Lesson video

Hi, welcome to today's lesson on rationalising a two term denominator.

By the end of today's lesson, you'll be able to use the technique of rationalising the denominator in order to write a two term denominator in a rational way and therefore, produce an equivalent fraction.

Our lesson has two parts to it and we're going to start with the first part on making integers.

We're gonna start with a little activity.

I'd like you to match the brackets on the left with an expanded form of them on the right.

In other words, when I expand these brackets, which of the following results do I get? Pause the video now while you do this matching activity.

Welcome back.

How did you get on? Did you notice that two of the expressions on the left went to the same result on the right? Look, the middle two, they both resulted in seven.

What can you see is the same about these two? Well, that's right.

It's the same two factors each time.

All we've done is just written it the other way round and we know with multiplication, it doesn't matter.

It still produces the same result.

What's interesting, of course, is that they both led to an integer product.

In other words, there was no surd present.

We've seen this happen before though when we were expanding binomial expressions.

Can you remember what we refer to this as? Even if you're not sure what we would've called that process or what we're referring to it as, perhaps there's a particular feature though that you are familiar with, maybe that will help you here.

Which of these four expressions would give us an integer answer? In other words, when I expand, I'd get an integer.

Pause the video now while you make your selection.

Welcome back.

Which ones did you pick? It should have been the middle two.

Can you spot what's common here and what's not? In other words, what's the same and what's different? That's right, it's the difference of two squares.

Remember, if we multiply a plus b by a subtract b, our result is a squared take away b squared.

For example, let's consider using three and five in place of a and b.

When I expand this, I get nine subtract 25 or in this case, negative 16 because I'd get a positive 15 term and a negative 15 term which would cancel each other out.

We can use this with surds.

For example, let's try in place of a, using four and in place of b, using root five.

I get four times four is 16 and negative root five times positive root five is negative five.

I'd also get four root five subtract four root five and they'd cancel.

So my result here would be 11.

What do you multiply seven plus root 15 by in order to make an integer? Which of A, B or C would you choose? Pause the video now and make your choice.

Welcome back.

Did you spot there was more than one correct answer.

That's right, it's A and C.

In the first one, the sign in front of the root 15 has changed.

Only one sign's changed.

So the sign in front of the seven is still the same but the other one is different, so we know that's going to work.

And then in C, although I've changed the order of the terms, one sign has stayed the same and the other has differed.

It doesn't matter which of these two I multiply seven plus root 15 by, I'll get an integer solution.

Feel free to try it out if you don't believe me.

What about this one? Root three subtract eight.

What would I multiply by to make an integer? Pause the video now and make your choice.

Welcome back.

Which ones did you go for? And did you spot there was two again? That's right.

We want one sign to stay the same and the other to change.

Because I have subtract eight, I'm looking for there to be a positive eight and for the root three to stay positive, which means I'm looking for something of the form root three plus eight and those terms could be in either order.

What about this one? What would you multiply four subtract three root two by in order to make an integer? Pause the video and make your choice now.

That's right, it's B, just one choice this time.

It'll be three root two plus four.

It's now time for your first task.

I'd like you to fill in the gaps to create final integer answer.

So in other words, what do we need to be there? Pause the video while you have a go at this first page.

Welcome back.

Let's look at the remaining questions.

So again, same thing.

This is just e and f.

Pause the video now while you complete them.

Welcome back.

Let's go through our solutions.

For a, we know that we need there to be a root three because otherwise we're not going to be able to rationalise the root three from the first expression.

This produces 16 subtract three or 13 is our final answer.

In b, we were given the root five subtract six.

So we need to have root five plus six.

This will result in five subtract 36 or negative 31.

In c, we were given the two root two plus seven so we need two root two minus seven.

In this case, we'll have eight subtract seven giving us one.

Then in a, in order to get the form a squared subtract b, I'll need to have a plus root b and a minus root b.

Let's look at e and f.

In e and f, you had quite a bit of choice here.

You know you needed it to be one.

So what I looked for here was something that was going to give me a result of one and here I'd end up with 10 squared, which is 100 and I'd be taking away nine lots of 11 or 99.

Well, 100 takeaway 99 is one.

And then for f, there were other solutions possible.

I've gone for this particular one but you could have had other options too.

It's now time for the second part of our lesson and in this, we'll be rationalising a two term denominator, using what we've just covered in section one, i.

e.

the difference of two squares.

To rationalise when the denominator has two terms, we'll use the difference of two squares and that's because if you remember from last lesson, we need our denominator to be an integer.

For example, if the denominator was a plus root b, I would multiply both numerator and denominator by a subtract root b because that will result in a rationalised denominator.

Let's try this out now.

I'm going to simplify eight divided by three takeaway root five.

What am I going to multiply both numerator and denominator by? I'm going to use my difference of two squares knowledge, so I'm going to multiply by three plus root five for both numerator and denominator.

On the numerator, I know this is going to give me 24 plus eight root five and on the denominator, I'll have nine subtract five.

Well, this will give me an expression that reads as 24 plus eight root five divided by four.

I'm then going to check to see if there's anything I can simplify.

I can see that I can take out a factor of four from both numerator and denominator, so I'll do that now to fully simplify.

This results in six plus two root five.

It's now your turn.

I'd like you to simplify the fraction four over three subtract root three.

Pause the video and do this now.

Welcome back.

Let's start with what you decided to multiply the numerator and denominator by to get an equivalent fraction.

You should have multiplied by three plus root three.

Remember, difference of two squares in order to rationalise the denominator.

For our denominator, that will result in nine subtract three and for the numerator, we'll have 12 plus four root three.

Well, let's tidy up the denominator.

That will become six and then we look to see if there's anything we can do to simplify.

We can see that we can divide each turn by two.

This will then result in six plus two root three over three and that's where I'm going to finish.

What about if it was written like this? Well, the first thing I'm going to do is write that as a fraction.

That looks better.

Now I know what to do.

I'm going to multiply both numerator and denominator and in order to know what to multiply by, I'm going to use the difference of two squares.

So I'm going to multiply by root three plus root two.

Now, this one's a little bit more complicated because the numerator of my original fraction has two terms, but that's okay.

I know how to expand when I have two binomials and I want to find the product.

My numerator will become five root three plus six root two and my denominator using the difference of two squares is three subtract two.

Let's tidy that up.

Well, luckily my denominator is just one, so I'm just left with the numerator.

A five root three plus six root two.

It's now your turn.

Calculate one plus root five divided by root five subtract root three.

Pause the video and do this now.

Welcome back.

How did you get on? Let's go through the working.

The first thing we do, remember is to write this as a fraction.

We can then multiply and we know what to multiply by because we're going to use equivalent fractions and the difference of two squares.

This results in a numerator that reads as five plus root three plus root five plus root 15 and that's all over five takeaway three.

Our denominator therefore, can be tidied up to be two.

And when we look at the numerator, we can't simplify any of those surds, and we have no like terms to gather, so we're finished.

It's now time for our second task.

I'd like you to rationalise both of these denominators.

Pause the video and do this now.

Welcome back.

Let's go through this.

For a, we know that we need to use the difference of two squares to identify what to multiply by.

Because the denominator in the original fraction is three plus two root two, we'll be using three subtract two root two.

Our numerator therefore is three takeaway two root two and our denominator becomes nine.

Subtract, well, two root two times two root two is four times two or eight.

In this case, we're just left with our numerator because our denominator evaluates to one.

Now let's consider b.

I'm going to multiply by root seven subtract three.

This leaves me with a numerator of two root seven subtract six and a denominator of seven takeaway nine.

Well, we know that seven takeaway nine is negative two, so I can now tidy this up and what I can do is I can divide each term by negative two.

Well, negative six divided by negative two is three and two divided by negative two is negative one, hence the result of three subtract root seven.

What about if I'm being asked to add or subtract fractions where we have different denominators and involving surds? Goodness me, looks like it's going to be a lot of work, except what I'm not going to do is I'm not going to fall into the trap of going straight into I'll make a common denominator.

No, I know with surds that simplifying can often make the calculation easier.

So instead of going straight for making a common denominator, I'm going to rationalise both of these denominators.

I may find then it's much easier to have a common denominator.

Let's see what happens.

I'll start with the first fraction.

So I know I'm going to need to multiply by three root three plus five.

Doing that gives me a numerator of six root three plus 10 and a denominator of 27 subtract 25.

Well, that leaves me with a denominator of two and then I can divide through, giving me three root three plus five.

Well, that fraction suddenly got a lot nicer, didn't it? The equivalent form is not even a fraction anymore.

Hmm.

Okay, let's look at the second fraction.

I have five subtract two root three over root three plus two so I know that I need to multiply by root three subtract two.

This gives me the following: a numerator of nine root three subtract 16, and a denominator of three subtract four.

Well, my denominator just evaluates to negative one.

Well, I can divide through by negative one.

That will give me 16 subtract nine root three.

Well, that fraction also reduced down and now I don't have a fraction there either because I simplified.

Let's clear up the screen so that we can actually carry out this subtraction.

Remember, these are the two final forms I reached when I fully simplified each of my two fractions.

So I know that my original question can be rewritten as three root three plus five and then subtract negative nine root three plus 16.

Now I can actually carry this out.

I'll have three root three subtract negative nine root three, which is a total of 12 root three, and five subtract 16, which is negative 11, so my final result is 12 root three subtract 11, which is of the desired form.

Now it's your turn.

Show that six root two over root six plus eight over root three subtract root seven can be written in the form a times root seven.

Pause the video while you have a go.

Welcome back.

How did you get on? Well, I started by rationalising both denominators.

We saw in the previous example that can make things a lot easier.

So I multiplied the first fraction by root six over root six and I multiplied the second fraction by root three plus root seven for both numerator and denominator.

Let's tidy that up therefore.

Well, the first fraction becomes six root 12 over six and my second fraction became eight lots of root three plus root seven, all over three subtract seven.

Well, we can tidy that up even further.

We can see that simplifies.

I can divide by six so that my first fraction is just root 12 and my second fraction I'm dividing by negative four.

Well, eight divided by negative four is just negative two, which is why the bracketed expression's coefficient has changed.

Let's go from there then and multiply out these brackets.

I've also simplified the root 12, so it's become two root three.

I've then got takeaway two root three and take away two root seven.

Well, we can tidy that up.

That just gives us negative two root seven, which is of the desired form.

It's now time for your final task.

I'd like you please to write both of these fractions in the form a multiplied by something root three plus something else, or in other words, a multiplied by b times root three plus c.

Pause the video and do this now.

Welcome back.

How did you get on? Let's start with a.

In a, we will have multiplied using the difference of two squares, so we're multiplying by two root three plus four over two root three plus four.

We'll have on the numerator eight root three plus 16 over 12 takeaway 16, which will give us negative two root three subtract four.

Remember, our denominator evaluated to negative four, so I simply then divided.

I then, of course, have to write this in the required form.

I've taken out a factor of negative two.

This leads me with root three plus two and that is of the form required in the question.

Let's look at b now.

Our numerator becomes six root three plus nine plus six plus three root three and this is divided by nine takeaway three.

We can tidy that up.

We'll have 15 plus nine root three over six.

I know that I can simplify this by dividing through by three.

That will give me five plus three root three over two but that's not gonna be the required form.

If I take out a factor of a half therefore, because dividing by two is the same as multiplying by a half, I'll have 1/2 multiplied by five plus three root three, and that is of the required form.

It's now time to summarise what we've covered in our lesson today.

The difference of two squares can make an integer from a surd expression.

Using equivalent fractions and the difference of two squares, the denominator of a fraction can be rationalised.

And we saw that in our second task today.

Well done.

You've worked really well today and I can tell you put in a lot of effort.

I look forward to seeing you for our next lesson.