Rationalize Denominator - Video Tutorials & Practice Problems

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Rationalizing Denominators

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Everyone. Welcome back. So we've multiplied the divided radicals. And one of the things that we should know about radicals is that they can never be left in the bottom of a fraction. This is one of those weird bad things that you just can't do in math. Now, you might be thinking we've already seen radicals in the bottoms of fractions like radical two over radical eight. But in that case, it was fine because usually those fractions reduced to perfect squares like 1/4. And then if it was a perfect square, the radical just goes away and you're left with a rational number. What I'm gonna show you in this video is that sometimes that doesn't happen, sometimes you might have an expression like one over radical three and you can't simplify that to a perfect square. So to solve these types of problems, we're gonna have to do another thing. We're gonna have to do something called rationalizing the denominator. I'm gonna show you what that process is. It's actually really straightforward. So let's just go ahead and get to it. So again, if we have something like radical two of radical eight, it's simplified to a perfect square and that was perfectly fine. So radicals can simplify to perfect squares and we don't have to do anything else because you're just left with something like one half. But if you can't simplify this radical over here to a perfect square, then we're gonna have to make it one. And the way we make it one is by doing this thing called rationalizing the denominator, it's actually really straightforward. Basically, we're gonna take this expression over here and we're gonna multiply it by something to get rid of that radical on the bottom. And so what you're gonna do is you're gonna multiply the top and the bottom, the numerator and the denominator by something. And usually that's something that you multiply by is just whatever is on the bottom radical. So in other words, we're gonna take this expression over here and I'm just gonna multiply it by radical three, but I have to do it on the top and the bottom, you always have to make sure to do it on the top and the bottom because then you're basically just multiplying this expression by one and you're not changing the value of it. So whatever you do at the bottom you have to do on the top. The reason this works is because let's just work it out what is radical three times radical three. Basically, once we've done this, we've now turned the bottom into a perfect square. It's the squared of nine, which we know is actually just three. So in other words, we've multiplied it by itself to sort of get rid of the radical. And now it's just a rational number on the bottom. All right. So what happens to the top? Well, again, we just multiply straight across and then we ended up with radical three over radical nine, which is just radical 3/3. So look at the difference between where we started and ended here, we had one over radical three, we had a radical on the bottom. And here when we're done, we actually have three on the bottom. And that's perfectly fine. We have a radical on top, but we can have radicals on the top and that's perfectly fine. So what I want you to do is I actually want you to plug in if you have a calculator handy, one divided by radical three. And when you plug this in, what you should get out of the calculator is 0.57. And now if you actually do radical 3/3, you're gonna get the exact same numbers, 0.57. So the whole thing here is that these two expressions are exactly equivalent. They mean the exact same thing. It's just that in one case, we've gotten rid of the radical in the bottom. So this is what rationalizing the denominator means. Thanks for watching and let's move on to the next one.

2

Problem

Problem

Rationalize the denominator. $-\frac{5}{2\sqrt7}$

A

$-\frac{5}{14}$

B

$-\frac{5\sqrt7}{2}$

C

$-\frac{5\sqrt7}{14}$

D

$-\frac{10\sqrt7}{14}$

3

Problem

Problem

Rationalize the denominator. $\frac{6+\sqrt{x}}{-\sqrt{x}}$

A

$\frac{6\sqrt{x}}{-x}-1$

B

$6\sqrt{x}+x$

C

$\frac{6\sqrt{x}}{x}+1$

D

$\frac{7\sqrt{x}}{-x}$

4

concept

Rationalizing Denominators Using Conjugates

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3m

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Welcome back, everyone. We saw how to rationalize the denominator by taking something like one over radical three, we multiply the top and the bottom by whatever was on the bottom and we end up undoing the radical and that was awesome. We could do that. So what I'm gonna show you in this video is that sometimes that won't happen. You might have a problem that looks like 1/2 plus radical three in which you actually have two terms in the denominator. And when a denominator has two terms multiplying by the same radical won't actually eliminate it. So what I'm gonna show you is actually, we're gonna need something else to rationalize the denominator. We're gonna need something called the conjugates rather than tell you, let me just go ahead and show you, let's just jump into our problem here. So why can't we just use two plus radical three on the bottom? Well, if you end up doing two plus radical three, then when you multiply across this actually ends up being a binomial multiplied by a binomial. So in other words, we actually have to foil and if you foil this, now, what's gonna happen is on the bottom, you're gonna get two times two. That's the first, we're gonna get two times radical three. Those are the outer terms, same thing for the inner terms. So in other words, you end up with four radical three and on the inner terms, radical three times itself will just be three. So in other words, I multiplied it by itself on the top and the bottom, but I still ended up with a radical on the bottom. I didn't get rid of that. And remember that's bad, you can't have radicals on the bottom. So multiplying it by itself is not gonna work here. So what do we do? Well, instead of multiplying by itself, we do something, we multiply it by what's called the conjugate of the bottom. Basically what the conjugate is is you're just gonna reverse the sign between the two terms. So for example, if I have two plus radical three, then the conjugate is just going to be two minus radical three. That's the conjugate. You just take the sign between the terms and you flip it. So the general formula is if you have something like A plus radical B, then the conjugate is gonna be A minus radical B and vice versa. Those two things are conjugates of each other. So what do we do here? Well, I'm just gonna rewrite this 1/2 plus radical three. Now we're gonna multiply not by itself, we're gonna multiply by the conjugate. So words two minus radical three, and we remember whatever we multiply on the bottom, we have to multiply on the top. It has to be the same thing. So why does this work? Well, if you notice here, we're actually multiplying the same exact terms two and radical three, except the signs are flipped between them. And this actually ends up being a difference of squares. So remember how a difference of squares works. Uh Basically what happens is that you square these two numbers. So two and two becomes four and you square these two numbers and and then you basically just stick a minus sign between them. So in other words, radical three, when you square, it just becomes three and this is just a difference of squares. And on the top, when you multiply straight across this ends up being two minus radical three. All right. So what do you end up with? You end up with two minus radical three divided by just one? Now, you won't always get one here. Uh We just gotten one because we had four and three. But basically what happens is that we just got rid of the radical. So in other words, we've rationalized the denominator here. So multiplying by multiplying a radical by its conjugate, in fact, always eliminates the radical. That's why it's super useful and it always just results in a rational number like one or something like that. So just let me summarize really quickly here. When you have a one term denominator, you multiply the top and the bottom by whatever is on the bottom. And when you have a two term denominator, you multiply by the conjugate of the bottom. All right. But these are just the two ways that you rationalize the denominator. Hopefully, that makes sense. Uh Let me know if you have any questions. Thanks for watching.

5

Problem

Problem

$\frac{\sqrt{7}}{5-\sqrt{6}}$ Rationalize the denominator and simplify the radical expression.

A

$\frac{\sqrt7}{19}$

B

$\frac{5\sqrt7+\sqrt{42}}{-11}$

C

$\frac{5\sqrt7+\sqrt{42}}{21}$

D

$\frac{5\sqrt7+\sqrt{42}}{19}$

6

Problem

Problem

Rationalize the denominator and simplify the radical expression. $\frac{2-\sqrt3}{2+\sqrt3}$

A

$7-4\sqrt3$

B

$\frac{1}{7+4\sqrt3}$

C

$7+4\sqrt3$

D

$\frac{1}{7-4\sqrt3}$

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