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Well done for taking the first step to learning by choosing to load this video today.

My name is Ms. Davies, and I'm gonna be helping you throughout this lesson.

Iteration is a fantastic topic with lots to explore and play around with.

There's some really fascinating results that's gonna come out of today, so I hope you have fun exploring things.

With that in mind, make sure you've got your calculator handy.

You will need that today.

And we'll talk about how to use that efficiently throughout this lesson.

Let's get started then.

This lesson is called the Signs of a Solution.

By the end of the lesson, you'll be able to deduce why change of sign may indicate a solution.

We're gonna look at iteration in the second part of this lesson.

So if you'd like a reminder of what iteration is, pause video and read through that now.

We're also gonna use the idea of lower bounds and upper bounds.

Again, if you need a reminder as to what those are, pause, video, and read through our keywords.

So we're gonna start by showing a solution that lies between 2 values.

Then the second part of the lesson, we're gonna bring in our iteration and use this to test our solutions so we can see roughly where a solution to an equation might be by testing some values.

So there's a cubic equation, and I can substitute some values for x and see if I get close to 0 'cause we've got our equation equal to 0.

Let's try negative 3, negative 2, negative 1, and 0.

Now, between x as negative 2 and x as negative 1, the value changes from being negative to being positive.

This means that the graph is likely to cross the x-axis somewhere between these values.

We're gonna look at why this might not mean it crosses the x-axis later.

For most graphs, where the sign changes, the graph is crossing the x-axis.

Now because our equation is equal to 0, we expect there to be a solution between these values.

'cause the x-axis is where the equation equals 0.

So if it's crossing the x-axis, that is a solution.

Now, if we want a more accurate idea of where a solution might be, we can look at values between negative 2 and negative 1.

So we can see there's a change in sign between x equals negative 2 and x equals negative 1 so expecting a solution between them.

Let's try x equals negative 1.

And we get 2.

125.

Between which values is the solution likely to be? Well done if you said negative 2 and negative 1.

5 because the sign changes from negative to positive between x as negative 2 and x as negative 1.

Our values go from negative 1 to 2.

125.

So between y as negative 1 and y as 2.

125, You're gonna get y 0 somewhere between the 2.

Okay, if I find the midpoint again, so I go halfway between negative 2 and negative 1.

5, so negative 1.

75, and I try that value.

The exact value is not important, which is why I've just truncated that.

The important thing is that it's positive.

So we know that the solution is between x as negative 2 and x as negative 1.

75.

Now if we draw the graph of y equal x cubed minus 3x plus 1, we can see there is a solution to x cubed minus 3x plus 1 equals 0 between negative 2 and negative 1.

75.

And we can see clearly now why going from a negative value to a positive value means a graph is probably crossing the x-axis.

Now we're also expecting there to be a solution between 0 and 1, and between 1 and 2.

So let's substitute x as 1 and x as two.

And again, we can see a change in sign.

So we're expecting there to be a solution between x as 0 and x as 1, and then another 1 between x as 1 and as 2.

That change in sign is indicating the graph crossing the x-axis.

Now, knowing the rough shape of the graph can be important.

So here's the graph of 1 over x.

If I substitute x as negative 1, I get negative 1.

1 over negative 1 is negative 1.

If I substitute x as 1, I get 1, 'cause 1 over 1 is 1.

Now there's a change in sign there.

we've gone from negative 1 to 1.

So using our previous rule, this will suggest there's a solution to the equation.

1 over x equals 0 between negative 1 and 1.

However, we can see from the graph that is not the case.

For this type of graph, the change in sign does not indicate the graph crossing the axis, and that's because there's those asymptotes as x equals 0 and y equals 0.

So even though there's a change in sign, the graph does not cross the axis.

For parabolas, or for cubic curves, or for a linear graph, You know that a change in sign indicates the graph crossing the axis.

But if you're looking at more complicated equations, then you need to be careful to make sure that you're not dealing with asymptotes where the graph doesn't cross the axis.

We've got a quadratic equation this time, and we're gonna show that has a solution between x as 3.

7 and x as 3.

So if I substitute 3.

7, I get negative 0.

11.

If I substitute 3.

8, I get 0.

24.

There's a change in sign, and that indicates the graph crosses the x-axis between those point.

Therefore there's a solution between x as 3.

7 and x as 3.

Now we can be confident with this because this is a quadratic, and it will form a parabola.

So we know that a change in sign will mean the graph crossing the x-axis.

Your turn.

I'd like you to show that this equation has a solution between x as 1.

5 and x as 1.

75.

So in x as 1.

5, we get 0.

75.

When x is 1.

75, we get negative 0.

0625 Now there's a change in sign, doesn't matter what values are, as long as there's a change in sign, and that's indicating that the graph crosses the x-axis between those points.

Again, this is a negative parabola.

We know that a change in sign is indicating that graph crossing the axis.

Andeep wants to show there's a solution to this equation between x as 3 and x as 4.

She's gonna substitute x as 3 and x as four.

Oh, that didn't work.

There was no change in sign.

Can you see why that didn't work? Pause the video.

What do you reckon? Right, his equation is not equal to 0.

A change in sign only indicates a solution to equation equal to 0, 'cause we're looking at where it crosses the x-axis, which is obviously when y is 0.

Now if we rearrange the equation, so if we rearrange it to x squared minus 2x minus five equals 0, then substitute 3 and 4, we do get a change in sign.

So just a reminder that if you're looking to use that change in sign rule, you do need your equation equal to 0.

Okay, this time Andeep wants to show there's a solution to this equation between x as 0 and x as 1.

He's gonna substitute.

Oh, he's definitely got equation equal to 0 this time, but it still didn't work.

Now, a graph is gonna help us see what's happening here 'cause this is a bit of an odd case.

So if I draw the graph of 3x cubed minus 3x plus 1, there you go.

And he chose to substitute x as 0 and x as 1.

So have a look at that part of the graph.

And what you'll see is there is a solution between x as 0 and x as 1, but there's actually two solutions.

So the values Andeep picked were too far apart because what had happened is that the sign had changed to negative but back to positive between 0 and 1.

So it's just another case to think about.

Just because there isn't a change in sign doesn't mean that there's not a solution between those values.

If you are using the change in sign rule, you want to make sure you're picking values quite close together or you're using a graph and thinking about the shape of a graph.

Right, Andeep says, "According to my calculator, there is a solution to this equation when x equals negative 1." So he's chosen to use negative 1.

1 and negative 0.

And these values are really close together this time.

But despite picking values just above and just below, there are still no change in sign.

I wonder why this happened this time.

Again, the graph is gonna be really important here.

So if we draw the graph of x cubed minus 3x minus 2, you'll be able to see that there is a solution there when X equals negative 1.

So Andeep has used his calculator correctly.

Can you see why the change in sign rule hasn't worked? Pause the video.

What do you reckon? Right, Andeep's solution is a repeated route.

So that means his graph just touches the axis at this point.

So it's not crossing the axis, so there's no change in sign between negative and positive 'cause it's not crossing the axis.

It's just touching the axis.

So actually testing a value either side and seeing if the sign changes is not gonna work.

So Andeep's got a good point here.

Looking for a change in sign is a useful way to indicate a solution, but it does have limitations.

So we are gonna use it 'cause it's gonna be a really good way of testing certain solutions, but just be aware that there's some times where a change in sign doesn't indicate the graph crossing the axis or where there is a solution but there's not a change in sign on both sides.

Right, quick check.

Here's a table of values for this expression.

Which of the following are true? Off you go.

Well done if you said B.

There doesn't look to be a solution between negative 1.

3 and negative 1.

2 'cause there's not a change in sign.

Now we have looked at examples where there is a solution, even though there's not a change in sign.

We definitely can't be sure that there is a solution.

We're not expecting there to be a solution.

Now there is a solution between negative 1.

2 and negative 1.

1 because we see that change in sign.

And because this is a cubic, we know that this will cross the axis or touch the axis if it has a repeated route where there's a solution.

Okay, so we can be pretty confident there's a solution there, but we don't know that it's exactly halfway between those values.

So we can't say the solution is negative 1.

15, We just know it's somewhere between those values.

Right, time for practise.

I'd like you to use that substitution and that change in sign rule to show that there's a solution to these equations between those values.

Off you go.

Well done.

For three and four, think about the form that your equation is in.

If you're using the change of sign rule, what do you need to make sure you do first with your equation? There's a bit of a hint.

Give this one a go.

Question five, I've got a cubic equation, so it's gonna be a cubic curve.

I want you to determine if there's likely to be a solution between the given values and write a sentence to explain your answer.

Off you go.

Question six, Aisha's checking if there's a solution to this equation between x as 0 and x as two.

She substitutes x as 0 and x as two.

There's no sign change between them.

She says, "There's no sign change, so no solution." I'd like you to give a possible reason why Aisha may be incorrect.

And can you suggest a way for her to improve her method to check whether there is a solution between those two points.

Off you go.

Let's have a look.

So we need to substitute and look for a change in sign.

So pause the video and check your values, and make sure you've written a quick sentence to say the change of sign so it crosses the x axis between these values.

It's a cubic? So we know that that is gonna happen if there's a sign change.

Once you're happy with your answers, we'll have a look at the next.

Okay, so we need to rearrange to equal 0 if we want to use the change in sign rule.

So we're gonna rearrange to x squared plus 2x minus 1 equals 0.

And then if we substitute those values, we see a change in sign between x as negative 3 and x as negative two, and again between x as negative 2.

5 and x as negative 2.

That's gonna be the same solution.

We're just getting closer to it.

For question four, again, we need to rearrange.

So we've got x cubed minus 2x minus 1 equals 0.

And then there's a change in sign, so there's a solution between x as 1.

61 and x as 1.

62, and there's another solution between x as negative 0.

61 and x as negative naught 0.

62.

That's two different solutions this time.

And for question five.

So if we substitute for A, there is a change in sign, so there will be a solution between those two values.

Now we know the general shape of a cubic, there are no asymptotes, so change in sign represents the graph crossing the x-axis.

For B, there is no change in sign.

So the cubic probably does not cross the x-axis between these values, so there's probably not a solution between those values.

We have seen some cases though where we've got a repeated route or where we've picked values too far apart where there is a solution without a change in sign, but there's probably not a solution between those values.

You could draw the graph on graphing software.

If you wanted to check.

For C, the same as before, there's no change in sign.

It probably does not cross the x-axis between those values.

But again, you need to check the graph to be certain.

And for D, there is a change in sign because this is a cubic, we know it's gonna cross the x-axis where there's a change in sign.

So there is a solution between those 2 values.

And for question six, so a possible reason why she might be incorrect.

Well, there could still be a solution between this values if it was a repeated route or if the graph crosses the x-axis twice so there's two solutions between those two values.

And she has picked values which are two whole integers apart.

It's quite possible that the graph crosses the x-axis twice between those two values.

So what could she do to improve her method? Well, she could draw the graph or she could try x values which are closer together so that she doesn't run that risk of the graph crossing the axis twice between those x values.

Well done.

There's lots of thinking going on there and lots of having to explain your answer.

If you want to pause the video, go back through some of those answers and check my wording, feel free to do that now.

Right, so now we're gonna use this idea to test some solutions.

So when we think we have a solution, we can substitute it into the equation to check.

So using iteration, we found a solution to this equation when x equals 0.

66 to two significant figures.

So when we substitute this into our equation, it should equal 0 because it's a solution to our equation.

We're gonna test this out and see if it works.

Pause the video if you'd like to do that now.

Let's have a look.

No, we don't get 0.

Why not? Well done if you notice it's because this is an approximate solution to two significant figures.

It looks to be close to the solution.

We're quite close to 0, but how can we check if it's definitely correct to two significant figures? It's definitely gonna be 0.

66 to two significant figures rather than 0.

67 or 0.

65.

What we can do is we can look at the upper and lower bounds.

If our answer's correct to two significant figures, then the exact solution will round to 0.

66, rather than 0.

65 or 0.

67.

So the solution is between the upper and lower bounds of 0.

66.

So our lower bound is 0.

655 and our upper bound is 0.

665.

If we substitute those into our expression, you can see that there's a change in sign between these values.

So there's gonna be a solution between 0.

655 and 0.

665.

Well, every value between 0.

655 and 0.

665 is gonna round to n 0.

66 to two significant figures.

So we can be confident now that we've got the correct answer to two significant figures.

All right, let's use this idea to show there's not a solution to this equation, which rounds to 1.

84 to three significant figures.

So what are the upper and the lower bounds of 1.

84 which has been rounded to three significant figures? Right, we've got 1.

835 and 1.

845.

So we substitute those in, and there is no change in sign this time.

So if there is a solution between 1.

75 and 1.

85, it does not round to 1.

84.

Right, quick check, what is the lower bound of x when x equals 1.

83 rounded to three significant figures? The lower bound is 1.

825.

It's the lowest value that would round to 1.

83.

The upper bound would be 1.

835.

So the smallest value that would round up to the next rounded value.

So we're gonna use these upper and lower bounds to test this solution.

Give it a go.

So there's the lower and upper bounds as we said before.

We substitute those in.

We see there is a change in signs.

So there's a solution between 1.

825 and 1.

835.

Therefore there is a solution x equals 1.

83, approximate to three significant figures.

So why are we looking at this? Well, this is a useful method to test the accuracy of a solution found by iteration.

So let's do a little bit of a recap of iteration.

So let's find an approximate solution to this equation.

Before we start, we're gonna see if we can find the nearest integer values.

So I've got a table of values, and I'm gonna substitute into my equation.

Right, without knowing the exact shape of the graph, we could not be certain.

This is gonna be a cubic, so it's possibly got as many as three solutions.

There seems to be solutions between x as negative 3 and negative 2, x as 0 and 1, and x as 2 and 3.

If we're gonna use iteration to find a solution, we need to rearrange into an iterative formula.

That means x must be the output and also the input.

There's lots of ways of doing this.

Here's one way.

So I've added 7x to both sides, subtracted 1, and cube root it.

I can write this as an iterative formula, x subscript n plus 1 equals the cube root of 7 lots of xn minus 1.

Now this iterative formula may work to find one solution.

It may work to find more than one solution.

It may not work at all, and it might depend on what values of x0 we use.

There's all sorts of things that mean an iterative formula might not work.

Hopefully, I've picked one that does.

We saw there was likely to be a solution between x as 2 and 3.

So let's use x0 equals 2, and see if we can find a solution.

Right, using our iterative formula, you might wanna pause the video and do this yourself first.

The differences between the iterations are decreasing so they are converging.

That's a good start.

The values are continuously increasing so we have to be careful to make sure we've got this to the right degree of accuracy.

Looks to be 2.

5 at the moment, it might not be.

So let's keep going and see if it stays as rounding to 2.

5 or if it gets big enough to round to 2.

Ah, looks like it's gonna be 2.

I've used my calculator really efficiently there.

I've kept evaluating iterations.

That is the 20th iteration, and I'm now fairly confident this is gonna round to 2.

6 rather than 2.

Okay, well, if that's my approximate solution, I can now test this by looking at the upper and the lower bounds.

If I substitute the upper and the lower bounds into my original equation, I can see that there is a change in sign.

So there's definitely a solution between 2.

55 and 2.

65.

All values between 2.

55 and 2.

65 round to 2.

Right, you're gonna use the same iterative formula but a different value of x0.

So see if you can find another solution to two significant figures.

Off you go.

Let's have a look.

Hopefully, you've got the solution x equals negative 2.

Again, using your calculator efficiently, you'll be able to generate lots of iterations really quickly to be confident that it rounds to negative 2.

Right, I'd like you to use the upper and lower bounds to check if x equals negative 2.

7 is a solution to two significant figures.

Off you go Right, there's our upper and lower bound, substituting in, and we have a change in sign.

So there is a solution between negative 2.

75 and negative 2.

65.

All values between them round to negative 2.

7 to two significant figures.

Of course if you had a calculator which can solve equations for you, you might want to try typing in x cubed minus 7x plus 1, and seeing if you get both the solutions that we found, and there might be a third solution as well.

'Cause if you remember our table of values, it did indicate that there was possibly a third solution.

So you could check using your calculator function.

Right, time to give this a go.

For this equation, I'd like you to test whether each solution is correct to the given degree of accuracy.

So you're gonna need to use the upper and lower bounds to see if there's a solution correct to that degree of accuracy.

Off you go.

For question two, I'd like you to show that there's a solution to that quadratic between those two values.

Then I'd like you to use the iterative formula, and I've given you a start value of 5 to find an approximate solution to two significant figures, and then use your upper and lower bounds to test your solution.

So bringing all our key skills together here to work through those parts of the question, and you should get to that final answer.

Give it a go.

For question three, exactly the same idea.

I'd like you to show that there is a solution between those two values and use the given iterative formula and the given value of x0 to find a solution.

Again, you want to use your upper and lower bounds to show your solution is correct.

Let's have a look.

So there, your upper and low bounds.

When you substitute in, there is a change in sign.

So negative 3 to one significant figure is a solution.

Right, I wonder whether negative 3.

2 to one decimal place is a solution.

It's probably the same solution, isn't it? So let's use the upper and lower bound.

No.

So there is a solution that rounds to negative 3 to one significant figure.

It doesn't round to negative 3.

2 to one decimal place.

Okay, this looks to be a different solution 0.

21.

Let's try it.

So there's our upper and low bounds.

Substituting in, no, no change in signs.

There's unlikely to be a solution between those two values.

And finally, there is a change in sign for that last one.

So we have got a solution correct to three significant figures.

That's quite an accurate solution this time, correct to three significant figures.

All right, so show the solution to this quadratic between those values.

So we substitute in and we look for a change in sign.

And this is a parabola.

It's a negative parabola, expecting the change in sign to indicate it crosses the x-axis.

If we use this iterative formula, there's x5 and x6.

You might wanna just pause the video and check you've got those right.

Right, and that's actually enough iterations to work out our solution to two significant figures.

It's going to be 5.

4 to two significant figures.

Now, we need to use the lower and upper bounds to just check.

Substituting in, we've got a change in sign, so there is a solution between 5.

35 and 5.

45, and all values between those two will round to 5.

So what we've done there is we've used iteration to find a solution.

We've been fairly confident that we were accurate 'cause we used our calculator to do many iterations, but we can still check it by using that upper and lower bounds.

Right, we did the same again.

So check your values and we've got a change in sign, and you should have approximate solution of negative 0.

37 to two significant figures.

Remember leading zeros do not count as significant figures.

Pause the video if you want to check any of that.

And then we just need to use our upper and lower bounds to check.

You've got negative 0.

375 and negative 0.

365.

Substituting them in, you can see there is a change in sign.

So we do have a solution negative 0.

37, correct to two significant figures.

Fantastic, well done.

You have worked incredibly hard today, and we've brought loads of key ideas together.

Because this is a new topic, you do need to make sure that you are practising , you're taking your time to think about what you're doing, and don't be too hard on yourself if you're finding parts of this a little bit tricky.

Graphing software can be a really useful tool to play around with this.

Whilst you're learning and practising , use your calculator, use graphing software, explore different ideas, see if you can work out what is happening.

Thank you for all your hard work today.

Hope you found some of that enjoyable and I look forward to seeing you again for another lesson.

I've finished the video