Lesson video

Hi, welcome to today's lesson on problem solving with surds.

By the end of today's lesson, you'll be able to use your knowledge of surds to solve a variety of problems. Our lesson has two parts to it and we're going to start with part one, which is all to do with comparing surds.

Manipulating surds into an equivalent form can be extremely useful when we need to compare them.

For example, which expression has the greater value? How can we compare these two fractions? We know that when comparing fractions, either a common denominator or numerator would be useful.

Well, if I look at the fraction on the right, I see that I could rationalise that denominator and in doing so, I'll in fact reach a common denominator across the two fractions.

This is a good start, except I'm not entirely certain which is larger, 7 root 300, or 56 root 3.

Luckily, of course, simplifying can help us here.

Let's do that now.

7 multiplied by the root of 300, well, root 300 can be simplified, let's do that now.

It becomes 70 multiplied by root 3.

Well, this is significantly easier to compare.

Since the denominators are the same I can compare the numerators.

70 lots of root 3 is larger than 56 lots of root 3, so I know that the fraction on the left is the larger of the two fractions.

As you can see, simplifying helped me greatly and ensured that it was easy for me to use my knowledge of comparing fractions in this particular example.

Let's have a look here at Aisha and Andeep.

They've both got a value that they think is bigger than 20, who's correct? You'll need to do some reasoning so you can compare each of their fractions to 20 and work out if you think it's bigger or not.

Pause the video and do this now.

Welcome back, let's start with Aisha.

Well, the first thing I did was I simplified the surd on the numerator.

This took me from 3 root 288 to 36 root 2.

I then rationalised the denominator by multiplying both numeration and denominator by root 3.

Doing that got me to 36 root 6 over 3.

I then divided through by 3, so I've got 12 lots of root 6.

Now, I can do some reasoning.

Root 6 has a value between the square root of 4 and the square root of 9.

I know this because the radicand is between 4 and 9.

Well, given that the value of root 6 must be between root 4 and root 9, I know that 12 root 6 must have a value between 12 lots of 2 which comes from square rooting the 4, and 12 lots of 3, which would be what I'd get if I square rooted the 9 and multiplied.

So 12 root 6 must sit between those two values.

Well, since the minimum value there is 24 I know that my fraction is definitely bigger than 20.

So Aisha is right with what she thought.

What's about Andeep? Well, here's his fraction and, again, I'm going to start by simplifying.

The numerator therefore, becomes 21 root 10.

I'm then going to rationalise the denominator and once again I'm going to simplify my surd on the numerator.

Once I've done that I've then divided both numerator and denominator by 5.

This gets me to 21 multiplied by root 2.

Well, since I know that root 2 is greater than 1, I've got 21 multiplied by something that's bigger than 1, so my result must be bigger than 20.

In fact, I know that's going to be bigger than 21 let alone being bigger than 20 so Andeep was also correct.

So, in this particular case, both of them were right.

Their fractions were indeed, bigger than 20.

It's now time for Task 1, you have here three sets of surds.

I'd like you please for each set to rearrange them so they are written in ascending order.

Once you've had a go feel free to use your calculator to check that they are in ascending order.

Pause the video and do this now.

Welcome back, how did you get on? Let's start by looking at a and b.

Here each of them are written in ascending order so there's a mixture of techniques you can do here either through writing equivalent fractions, or through reasoning.

For example, I didn't need to do anything to root 2 and root 5, I know that the square root of 2 is smaller than the square root of 5 so that one was fairly straightforward.

For others you may need to have written equivalent fractions and don't forget, an equivalent fraction doesn't necessarily mean we have to simplify.

For example, 1 over root 3, I might well have started by rationalising that denominator so I have root 3 over 3.

I could then, of course, have multiplied any other fraction by 3 over 3 so that I could have reached a common denominator and that's in the case of root 2 or root 5.

There are lots of different methods you can use here for your fractions to help you write these in ascending order.

Here is c, as you can see, you may well have done a mixture in this particular question of either reasoning or writing equivalent fractions.

It's now time for the second part of our lesson and this is where we're going to look at a variety of problems involving surds.

Now, we've seen that manipulating surds is a useful set of skills.

We're now going to give you some opportunities to apply these skills to various problems. In this first task, you're going to be using your skills of manipulation.

In part a, you are told what a, b and c stand for.

You then need to do some substitution and simplify.

In b, you have two fractions and we'd like you to add them please.

I'm giving you a hint, you might want to simplify first.

Just in case you don't remember, when we covered adding fractions like this in one of our previous lessons.

Pause the video now while you have a go at a and b.

Welcome back, let's start by going through a.

Well, we can substitute either straightaway or we could simplify first.

I've gone for simplifying first because I know that tends to make the problem easier, especially given I've got to simplify at the end anyway.

Root 75 has therefore become 5 root 3, b is as simple as it can be so it stays as it is.

And root 48 becomes 4 root 3, I've then substituted, a is my numerator so that's 5 root 3, b plus c, well, 2 root 3 plus 4 root 3 is 6 root 3.

So if I hadn't simplified first I would've substituted and then in order to add those surds I would've had to have simplified c.

I'm therefore left with 5 root 3 over 6 root 3, the root 3s cancel, leaving me with 5 over 6.

Let's now look at part b.

Well, the first thing I'm going to do is I'm going to simplify both fractions and then consider how I can add them.

For 2 over 3 root 2, I've started by rationalising the denominator so I multiplied by root 2 over root 2.

I then tidied up by going, well, 2 over 6 is the same as 1 over 3.

Hence, I have root 2 divided by 3.

For the second fraction remember, I multiply by using the difference of two squares to work out what I should multiply by.

So I've multiplied both numerator and denominator by 2 plus root 3.

This gave me a numerator of 4 plus 4 root 3 and a denominator of 4 subtract 3.

Well 4 subtract 3 is just 1, so I'm just left with my numerator and suddenly, this got a lot easier.

I'm left with 1/3 of root 2, or root 2 over 3, plus 8 plus 4 root 3 and actually I can't combine that because there are no like terms there and everything is written as simply as possible, so it's done.

Now, we know that surds don't just appear in algebra.

They can appear in a variety of mathematical topics.

For example, I want to find the distance of the straight line between the two points you see here.

In other words, how long is that line? Now you may not spot it, but actually I think I can see a triangle here.

In fact, I can see a right angle triangle here and I can use the fact that I have coordinates to work out the distance of both the vertical and horizontal lines.

Let's start with the horizontal line.

My x coordinate for the top point is 3 and for the bottom point is 7.

Well, the difference between 3 and 7 is 4 so therefore, that horizontal distance must be 4.

Let's consider the vertical.

The difference between 10 and negative 2 is 12 so therefore the height of my triangle must be 12.

Now, if I want to find the distance i.

e.

the length of that straight line, what am I going to use? That's right, I'll use Pythagoras' theorem because I'm trying to find the hypothenuse, or the longest side, I know that that's equal to the square root of the sum of the two shorter sides both squared.

In other words, 12 squared plus 4 squared and then square root that answer.

Well, 12 squared plus 4 squared is 160.

Now, I know that the root of 160 is not a rational number so what I'm going to do is I'm going to take that surd and simplify so it becomes 4 root 10 and I know, of course, I can leave my answer in that exact form.

Quick check for you, because it's been a while since we looked at Pythagoras.

Pythagoras' theorem can be written as b equals the square root of c squared minus a squared and I've given you a right angle triangle on the right-hand side of the screen so you can see what I'm referring to with a, b, and c.

Now, is it true that we can write the theorem like that, or is it false? Don't forget, pick either A or B to justify your answer.

Pause the video now while you make your selection.

Welcome back, what did you pick? That's true of course, all I've done is rearrange the formula that you can see at the bottom of the screen.

a squared plus b squared equals c squared.

By rearranging that I can reach the form we saw in the statement at the top of the screen.

It's now time for our third and final task.

There are 3 parts to this, so we'll start with part a.

The perimeter of the rectangle is 6 root 2 subtract 2.

Given that that's the perimeter I'd like you to find the area please.

Pause the video while you do this.

Welcome back, time for part b.

In part b, you're told the coordinates of two points and you're also told the distance of the line between them.

I'd like you please to find the value of a.

Pause the video and do this now.

Welcome back, time for part c.

I'd like you please to find the perimeter and the area of this triangle.

Pause the video and do this now.

Welcome back, it's time to go through our solutions.

Let's start with a, you were told the perimeter of the rectangle was 6 root 2 subtract 2 and I asked you to find the area.

Well, that means the first thing we're going to need is we're going to need to identify the length of the rectangle.

Now, I've called that missing side b, just so I'm clear what it is I'm finding.

We know of course that half of the perimeter is equal to the two perpendicular sides summed together.

In other words, length plus width equals 1/2 of the perimeter, so 1/2 of 6 root 2 subtract 2 is equal to the side we do know, root 2 plus 1, plus our missing side.

Well, by expanding the brackets, we have 3 root 2 subtract 1 equals root 2 plus 1 plus b.

We can now, of course, gather our like terms and that tells us that b is equal to 2 root 2 subtract 2.

That's as simple as I can write that.

Let's use that now to calculate the area.

To find the area we, of course, multiply the two sides.

I'll need to expand those brackets and that leads me to the expression 4 plus 2 root 2 subtract 2 root 2 subtract 2.

Oh, look at that.

the middle two terms will cancel out leaving me with 4 takeaway 2, or 2.

In part b, you were given the coordinates of two points and the distance of the line between them.

If I remember from earlier, this is a great time to draw that right angle triangle.

I can work out the coordinates of the corner of my triangle.

At that point, I know that if I've moved from the bottom left coordinate straight across although my x value has changed, my y value has not so the y coordinate must be 3 for the corner.

Likewise, if I consider the top point, 7 and 5, I know that if I move straight down from that point the x coordinate has not changed, but the y has which means the x coordinate for that corner will be 7.

In other words, the coordinates of the corner are 7, 3.

I'm going to use those coordinates so that I can write both the length and the height of this triangle.

The difference between 3 and 5 is 2 so that's the height of my triangle.

I don't know that length of course, so that's what I'm going to start off by calculating because I know a theorem that connects the three lengths of a triangle.

It's Pythagoras, of course, c squared, so in other words 2 root 10 multiplied by itself is 40.

That's equal to the sum of the squares of the two shorter sides.

In other words, 2 squared plus x squared, or 4 plus x squared.

I can take away 4 from both sides and then square root so that distance must be 6.

If I'm going for an x coordinate of a to an x coordinate of 7 and there's a difference of 6 between those coordinates then I know a must be 1 because 1 add 6 is 7.

In part c, we wanted to find the perimeter and the area of the triangle.

I've started by saying the missing side length I'm going to refer to as x.

What I've then done is squared both sides of my triangle.

I know I'm going to need to do this because I need to do Pythagoras to find the missing side length.

Having squared both sides, I'm going to use Pythagoras.

When I sum these, I end up with 24 therefore x squared would be equal to 24 or x equals root 24, which I can simplify to 2 root 6.

Now that I've got my three side lengths I'm capable of calculating perimeter.

I could of course calculated areas straightaway without doing any of this because the two sides I need for the area are already there.

Here's my expression for the perimeter and I got that by summing the three side lengths.

I then gathered the like terms to give a result of 6 plus 2 root 6.

The area, of course, I multiplied the base by the perpendicular height and then divided by 2.

Or I could of course, say base times perpendicular height multiplied by 1/2.

Did you spot that my two shorter sides are actually the difference of two squares, so I'd end up with 9 subtract 3, which is just 6, and 1/2 of 6 is 3.

It's now time to sum up what we've done in our lesson today.

The key surd rules that we've been using are that the square root of a times b is equal to the square root of a multiplied by the square root of b.

We've also used the square root of a over b is equal to the square root of a divided by the square root of b.

We've also used that if we square a surd we get a result that is equal to the radicand of that surd.

We can use these rules to solve a variety of problems that involve surds.

Well done, you've worked really well, not just in this lesson but across the whole unit.

I look forward to seeing you in one of our future units.