Lesson video

Hi, welcome to today's lesson on the distributive law with surds.

By the end of today's lesson, you'll be able to apply the distributive law to expressions involving surds.

Our lesson today has three parts and we're going to start with part one, which is the distributive law with a single term.

We're going to start by reminding you of the distributive law, and I've chosen to use an area model for this.

It's one of the nice visual ways of showing what the distributive law actually looks like.

If we have a rectangle with a length of a and another length which is written as b add c, so it's like a combined total there, then I can draw my rectangle as one with one length of a and the other length I can break the rectangle up and say that the length of one of the rectangles would be b and the other length c.

Well, if I then cut along that dotted line and break the rectangles into two, then I now have two of them, one rectangle with measurements of a and b and the other with measurements a and c.

If I consider the area of these rectangles, then what do I actually have? I have the following, a multiplied by b plus c, but the area of the larger rectangle is equal to the sum of the areas of the two smaller rectangles, and we know that's true right back from year seven when we were looking at the area of composite rectilinear shapes.

We can express the area of the other two rectangles as follows, a times b plus a times c, and that's our distributive law.

It's showing us how we expand a bracket.

In other words, a multiplied by b plus c is equivalent to a multiply by b add a multiplied by c, and we're going to use this law today with expressions involving surds.

We're going to expand the following expression.

We know that we can therefore write this using the distributive law as three lots of the first term, so two root five, and then three lots of the second term, so three lots of root 27, and then sum this together.

What does this give us? Well, it gives us six root five.

Now, the three root 27 I've simplified, because root 27 can be written as three root three.

Three root three multiplied by three is nine root three.

I can't simplify any further because both surds are in their simplest forms and they are not like terms. It's now your turn.

Please expand the following expression.

Pause the video and do this now.

Welcome back.

Let's see how you got on.

We can write four lots of this expression as four lots of three root three add four lots of root 50.

Well, four times three is 12, so we have 12 root three, and then root 50 can be simplified to five root two.

Five root two multiplied by four means we have 20 root two, and again, I can now check.

Both surds are in their simplest form and they're not like terms, so we cannot simplify any further.

Let's consider this one, two lots of three root three minus root six.

Well, we can write this as follows.

Three lots of root two times root three is root six, and then subtract root six times root two, which is root 12.

Now the root six is fully simplified, but we know the root 12 isn't, so this can be written as three root six subtract to root three, and now I know that I'm fully simplified, because both of my surds are written as simplest form and since we don't have any like terms, we can't simplify any further.

And now it's your turn.

Expand the following, please.

Pause the video while you do this.

Welcome back.

Let's see how you got on.

Root two multiplied by three root two is three times root four, or just three times two, which will be six.

And then root two multiplied by negative five root five gives us negative five root 10, or in other words, we're subtracting five root 10.

Let's do a quick check.

Expand the following and simplify if possible, please.

What result do you get? Pause the video now and make your choice.

Welcome back.

Which one did you pick? If you've picked B, then you are correct.

We know that two root five multiplied by root 20 is just two times a square root of a hundred.

Well, two times a square of a hundred is two times 10, so that's 20.

And then we have two root five multiplied by three root five, well, that just becomes six times five, which is 30, and 20 add 30 is 50.

What about this one? Again, pause the video while you make your selection.

Welcome back.

Which one did you choose? If you chose C, well done.

Now, A and D were partially correct, but they weren't fully simplified, so you have to be careful of those.

We know that we had two root of three y multiplied by two times the square root of xy.

Well, two times two is four.

When we multiply the radicands together, we'll have three xy squared, and the y squared can be simplified.

So it would've got four y and then the square root of three x.

Well, that first term means it would either be C or D, but now we need to consider the second term.

Two multiplied by the root of three y, and then multiply by negative root of six y.

Well, we know we're going to have two as a coefficient and then inside, the radicand, in other words, will be three y multiplied by six y, which is 18 y squared.

So we know the y squared will simplify and will become a coefficient of y, so at the moment we'd have two y root 18, which is what D shows.

However, root 18 is not fully simplified.

We know that simplifies to three root two, and that means we'll be multiplying our coefficient by three, which will give us six y root two, thereby showing that C is the correct answer.

We're getting ready to begin our first task, but before we do, I want to introduce this diagram to you.

You may have seen this before, but in case you haven't, I'm going to do a quick recap.

We're referring to this diagram as a tripod, and it's a triangle in which the values on each edge of the triangle are the product of the values at the adjacent vertices.

In other words, seven times six is 42.

It's now time for our first task.

I'd like you please to complete the tripods you see below.

Remember, the values on each edge of the triangle are the product of the values at the adjacent vertices.

Pause the video while you do this.

Welcome back.

Let's see how you got on.

Looking at the tripod A, we know that if we multiply three plus root five by root 20, we'll get six root five plus 10.

Multiplying root 20 by negative root five means we'll end up with a negative root of a hundred, which is negative 10.

And then negative root five multiplied by three plus root five gives us negative three root five subtract five.

In B, we were able to work out two values because they were on the edges and so we multiplied the adjacent vertices.

In order to work out the value of the bottom left vertex, I had to do some division.

Personally, I did two b root a divided by the root of ab.

I found that was a little bit easier, because I knew therefore the root a's would cancel and I'd be left with two b divided by root b.

Well, what I did there was I multiplied both the numerator and the dominated by root b and that got me to two b root b divided by b, and so the b's cancelled to leave me with two root b.

It's now time for the second part of our lesson, and this is where we're going to use the distributive law with multiple terms. Let's see what we mean.

This is what I mean.

In other words, there's going to be more than one expression here.

We're going to need to use the distributive law on both expressions to fully expand and then simplify.

You can see here that I've expanded the first bracket.

Two root two times three is six root two, and then two root two multiplied by negative four root three gives us negative eight root six.

I've then expanded the second bracket, and you can see the terms there.

After that, I gathered the like terms, and that produced my final line of working.

Now it's your turn.

Be careful, expand each bracket separately, and then look to combine like terms. Pause the video and do this now.

Welcome back.

Let's see how you got on.

When we expand the first bracket, I have root five multiplied by three root 15, which gives me three root 75.

When I do the second term inside that bracket, I have two root five multiplied by root five.

Well, that's just two times five, which is 10.

The second bracket leads to three multiplied by two root two, which is six root two, and then three multiplied by negative four, which is negative 12.

I then check, I don't see any like terms for the surds, but I do for my integers.

10 subtract 12 is negative two.

Now, in terms of the surds, I know I can simplify the square root of 75.

That's five root three, so I end up with 15 root three plus six root two subtract two.

And at this point, I know I'm finished, because my surds are fully simplified and there are no remaining like terms. It's now time for our second task.

In it, you'll be asked to use the distributive law.

In the first one, part a, with just a single term, and then in b and c, we have multiple terms. I've asked you to give your answer in a particular way, and you may find that helps when you are considering whether or not you need to simplify any further.

Pause the video while you have a go at this task.

Welcome back.

Let's see how you got on.

Well, with a, we had to first expand the brackets.

Root five times root eight is root 40, and then root five multiplied by root 18 is the square root of 90.

Now, root 40 and root 90 are not like terms, but we know we can simplify here.

The square root of 40 is the same as saying two root 10, and root 90 is the same as three root 10.

Ah.

Now we have like terms. Two root 10 add three root 10 is five root 10, and when I look, I was told to give my answer in the form something multiplied by root 10, so even if I wasn't sure how to simplify, I'd have known that 10 had to be the radicand, which would've helped me realise when I reached root 40 add root 90 that I wasn't finished there, that there was a way that I was going to be able to combine these.

Now let's look at b.

We'll start by expanding the brackets.

Three root five multiplied by two root two gives me six root 10, three root five multiplied by three root three gives me nine root 15.

I then move on to my second bracket, negative root five times root two is negative root 10, and negative root five multiplied by two root three gives me negative two root 15.

Well, although 10 and 15 as radicands cannot be simplified because they contain no perfect square factors greater than one, I do have some like terms here.

Six root 10 take away root 10 is just five root 10, and nine root 15 subtract two root 15 is seven root 15.

I couldn't think I can do any more here, and then when I check, I was told to give my answer in the form something multiplied by root 10 plus something multiplied by root 15, and that's what I have here.

So I've reached the desired form and I'm done.

Let's look at c now.

Let's start by expanding our brackets.

Root 20 times root five is root a hundred.

Root 20 multiplied by negative three is negative three root 20, root two multiplied by six root 10 is six root 20, and then root two multiplied by root 18 is the root of 36.

Now, what you could have done, of course, is you could have simplified some of these surds before you started expanding the brackets.

The root 20 could have been turned into two root five, and root 18 could have been turned into three root two.

And you're welcome to have done that if you chose that method to go with.

I'll continue though with the working I have on the right.

The square root of a hundred is 10, I've got six root 20 take away three root 20, well, that's just three root 20, and then the square of 36 is six.

10 add six is 16, and root 20 we know simplifies to two root five, so three lots of two root five is six root five.

I check and this is the required form.

Remember, I wanted something add something multiplied by root five, and that's what I have.

It's now time for the final part of the lesson, and this is where we apply the distributive law.

Let's have a look.

We need to find the area of this triangle.

Well, we know to find the area of a triangle, I need the base multiplied by the perpendicular height and then divide by two.

In other words, I need two root two multiplied by three plus four root eight, and then I either divide by two, or I can say I multiplied by a half.

So let's start to simplify.

I'll first deal with the coefficient.

Half of two is just one, so I can actually write root two multiplied by three plus four root eight.

Well, let's expand those brackets.

This gives me three root two, and then root two multiplied by root eight is root 16.

Root 16 is just four, so I have four times four, which is 16.

What about if I apply my knowledge of surds to geometric sequences? We're going to find the first three terms of a geometric sequence, and the first term is written as three take away two root 10 and we have a common ratio of two.

Remember, to generate a geometric sequence, we multiply the previous term by the common ratio to get the next term.

So if our first term is three take away two root 10, then we would have to multiply this by our common ratio of root two in order to get our second term.

Well, three multiplied by root two is three root two, and negative two root 10 multiplied by root two would be negative two root 20, but we know that the square root of 20 simplifies to two root five, so we end up with negative four root five.

To find the next term, i.

e.

, our third term, we must multiply the second term by root two.

Three root two times root two means three times two, or six, and then negative four root five multiplied by root two would be negative four root 10.

It's now your turn.

This task begins by asking you to calculate the area of the two shapes you can see on the screen.

The first shape is a trapezium and the second is a composite rectilinear shape.

Remember, composite rectilinear shapes are those shapes that are made solely of rectangles, and you can use whatever method you like in order to find that area.

Pause the video and have a go at this task now.

Welcome back.

Let's look at the third part of the task.

In this part, you're asked to find the missing terms of the geometric sequence.

You've been given the first term and the common ratio so you should be able to generate the second and third terms. Pause the video and do this now.

Welcome back.

Let's go through our solutions.

Starting with our trapezium.

Remember, the area of the trapezium is the sum of the parallel sides, so that will be three root five subtract root 30 and then add five root five.

Well, before we go any further, we can see we have two terms involving root five, so let's sum them.

That will be eight root five subtract root 30.

Now, once we've summed our parallel sides, we then multiply by the perpendicular distance between them and that's root 20, so in other words, our eight root five subtract root 30 has to all be multiplied by the square root of 20.

You can see that as the numerator of our fraction.

We can then expand this.

Eight root five multiplied by root 20 is eight root a hundred, and then we're subtracting root 30 times root 20, which is the square root of 600, and we can simplify that.

The square root of a hundred is just 10, so we have eight times 10 is 80.

Now, 600 could be written as a hundred multiplied by six, so the square root of 600 simplifies to 10 root six.

That's as simple a numerator as we can make it.

However, remember it's a trapezium, so we must divide by two.

We're going to divide all the terms on the numerator by the denominator, so 80 divided by two is 40 and 10 divided by two is five, giving us a final result of 40 take away five root six.

Next, I'll look at shape b.

In b, we have a composite rectilinear shape.

Now, remember, I said it was up to you how you chose to calculate this area.

This is the method I've gone for.

I broke my shape into two parts.

I then found that missing height.

If I know the height of the entire shape is three lots of root five and I already have one lot of root five shown on my diagram, the parallel side to the left at the bottom, then what remaining length is there? That's right, two root five.

I now have the measurements I need to work out the area of each smaller rectangle.

Once I've done that, I can sum them to find the area of my composite rectilinear shape.

The first part of working shows me working out the area of the first rectangle, so root five multiplied by three lots of two root five plus root seven, and then the second part I'm adding on the area of my smaller rectangle, so two root five multiplied by root five minus two root seven.

Expanding those brackets leads me to 30 plus three root 35 add 10 subtract four root 35.

I can then simplify this to 40 take away root 35.

Let's now look at our geometric sequence.

Remember, to generate the second term, I needed to multiply the first term by negative two root two, so negative two root two multiplied by negative two root 14 would've given me four root 28, but root 28 can be simplified into two root seven, and four times two root seven is eight root seven.

Multiplying three root two by negative two root two meant I had negative six times two, or negative 12.

I then multiplied my second term by negative two root two in order to get to the third term.

It is time mow to sum up what we've done in our lesson today.

Brackets involving surd terms are expanded using the distributive law.

It's important to fully simplify wherever possible, and we saw much simplification today.

Remember, you can simplify before multiplying or dividing, and sometimes it's easier to simplify afterwards.

It's really a personal choice, so it's entirely up to you.

You decide which one you find easiest, and sometimes it will depend on the numbers involved.

Well done.

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

I look forward to seeing you for our next lesson.