This square root of a, divided by the square root of b, is equal to the square root of the quantity, a divided by b. Let’s use that rule on an example. This rule says we can rewrite the square root of 32, divided by the square root of 8, as the square root of 32 divided by 8. Why do we care? Why does this help?

Well now that we have that whole division problem underneath one radical, that means we can divide first to simplify. Divide 32 by 8, and now you have the square root of 4, which is simply 2. Let’s use it on another problem. The square root of 56, divided by the square root of 18. First, rewrite it as the square root of 56 divided by 18, then simplify.

You can divide your numerator and denominator both by 2, so this becomes the square root of 28 divided by 9. Now since you can’t simplify that fraction anymore, go ahead and break it back up into two separate radicals that are being divided, so the square root of 28, divided by the square root of 9.

You can take the square root of 9, that’s 3. 28 is not a perfect square, but it does have a perfect square factor, so let’s rewrite the square root of 28 as the square root of 4, times the square root of 7, (because the square root of 4 times the square root of 7 is this square root of 28) and now you can take the square root of 4, which is 2, so we have 2 square roots of 7, divided by 3.

Here’s where it gets a little trickier. We start out the same, rewrite it, square root of 24 divided by 30. Simplify it. You can divide the numerator and the denominator both by 6, so we have the square root of 4/5. 24 divided by 6 is 4, and 30 divided by 6 is 5. Break it up, again, since we’re done simplifying, square root of 4, divided by the square root of 5.

This square root of 4 is 2, but you can’t take the square root of 5, so that answer looks fine, right? Unfortunately, we can’t leave a radical in the denominator because the square root of 5 is an irrational number, and our denominator can’t be irrational, so we have to rationalize the denominator, meaning take this irrational number and create a rational number.

The way we do that is to multiply our numerator and denominator by whatever is underneath our radical in our denominator, so in this case the square root of 5. Notice that what you’re actually multiplying by is 1 because anything divided by itself is 1, so we’re not changing the value of this expression, we’re simply rationalizing the denominator.

Multiply straight across, 2 square roots of 5, divided by, the square root of 5 times the square root of 5 is the square root of 25, which is simply 5. Now we have a rational denominator. If you could simplify this further, for instance if this said 2/6 instead of 2/5, then you would, but we can’t, so there’s our answer.

Now this last example I have up here is a cube root instead of a square root. It works in a similar way, though, to the square root. The idea here is we need to be able to square root our denominator, so we multiplied our denominator by what we needed to in order to be able to take the square root of our denominator.

Now we need to be able to take the cube root of our denominator, so we’re going to multiply our denominator, and thus, our numerator by what we need to in order to take the cube root of it. To be able to take the cube root of our denominator we need to multiply by 5 squared and x squared, because 5 times 5 squared is 5 cubed, which we can take the cube root of, and x times x squared is x cubed, which we can take the cube root of, but we can’t just add things to our denominator, or multiply our denominator by new things, without doing the same thing to our numerator, so we multiply our numerator by 5 squared, x squared as well.

Now simplify. 5 squared is 25, times 3 is 75, so this is the cube root of 75x squared, divided by, 5 times 5 squared is 5 cubed, and x times x squared is x cubed. Now you can break it up, so we have the cube root of 75x squared, divided by, the cube root of 5 cubed x cubed, and you can’t take the cube root of your numerator, go back and look where it was broken up we have 3 times 5 squared, times x squared, so there’s nothing there that’s cubed to take the cube root of.

We’re going to leave our numerator as the cube root of 75x squared, but in our denominator the cube root of 5 cubed is 5, and the cube root of x cubed is x, so it’s simply 5x, so there’s our final answer with a rational denominator. Good luck.

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by Mometrix Test Preparation | Last Updated: June 13, 2019