will only explain rationalizing the denominator and if you need to rationalize a numerator you do the same thing with the numerator.

There are 3 cases that commonly come up in algebra classes.

1. single square root
2. single higher root
3. sums and differences of square roots

1. When you have a single square root in the denominator you just multiply top and bottom by it.

2. When you have a single nth root in the denominator, multiply by something so that you will get a perfect nth power. You can always do this by multiplying by the denominator to the n-1 power, but that isn't always necessary. For example if the denominator is the cube root of 3 you can multiply by the cube root of 9, because 9 is 3 to the 2nd power and 2 is 1 less than 3, but if the denominator is the cube root of 9, you don't have to multiply by the cube root of 81, you can multiply by the cube root of 3.

3. Multiply top and bottom by the difference if the original is a sum and the sum if the original is a difference. That way you will get the difference of squares and get rid of the square roots that way.

Examples:

For the following problems the instruction is to rationalize the denominator, which means to write an equivalent expression for it that doesn't have any radicals in the denominator.

Example 1:

Solution:

Explanation:

There is just one square root in the denominator, so this is case 1. So to get rid of it we just multiply top and bottom by it. The square root of 3 times the square root of 3 is 3 simply from the meaning of square root, you don't have to write it first as the square root of 9. Remember that the square root of 3 means the number you can square, that is multiply by itself, to get 3, so if you squared it and didn't get 3 it wouldn't be the square root of 3. So by multiplying the top and the bottom by the square root of 3 we get rid of the square root in the bottom. In punishment for this the top got more complicated, and in fact the whole thing does in fact look more complicated after this 'simplification', but many times, like  for example when you are adding and need to find common denominators, this sacrifice is worth it, because simplicity is more important in denominators than in numerators.

Example 2:

Solution:

Explanation:

Here we first need to split up the quotient into two separate radicals. (See  Square Roots.) Then just like the last one we multiply top and bottom by the square root that is downstairs, the square root of 6, to turn the square root of 6 into 6. The numerator again suffer from it. Here since there was a square root upstairs it made it look nicer to pull the radicals together under one radical and multiply. (See  Square Roots.)

Example 3:

Solution:

Explanation:

This one is similar except that there is something multiplied by the radical in the denominator. Here we don't have to multiply by the whole denominator, just the radical part of it. The 5 just sits there and comes along for the ride. another interesting thing that happened is that after we multiplied the square roots of two to get 2, we were able to simplify further, because that 2 canceled with the 2 we already had upstairs.

Example 4:

Solution:

Explanation:

This one is an example of case 2 where we have a higher order root in the denominator. Don't let the variables bother you, we deal with them just like we deal with numbers, of course we do, since they are just place holders for numbers that we haven't yet decided on. Since this is a 3rd root, in order to get a total of 3 of them multiplied together in the bottom we need to multiply by 2 more, that is the square of the radical. When you do this it doesn't matter whether you put the square on the inside or the outside of the radical, because they are equivalent. Putting it on the inside makes it easier to multiply in the top.

Example 5:

Solution:

Explanation:

Here we have the same sort of thing except for a couple of difference. First we have to split the quotient into two separate radicals. Then this is a very good example of the exception that I spoke of in the description of case 2. Going by the rule of multiplying by 1 less power we would have multiplied by the cube root of 49b⁴. We could have done that, but the way we did it was simpler. Since there was already a b², we only needed to multiply that part by one more b to get a perfect cube, so we just multiplied top and bottom  by the cube root of 49b. You could make all kinds of elaborate rules for this, but the easiest thing to do is just think in terms of trying to multiply the thing inside the cube root by something so that it will make a perfect cube. In this case, since there are already two b's multiplied we only need one more to make the required 3.

We didn't really have to multiply out the 7 times 49. Since we chose 49 specifically so that it would multiply with 7 to make the perfect  cube 7³, it would be easier to just write it is 7³ and not worry about what number it was. We are only interested in what its cube root is anyway, and we can find that easier by leaving it as 7³.

Then just as with the square root problems, we need to put things under one radical upstairs and multiply out, but this time we can do some further simplifying afterward. (See  Square Roots.) We can re-split it up so that the perfect cube a³ is under a separate radical, and then we can take its cube root and get a, which goes on the outside of the radical.

Example 6:

Solution:

Explanation:

Here since this is a 5th root, we want to multiply by something that gets it raised to the 5th power. Since there is already one 5th root of 2, we need 4 more to multiply, so that would be the 5th root of 2 to the 4, the 5th root of 16. One like this doesn't come up quite so commonly in algebra class, but I did it so that you could see that the idea will generalize. In my classes we normally only go up to cube roots.

Example 7:

Solution:

Explanation:

Even though there is only one radical in the denominator this is an example of case 3, because there is something added to it. If we simply multiplied top and bottom by the square root of 10 it wouldn't work, because we would have to multiply it by both terms, so there would still be a square root of 10 multiplied by the 9. If we multiplied top and bottom by 9+sqrt(10) that wouldn't work either because there would be a middle term from the inners and outers. (See How to Add, Subtract, Multiply, and Divide Polynomials.) But there is a special multiplication formula that comes to rescue. When you multiply a sum and difference of the same thing you can a difference of squares, because the outers and inners cancel out. (See How to Add, Subtract, Multiply, and Divide Polynomials.) So the thing to do is to multiply by the difference instead of the sum, so that all together we get the product of a sum and difference. Then this gives us 9 squared minus the square root of 10 squared. To square 9 we just do it, we multiply it by itself to get 81. To square the square root of 10, just like in the previous examples, we use the meaning of the square root. Since the square root of 10 means the number you can square to get 10, if when you squared it you got anything other than 10, it wouldn't be the square root of 10, so when you square the square root of 10, the only thing  you can get is 10. Then to finish up the problem we distribute the 7 in the numerator and subtract in the the denominator.

Example 8:

Solution:

Explanation:

This one is also case 3, but now we have a difference, so we need to multiply by the sum. Then again we get the difference of squares, and when you square the square root of 3 you get 3, and when you square the square root of 5 you get 5. After we subtract there are minuses both upstairs and downstairs, so we can simplify by getting rid of both of them. Then we just have to multiply out in the top to get the final answer.

Example 9:

Solution:

Explanation:

This one is similar to the last two, but just a bit more complicated because of the difference in the numerator. Since there is a difference in the denominator, to get a sum and difference, we need to multiply by the sum, and then that sum has to get multiplied by the difference in the numerator, but in the numerator they are not sum and differences of the same thing, so there is not shortcut to multiplying them. We just have to use FOIL , (see How to Add, Subtract, Multiply, and Divide Polynomials), and this time there isn't even any combining of like terms. Also it is important to realize that when we square a product like 2sqrt(7), we have to square both the 2 and the sqrt(7). To square the 2 we just do it, multiplying it by itself, and to square the sqrt(7) we get 7 because sqrt(7) is by definition the number we can square to get 7.

Example 10:

Solution:

Explanation:

This one is similar except that it has variables instead of numbers. Since we have a difference in the denominator, we multiply by the sum. Then we something multiplied by itself in the numerator, so we can use the squaring formula. (See How to Add, Subtract, Multiply, and Divide Polynomials.)

Extra for Experts

What if you have a sum or difference of a higher root in the denominator? Is there a way to rationalize the denominator then? Like, for example, what if you have a difference of cube roots or 4th roots. Well, I have never seen this taught in algebra classes, but you can rationalize the denominator in such cases, and the way to do it is to use such things as the sum and difference of cubes factoring formula, and there are such formula for higher degrees as well. (See  Factoring Polynomials.)

Example 11:

Solution:

Explanation:

Here we can use the sum of cubes factoring formula,

A³+B³=(A+B)(A²-AB+B²),

because we already have the A+B part in the denominator.

Example 12:

Solution:

Explanation:

Here what we need is something to multiply A-B by to get at least A raised to the 4th power, so we can apply it here to get rid of the 4th root. There isn't really a specific formula for factoring the difference of 4th powers, but we can come up with one by factoring it as a difference of squares twice.

A⁴-B⁴=(A²-B²)(A²+B²)=(A-B)(A+B)(A²+B²)

Since our denominator already has the difference what we need to multiply top and bottom by is the other two factors here, (A+B)(A²+B²), where A is the 4th root of 5 and B is 1.