Lesson video

Welcome to today's lesson.

My name is Ms. Davies and I'm gonna help you as you work your way through these exciting algebra topics.

Thank you for choosing to learn using this video.

The great thing about that is that you are gonna be able to pause things and have a real think if you come across anything you're finding a little bit trickier.

I'll help you out in any way I can as we work our way through.

Let's get started then.

Welcome to this lesson on advanced problem solving with algebraic manipulation.

We're gonna use all your algebraic manipulation skills today to solve a range of problems in a range of different contexts.

We're gonna have lots of fun and we're gonna explore the limits of what we can do with our quadratic skills.

There are a few keywords I'm gonna be using in today's lesson.

Please pause the video and check you are happy with those before continuing.

So we're gonna start by looking at using algebra to solve number puzzles.

Whenever we have an unknown value, we can use a letter to represent this.

So if Aisha is thinking of a number, we can use a letter to represent her unknown number.

You can use whatever letter that you like to stand for that number.

Jacob says, "If you double your number and subtract 17, what do you get?" Aisha says, "I get negative 32 as the final value." How could we write this using algebraic notation? We have a statement of equality so we can form an equation.

So Aisha is thinking of a number.

We're gonna let M be the number Aisha is thinking of.

It might seem obvious for this example, but it's really important that you write down what it is your letter is representing.

We're going to double and subtract seven, which we can write as 2n subtract seven and we know this equals negative 32.

So we've got an equation.

We could now solve this equation using inverse operations to find Aisha's number.

A quick check.

Alex's grandfather is three times as old as Alex's sister who's two years younger than Alex.

Which of the following is a correct expression for Alex's grandfather's age? We are using a to represent Alex's age.

You may wish to write this down in steps so that you can see which of these is the correct expression.

Off you go.

Well done if you said 3a minus six.

You might have started by thinking about Alex's sister whose age could be written as a minus two, and then Alex's grandfather is three lots of that.

So it's three lots of a minus two.

Of course, if we've got three lots of a minus two, that's the same as 3a minus six.

Okay, a little bit of reasoning this time.

Two consecutive evens integers can be expressed algebraically as n and n plus one.

Do you think that's true or false? And have a look at the justifications.

Well done if you noticed that that was false.

N and n plus one could stand for consecutive integers, but we wanted consecutive even integers.

So if n is an even number, the next even number will be n plus two.

Time to put that into practise.

So for each question, I'd like you to form an equation and use that to find the solutions.

A quick reminder that if your expression is quadratic, you've got options of how to solve that equation.

Often the simplest will be writing it equal to zero and then factorising.

This might not always work.

Have a go at those two problems and come back for the next one.

Well done.

Third question.

Sofia's older sister recently had a birthday.

Sofia pointed out that in 15 years' time, her sister's age will be the square of her age 15 years ago.

Work out how old Sofia's sister is.

So you're gonna need to form an equation and solve and then think about your solutions 'cause remember, Sofia is in secondary school.

So think about whether your solutions are valid.

When you're done, have a go at creating your own birthday number puzzle.

Play around with the structure.

Are there values that work? Are there values that don't work? Have a go and you could even get somebody else to try it out once you've figured out a puzzle.

Off you go.

Let's have a look at our answers then.

So we need to form an equation for Aisha's number.

So we should have n subtract six multiplied by n is negative five.

It's entirely up to you how you want to solve this, but the easiest way is going to be expand, make equal to zero and solve by factorising.

So we get n squared minus 6n plus five equals zero, and that factorises nicely to n minus five, n minus one equals zero.

So the number Aisha is thinking of is either five or one.

And then we can check that works.

So if Aisha's number was five, five subtract six is negative one, multiply by five is negative five.

Try the same with one.

One subtract six is negative five multiplied by one, also negative five.

So we know that they work.

Okay, let's think about our two consecutive even integers.

We said earlier that if the first even integer was n, then the next one would be n plus two.

The product of those is 624, so we can write that as n lots of n plus two equals 624.

Again, our easiest way to solve will be to expand, make equal to zero and then factorise.

So we've got n squared plus two n minus 624 equals zero.

If we're looking to factorise this, we want values that multiply to negative 624.

It also helps to be aware that their absolute values are only gonna have a difference of two because they'll have to sum to two.

And that means they're gonna be pretty close together as far as integers go.

If you think about the square root of 624, that'll give you two values that are the same that multiply to 624.

Then you want something around that.

Now, if you think about 625, the square root of 625 is 25.

So we're roughly looking around 25.

Well, let's think about 24 and 26 then because they have a difference of two.

24 times 26 does give you 624.

Of course, we wanted negative 624, so that's gonna be positive 26 and negative 24.

Just a useful skill to use if you're struggling to find factors.

Well, that means n is negative 26 or n is 24.

And remember, n was the smaller integer.

So if the n was negative 26, the other number would be negative 24.

Or if n was 24, the other number will be 26.

Okay, slightly more complicated problem this time.

So let's say x is Sofia's sister's current age.

In 15 years' time, her age would be x plus 15.

15 years ago, her age would be x minus 15.

Sofia's statement was that her age in 15 years' time is the same as her age 15 years ago squared.

So that's how we formed our equation.

Now we're gonna need to manipulate this 'cause we've still got xs on both sides.

So if we expand x minus 15 all squared, that gives us x squared minus 30x plus 225.

Of course, that needs to be rearranged to equal zero.

So we've got x squared minus 31x plus 210.

And the easiest way to solve that is going to be to factorise again.

So x minus 21 or x minus 10, which gives us the solutions x equals 21 or x equals 10.

Of course, you could use your calculator to check that those solutions are correct, especially if your calculator has a function for solving quadratics.

So you could use A as one, B as negative 31 and C as 210 and then either use the quadratic formula or your calculator's equation function to check that the answers are 21 and 10.

Right, from that point, remember, Sofia is in secondary school and her sister is older, so her sister being 10 does not make sense in this question.

So Sofia's sister must be 21.

In addition to that, if her sister's age is currently 10, 15 years ago would be before she was born.

So that also does not make sense.

So just check for those valid solutions.

Well done.

I wonder if you then had a play around with the birthday number puzzle.

Here's an example I had a go at.

So in 21 years' time, my age will be the square of my age 21 years ago.

So that means x plus 21 equals x minus 21 all squared.

Rearranging gives me x minus 28, x minus 15.

So I'm either 28 or 15.

15 would not work because 21 years ago would be before I was born, so I'd have to be 28 in that case.

Now, I wonder if you sort of played around and found values that did work and didn't work.

If you started with x plus one equals x minus one all squared, you'll have found that that one did work.

However, x plus two and x minus two all squared didn't work.

You might wanna pause the video and play around a little further.

See if you can find any patterns with the numbers that work.

We've looked at 15 and 21.

I've just told you that one works but two doesn't.

Maybe you can think about what you think's happening and then explore that a little bit further.

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

Right, I hope you had fun playing around with that if that's something you wanted to do.

You might have noticed that it's the triangular numbers that work for this problem.

If you're not convinced by that already, maybe play around with those triangular numbers, see if you can work out what is happening.

We're now gonna apply our quadratic skills to geometric problems. So we can use expressions to represent the lengths of different shapes.

So we've got a triangle with expressions for length and remember, the fundamental properties of the shape are the same as they always are.

So anything you know about triangles applies when our side lengths are expressions as well.

Can you write an expression for the area of this triangle and the perimeter of this triangle? Okay, so our area is 1/2 times the base times the height and we can write that in a variety of ways.

You might have decided to multiply the binomials first to get 14x squared plus 19x minus three and then it's half of that.

So you could halve each term if you wished.

Of course, you could halve one binomial before multiplying it by the other.

It does not matter.

Of course, if all you wanted was an expression for the area, 1/2, 2x plus three multiplied by 7x minus one is fine as well.

The perimeter is even easier because we've got a base of 2x plus three.

Then 'cause it's an isosceles triangle, we've got two lots of 5x minus six as well.

We could expand that and write that as 2x plus three plus 10x minus 12 or 12x minus nine.

Okay, a quick check of your volume knowledge.

So to calculate the volume of a prism, you always multiply the height, the width, and the length, true or false? And think about your justifications.

Well done if you said that was false.

If you said that was true, you're probably thinking about a rectangular prism.

For a triangular prism, for example, you'd need the area of the cross-section, which would be 1/2 times the base times the height, then multiplied by the length.

If you're looking at other prisms, again, you need to find the area of the cross-section and times by the length.

That's not always just gonna be the height by the width by the length.

Okay, we're gonna have a look at some area puzzles now.

These are one of my favourite type of puzzles to have a play around with.

So we're gonna use the expressions for the lengths and the areas to work out the length of a.

You'll see that we've got a couple of side lengths and we've got the areas and the areas are given to us as quadratic expressions.

So I'd want you to think about what piece of information would be helpful to work out first to help us get a and how can we use the information we already have? Once you've got an idea, we'll go through it together.

Okay, so how can we calculate an expression for the length of the dashed line? What do you think? Okay, in this case, we haven't got the right information to just subtract one length from another length.

The only piece of information we can use is the area of that rectangle.

So what we're gonna need to do is factorise 6x squared plus 8x minus eight, and that will tell us the two side lengths.

By factorising, we're writing two things that multiply to give that area, hence we found the side lengths or possible side lengths.

So let's look at factorising 6x squared plus 8x minus eight.

So if you can't see the factor straightaway, we can look at decomposing the x term.

To help us see how that decomposes, we can do six multiplied by negative eight is negative 48.

So we know that the coefficients of x must add to eight and multiply to negative 48.

That's gonna be 12 and negative four and then you can factorise those separately.

If you do that, you get 3x minus two, 2x plus four.

Just check your method for factorising before you carry on.

Now, we know that one of the expressions is gonna be 3x minus two because we can see that on our rectangle.

The other one then must be 2x plus four.

We could have used that to help us find our binomials.

We might not have needed to use a method to find those factors if we could see that one was 3x minus two.

Okay, how could we write an expression for the length of that line? Think about how you can find that length and then what your expression would look.

Like off you go.

So we can find that length by doing 3x plus two.

Subtract 2x plus four, remembering we're subtracting the whole of that expression.

So that gives us x subtract two for that length.

So now we can work out the other unknown lengths.

We can use the area of the left-hand rectangle to work out the height.

So we need to factorise 4x squared minus 6x minus four.

However, we know one of the binomials has got to be x minus two.

So that should help us find the other binomial.

The other binomial then must be 4x plus two.

Now, if you just factorised 4x squared minus 6x minus four, you might have ended up with other binomials.

Because 4x plus two is the same as two lots of 2x plus one, it's possible that you could have had that as 2x plus one, 2x minus four.

But in this case, we know that's not going to work because we know the bottom length is x minus two.

That means the height has to be 4x plus two.

Just to be aware that when you're factorising, if one of your factors still has a common factor, so for example, 4x plus two still has a common factor of two, it can be factorised further.

That means that could be written as two lots of 2x plus one, x minus two.

In the case of this problem, that's not going to help us.

It's just something to be aware of.

Okay, which of these are valid ways to express a? Off you go.

Hopefully you spotted it's the bottom one.

4x plus two, subtract the whole expression 3x minus two, but that simplifies to x plus four.

Our final value for a is therefore x plus four.

Time to put all of this into practise.

So question one, the volume of this prism is 88 centimetres cubed and I want you to work out the surface area.

Give yourself plenty of space to do your working.

Think about what information you know, what information you can work out and how that'll help you find the surface area.

When you're happy with your answer, don't forget to check it and then come back for the next question.

Question two, we've got another area puzzle for you to have a go at.

Can you find the value of a? Okay, lots of problem-solving skills in this one.

I suggest you sketch out or you use a diagram to help you think about how you'll solve this.

So we've got a rectangle that sits inside a circle.

Each vertex of the rectangle sits on the circumference of the circle and that means the centre of the rectangle is in the same location as the centre of the circle.

You could start by marking that on.

The circle has a radius of six, the perimeter of the rectangle is 28.

Can you calculate the area of the rectangle? You don't need to be able to get to the right answer straightaway.

As always, think about what you know, what things you don't know and could use algebra to express.

And then how you can manipulate that to find out what you want.

Give that one a go.

Come back when you're ready for the answers.

Okay, so for this first question, we need to start by forming an equation.

The volume is 88 centimetres cubed.

Then 1/2 times the height, which is two times the base, which is x minus six times the length, which is x plus 12 and that must equal 88.

It's entirely up to you how you then want to rearrange.

1/2 multiplied by two is one, so I've written that as x minus six, x plus 12 equals 88 and then expanded and rearranged.

You get x squared plus 6x minus 160 equals zero and that factorises nicely.

So we have x minus 16 and x plus 10 as our two solutions.

Again, you could use your calculator to check your solutions either using the solve function or using the quadratic formula or just checking by substituting into the original equation.

Right, we've got our two solutions, but x equals negative 16 is not going to work in the context of our problem.

That gives us negative side lengths, which is just not going to work.

Therefore, x has to equal 10.

Well, don't forget that we were looking for the surface area.

So now we found x, got a bit more work to do.

So we can work out the expressions of the side lengths.

So the base is going to be four and the length is going to be 22.

You could do a quick check here to make sure you have a volume of 88, so four times 22 times two times 1/2, and check that equals 88, which it does.

So now we need to find the surface area.

So we need the area of all the faces and then add them together.

Check that your working out looks like mine and you should have a final answer of 316 centimetres squared.

It's okay if you didn't get the right answer first time round.

What's really important is that you've laid your working out in a way that you can follow it.

So you can now go back through and see where you've made a mistake.

We had lots of methods for checking our answers as we went along, so we should be fairly confident at the end that if we made a mistake, it's probably in those last couple of lines of working.

And now we're gonna have a go at this problem.

So we're gonna start by working out what we can.

So we can work out the height of the whole shape by factorising 18x squared plus 6x minus 40.

We know that one of the factors is gonna be 6x minus eight.

So the other factor must be 3x plus five.

Now, if you just factorised 18x squared plus 6x minus 40 without thinking about what one of the binomials had to be, you may have ended up with different binomials.

because 6x minus eight is two lots of 3x minus four, it's possible that when you factorise this, you ended up with 6x plus 10 multiplied by 3x minus four 'cause that would be the same thing.

Obviously, if that happened, you wouldn't have had the correct binomials to make up the length of this rectangle because we know one of them has to be 6x minus eight.

It's just something to be aware of when you're factorising.

If one of your binomials still has a common factor, then it is possible it can be factorised even further.

So this could be factorised as two lots of 3x plus five, 3x minus four.

And then, of course, e, the binomial, could be multiplied by two and that would stay equivalent.

Just something to be aware of.

In this case, we know that one of the binomials is 6x minus eights, so the other is 3x plus five.

Now, we can also work out the height of the smaller rectangle.

The area is the difference of two squares.

So if that's x plus one, the height is gonna be x minus one.

Now we've got those two lengths, we can work out the height of the other rectangle.

3x plus five subtract x minus one, making sure we're subtracting the whole expression, gives us 2x plus six and then we're gonna factorise 14x minus 4x squared minus 78.

This might be easier if we factor out negative one before we start.

So we could rearrange this and write it as negative 4x squared minus 14x minus 78 and that might be easier now to factorise.

We know that one of the factors is 2x plus six.

So with a little playing around with some of our values, we can see that is negative 2x plus six multiplied by 2x minus 13.

We know one of our factors is 2x plus six, so the other factor must be negative 2x plus 13 if we multiply that second factor by negative one.

If you want to do a quick check that that expands to give us 14x minus 4x squared minus 78, that would be great.

And it does.

Now we have all the information we need to work out a.

So we've got negative 2x plus 13, subtract x plus one, which gives us negative 3x plus 12.

So a is negative 3x plus 12.

It's not always going to be the case that our final answers are numerical values.

In this case, we were just asked to express a in terms of x, which is what we've done.

There was loads of skills you used there, so well done.

If you haven't factorised with a negative coefficient of x squared before, that might be something you wish to practise.

That method of factorising out negative one to then make the x squared positive might help you with those.

And finally, well done if you managed to get yourself started.

Even better if you got to a final answer for this problem.

We're gonna work through it step by step.

Notice that I've started by drawing on the diameter of the circle.

We know that that goes through the centre of the rectangle and through the centre of the circle, so that must be 12 'cause we're told that the radius of the circle was six and then I've written the length as l and the width as W.

Okay, so now that I've started by labelling the things I don't know and the thing I do know, I can think about what relationships I have.

Now, if we use Pythagoras' theorem, we know that l squared plus w squared equals 12 squared because that's a right-angled triangle that we've got.

So l squared plus w squared equals 144.

Right, I'm just gonna leave that there for the moment and think about what else I can write.

Well, I'm told that the perimeter of the rectangle is 28.

If you go back to the question, think about what information you haven't used yet.

We haven't used that the perimeter of the rectangle is 28, so I know that 2l plus 2w equals 28, so that means l plus w equals 14.

Right, now we've got two equations.

We've got l squared plus w squared equals 144 and l plus w equals 14.

So well done if you've got to this point where you've formed these two equations from things you know.

Now, there are options here as to what to do next.

You might have decided to rearrange to make l the subject and written that as l equals 14 minus w.

Then you can substitute that in for l in your equation l squared plus w squared equals 144.

That's absolutely fine.

If you're not sure about that, maybe if you've not used that method of substituting expressions in before, the other thing you can spot is we actually want an expression for l squared plus w squared, and we're gonna get that by squaring both sides of the equation.

So if we square l plus w, we get l plus w all squared.

If we square 14, we get 196.

Now, if we expand those brackets, we get l squared plus w squared plus 2lw.

Don't forget, we also have those two partial products, which sum to 2lw, but we know that l squared plus w squared is 144, so we can write that as 144 plus 2lw equals 196, subtract 144 and divide by two to get lw equals 26.

So a couple of methods we've talked about there, which do get you to slightly different stages of this problem.

Now, at this stage, we've got lw equals 26, and actually, we've solved it.

The question we were asked in the first place was to find the area of the rectangle, and the area of the rectangle is the length times the width.

So our area is 26 centimetres squared.

For this problem, we did not need to find the length and the width separately.

However, if your method led you to find the length and then the width, that's absolutely fine.

Well done.

There was lots of elements to that problem.

At this stage, if you managed to write yourself out some equations but got a little bit stuck about how to solve it, that's absolutely fine.

The more you look at these problems, the more you can see some of the underlying structures and then you can spot some of the things that are gonna help.

For example, that step of squaring both sides wasn't the easiest thing to spot that that was going to help us in this case.

We've looked at lots of exciting problems today and really stretched our knowledge of solving quadratic equations, especially using our factorization skills.

Don't forget that when things don't factorise, we have got other methods to solve and those methods can help us check our answers, especially using a calculator where available.

We've talked about how not all solutions are numerical and also how we don't always need to find the value of each unknown.

Think carefully about what the question is asking you and don't do extra work if it's not necessary.

Fantastic.

I hope you're now gonna go away and find some more problems that you can play around with and use these skills for.

I look forward to seeing you again.