Rationalizing the Denominator

Now multiply straight across the top, little 8 times the square root of 5. Now that doesn't seem like that helped us much, but when we look at the denominator and we do the square root of 5 times the square root of 5, that will give us just plain old 5. And then we can look to simplify. Now, when I look to simplify, I'm going to look at the numbers here, not anything inside the radical. And 8 and 5, you know what? They don't simplify. There's no way to reduce that fraction. So that's your answer. Doesn't look all that much prettier, but there's no radical in the denominator. Let's check out this one here. We're going to use the property that's says we can split this into radicals in a fraction instead of one and one instead of one fraction in a radical. So I've just split those up. It's the exact same thing. But that gives me the radical in the denominator. So let's get rid of that. By taking the square root of 15, or the square root of 15. And again, this is just one, so multiplying this whole thing by one. It doesn't change what it is. It just changes its form. 

Multiplying straight across the top, we're going to get the square root of 35 times 15. 35 times 15. Is the square root of 5 25. And then we're going to multiply straight across the bottom here. And again, the square root of 15 times the square root of 15 is just 15. We'll carry that over. Now no more radicals in the denominator. The good news. Is that we can still simplify our 5 25 radical. Good news, yes, because math is fun. So let's take this square root of 5 25 kind of off to the side here. Actually, let me do it over here. I'm going to slide over and get more space over here. And let's simplify this by factoring it out. So let's take out the 5 first. So 5 25 divided by 5, one O 5. Again, we can take a 5 out again. You can probably hear me, I'm just clicking along as we go. 21 breaks into 7. And three. So any end of my tree here, I would rewrite. Just like this. And then in the last video we covered how any pairs, just becomes that self number out front. So this numerator here became this. Or when I rewrite it, 5 times the square root of 21. At 7 and three, I can multiply it back together. I can't simplify that anymore. And my denominator 15 comes right along. 

The last step, look at this fraction, we can simplify that. Take a 5 out from each, so we'd have one. Over three. Or don't bother writing the one. And just 21, square root of 21 over three. And that would be your final answer. Let's practice some other ones here. When we have some more complex, I didn't have any more down here, did I? No, okay. When we have some more complex denominators. So this was the little bit trickier part. It's the same idea. We're going to multiply this by a form of one to get rid of this. Radical and the denominator can't have him there. But we have to include everything that's in the denominator. When we write our form of one. So when I write my convenient form of one, I'm going to have three, square root of 5, over three, square root of 5, and the trick. So these will cancel out is to give this the opposite operation. So I've got that kind of over here on the side. So if our denominator is an addition problem, we're going to make it a subtraction problem. When we write the form of one, if it was a subtraction problem, start off with, we're going to make it an addition problem. And you'll see here that that enables us to cancel things out. 

So let's multiply straight across the top here. We'll do that in red. So that I end up with three times three minus the square root of 5. So all we did, we're just going to multiply straight across. Then I'm going to distribute. So I'm solving just the numerator here first. Three times three is 9. Three times negative square root of 5 is just negative three square root of 5. I can't really solve that or combine it. So here in red is my numerator. My denominator, we're going to solve in green. I'm going to do it over here. So my denominator is three plus the square root of 5. My original denominator. Times. The opposite sign. And take your time with these and be diligent. You're going to take that first number and multiply it by the first number in there. So three times three, 9, then you're going to take the three and multiply it by the back number. So it'll be, again, negative three square root of 5. Then you have to do the same process with the other number from the first one. So this is a positive three times the square root of 5. And square root of 5 Times Square root of 5. Is just plain old 5. Be careful though. I almost did it. I almost lost my negative here. It should be a negative 5. 

Now, the whole point of doing this is because right here in the middle, this cancels each other out. Negative three times the square root of 5 plus three Times Square root of 5. It's like saying negative three plus three, right? Equals zero. Cancels out. These cancel each other out, and we're left with 9 plus negative 5. Simply four. That's my denominator. So I take my numerator that I found and put that over my denominator here, and that's your final answer. Fun, right? Let's try this one. So following our same trick. We're going to multiply by a form of one. That mirrors the denominator, but has that opposite operation. So let's solve this numerator first. So we're going to do 5 times two plus the square root of 6. I'm going to distribute. To get ten plus 5 times the square root of 6. Then I'm going to come back here and solve for the denominator. So here I would have two minus the square root of 6 times two plus the square root of 6. And again, we're going to take our time and distribute each piece. 

So two times two, four, two times the square root of 6. Is just two times the square root of 6. And then I'm going to come and do negative square root of 6 times two. So negative square root of 6 times two gives me that. And negative square root of 6 times the square root of 6. It's just negative 6. Again, right in the middle, all that cancels out. They cancel each other out. We're left with four -6 negative two. And so our denominator, all that work here to solve our denominator, negative two. Now many students, we said this in class, many students are tempted to try to simplify this. If that were just a fraction of myself, we could. However, we can not simplify that because we have all this, this is an addition problem up here. So it's being added into this. It's not being multiplied. If it were being multiplied, we could simplify. But this addition prohibits that. So this is the whole thing, our final answer. Last one, hopefully by now you're getting a little bit of the pattern. Again, we're going to multiply by a form of one so that the radical will leave the denominator by choosing the opposite operation from the denominator. And if you'll forgive me, I'm just going to kind of multiply it in that way, maybe that helps you visually. 

So we'll have two times one is two. Plus two times the square root of ten. That's my numerator. Then the denominator, I'm going to do down here. In red, so we have the original denominator. Times my new denominator. I'm going to do one times one is one. One times the square root of ten. Is the square root of ten. I'm going to do each piece of different color here. Negative square root of ten times one is the negative square root of ten. And negative square root of ten times the square root of ten. Is negative ten. Again, in the middle, these guys cancel each other out. We left with one minus ten. That's negative 9. That's my denominator. And that's the final answer. One more here, try to pause the video and solve it on your own. And I'll hop up on the screen here in a second, show you the final answer. All right, and over here. Is your final answer. So 30 -6 times the square root of 11 over 14. All right, I hope that helped. Algebra, stay tuned for Pythagorean theorem.