Lesson video

Well done for deciding to learn using this video today.

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

Now that you've got the basics, we're gonna start bringing together all the things you know about algebra to tackle some really interesting problems. So I hope you're looking forward to it and I hope that you're spending the time to make sure you really understand what is going on.

The more you understand with mathematics, the easier it is to replicate things in the future even if situations are slightly different to ones you've seen before.

With that in mind, make sure you pause the video when you need to, check your answers, and you can use calculators for bits and pieces, especially to check your answers as well.

Make sure then that you've got everything you need and let's get started.

Welcome to today's lesson.

In this lesson we're gonna be looking at solving quadratic equations by factorising where rearrangement might be required first.

With that in mind, you need to be confident with how to factorise an expression and understand how we find solutions from a factorised expression.

Have a look at a lesson on that before moving on with this video.

So we're gonna start by looking at rearranging quadratic equations in order to solve.

Andeep and Izzy are trying to solve the equation, x squared plus 5x plus 6 equals 2.

Andeep says, "I think we should start by subtracting six from both sides and then solve by doing inverse operations.

Izzy reckons that we have to factorise.

What do you think? Well, let's have a look at Andeep's idea.

It might seem tempting to start by subtracting six from both sides and that would leave us with the equation x squared plus 5x equals negative 4.

But then we get a bit stuck, 'cause we could subtract 5x from both sides, but that doesn't eliminate the x term.

We've still got x squared equals negative 4 subtract 5x, and that doesn't bring us closer to finding the solution.

This method works well for linear equations, but not for quadratic equations, which have x squared and x terms, is we can't isolate x using inverse operations.

Let's try what Izzy was thinking.

X squared plus 5x plus 6.

Factorise this to x plus 3 multiplied by x plus 2 and that's equal to 2.

Oh, now Izzy's a bit stuck.

The brackets have to be values that multiply to two, but that could be loads of things, two and one, one and two, 4 and 0.

5, negative 10 and negative 0.

2.

The options are endless.

So Izzy was half correct.

Factorising is a method to solve quadratic equations, but it's only useful when the equation is equal to zero.

Then we can use the special properties of the value zero.

So let's try again.

We must start by rearranging the equation so it equals 0.

We can do that by adding negative 2 to both sides 'cause 2 and negative 2 are a 0 pair.

Then the left-hand side becomes x squared plus 5x plus 4.

The right-hand side becomes zero.

Now we can factorise.

We've got x plus 4, x plus 1 equals 9.

And remember if the two values have a product of zero, one of them must equal 0.

So x plus 4 must be 0 or x plus 1 must be 0.

That gives us our two solutions of x equals negative 4 or x equals negative 1.

We could substitute them back into the original expression if we wished to see if that equaled 2.

Let's try that then.

(keyboard typing) Negative 4 squared plus five lots of negative 4 plus 6 gives us 2 and that is equal to 2.

Let's try it with negative one.

And we get two again, so both those solutions work for our equation.

Okay, I'm going to run through one on the left-hand side.

I'd like you to have a go at a similar one on the right.

Pay attention to the steps in my explanation as we go through.

So we've got a quadratic equation.

We've got x squared terms and x terms, so we're gonna want this equal to 0 so we can factorise.

I'm gonna start by adding negative 2x to both sides.

That gives me x squared plus x plus 5 on the left-hand side and 25 on the right.

Now I can add negative 25, which gives me x squared plus x plus negative 20 equals 0 or x squared plus x minus 20 equals 0.

I can now factorise.

So if I've had got x plus 5, x minus 4 equals 0, that would be x squared plus 5x minus 4x minus 20, and that simplifies to the expression I wish.

And now I can use that fact to find my solutions.

If x plus 5 is 0, then x equals negative 5, or if x minus 4 is 0, x equals positive 4.

Hey, time for you to have a go.

So you might wish to add negative 3x first, which gives you x squared minus 3x plus 2 is 20.

Then subtract 20 to get x squared minus 3x minus 18 equals 0.

From here, we need to factorise.

We've got x plus 3 and x minus 6 equals 0.

That means that x plus 3 is 0 or x equals negative 3 or x minus 6 is 0, so x equals 6, well done.

Aisha is solving x squared plus 4 x plus 5 equals 5.

She starts by adding negative 5 to both sides and ends up with x squared plus 4x equals 0.

Then she divides by x to get x plus 4 equals 0, so x equals negative 4.

Right, Izzy's not sure about this.

I was expecting there to be two solutions.

Let's look at the graph.

Right, and the graph supports what Izzy is saying.

We're expecting there to be two solutions.

Aisha's got this solution but she's missed a second solution when x is 0.

This is where she's gone wrong.

By dividing the equation by x, she's lost a possible solution.

We should not divide our equation through by x.

It causes problems in that if x is 0, that would've worked.

So we're losing a solution there where x is 0.

Instead, to avoid that, she needs to factorise at this stage.

So her first step was good, but then when we've got x squared plus 4x equals 0, we can factorised x lots of x plus 4.

Remember some expressions will factorise to a single bracket, some will be the product of two binomials.

You've just gotta keep your eye out for what the factors will be.

This is going to be a single bracket x multiplied by x plus 4.

That tells us that either x is 0 or x plus 4 is 0.

If x plus 4 is 0, then x is negative 4.

If you look back at your graph, you'll see that that was the missing solution we didn't have before.

Okay, check for understanding then.

Andeep is solving the equation x squared plus 2x minus 36 equals 2x.

He gets a solution of x equals 6.

I'd like you to check that this is correct by substituting.

Off you go.

Let's have a look.

So 6 squared plus 2 lots of 6 minus 36 equals 2 lots of 6.

That gives us 36 plus 12 minus 36 equals 12 or 12 equals 12.

So x equals 6 is a solution to our equation.

Right, then he draws the graph of y equals x squared minus 36.

How many solutions should there be? The equation x squared minus 36 equals 0.

Make sure you explain your answer.

Of course there's gonna be two solutions.

We can see it crosses the x-axis when x is 6 and x is negative 6.

So why has Andeep missed the solution? We're gonna need to look at his working out in more detail now.

Can you see where the missed solution has come from and what he needs to do instead? Off you go.

Okay, when he square rooted the 36, he forgot that there's also a negative root.

If x squared is 36, then x could be 6 or negative 6.

He's got two options of how he can avoid this in the future.

The first is to remember that when you square root a value, there's a positive and a negative solution.

Or he could've factorised at this earlier stage.

X squared minus 36 is the difference of two squares, so he could've factorised that into x plus 6, x minus 6 equals 0, and then he is less likely to miss that second solution.

So x equals negative 6 or x equals 6.

Time for a practise.

For each of these, I would like you to spot the mistake and correct each attempt to solve a quadratic equation.

Once you're happy with those, come back and we'll look at the next bit.

Your turn this time.

I know you're not gonna make any of the same mistakes as on the previous slide.

Can you find these solutions? Off you go.

And there's another set for you to have a look at.

Pay particular attention to your rearranging before you solve.

Off you go.

Let's have a look then.

So the first one, there's a mistake in our rearranging step to start with.

We're adding negative 24 to both sides.

We need a 0 pair with 24.

So we add negative 24 to both sides.

We get x squared minus 10 minus 24 is 0 and then that factor rises to x minus 12 x plus 2, which gives our solutions of positive 12 and negative 2.

For the second one, it was just the solutions that are incorrect.

The rearranging and the factorising was spot on, but the final answer is incorrect.

It's the brackets that should be put equal to zero.

If x minus 15 is 0, x is 15.

and if x plus two is 0, x is negative 2.

Solutions have the incorrect sign.

And there was madness going on in this last one.

This equation is not equal to zero.

What we need to do is we need to expand the brackets and rearrange so it is equal to zero.

We're gonna explore that more in the later part of our lesson.

But we cannot just put x equal to 0 and x plus 3 equal to 0, 'cause our equation is not equal to 0.

Have a they look at your answers for this part.

You'll see I've set my working out with our rearrangement, then our factorization, and then our final answer.

Check your answers before you move on to the next one.

Okay, for question two, we have to be a bit more careful with our rearranging.

With f and h, notice that once we rearranged, we needed to factorise into a single bracket rather than into the product of two binomials.

So we could factorise out x, because both terms had a factor of x.

So with f, you got x lots of x minus 2 equals 0, and with h you got 2x lots of x plus 4 equals 0.

Just a reminder that if 2x is 0, then x is also 0.

Make sure you're happy with those answers before continuing with the remainder of the lesson.

Well done, we're gonna step it up a notch now.

In this second part of the lesson, we're gonna have a look at solving some more complicated quadratic equations.

However, we have all the skills that we need and we know in mathematics that the rules that work for manipulating algebra will work for anything even if it looks more complicated.

The same with our number skills, we can apply those to our algebra skills too.

So let's have a look.

We've got 2x squared plus 10x equals 5x minus 3.

Doesn't matter how complicated this looks, we've got an equation with an x squared term and x terms, so we're gonna need to rearrange to equal 0.

So first, I have added negative 5x to both sides and then added 3 to both sides.

So I've got my equation equal to zero.

Just as before, the method is the same.

We need to write this as a product of factors equal to 0.

The only difference is that it might be a little bit more complicated to factorise.

In this example, you might notice that the coefficient x squared and the constant are prime and all the terms are positive.

So this is an easier one to factorise.

That means there's not as many combinations as there might otherwise be, so you might be able to spot the factors that are going to work.

If not, you can look at decomposing the x term and again, we know they're gonna be positive values that add to 5x.

So that reduces the difficulty in finding the possible combinations.

If you remember the method for finding that decomposition of the x term, if you multiply the coefficient of x squared by the constant, so 2 times 3 is 6, then we're looking for values that add to 5 and multiply to 6.

That gives us 2x squared plus 3x plus 2x plus 3.

Then we can factorise them separately, x lots of 2x plus 3 plus 1 lot of 2x plus 3.

Because the brackets are the same, we can write that as x plus 1 times 2x plus 3 equals 0.

As I said, you might've been able to spot those factors just from the original quadratic because those numbers were fairly nice.

If you're not sure about factorising quadratics with a coefficient of x squared greater than 1, you may want to practise that a little bit to help with the next bit of the lesson.

So Izzy says that means x is negative 1 or x equals negative 3.

Do you agree? I think Izzy has rushed herself a little bit here.

Absolutely x equals negative 1, 'cause if x plus 1 is 0, negative 1 plus 1 is 0.

But that second bracket is gonna take an extra step.

First we need to add negative 3, so 2x equals negative 3.

And then 1/2, so x equals negative 3 over 2.

When solving quadratics with a coefficient of x squared greater than one, there's often going to be fractional answers, so absolutely leave your answers in terms of a fraction.

Let's have a practise then.

So I'm gonna start by rearranging so this is equal to 0.

Adding negative 2x squared gives me 8x squared minus 5 equals negative 18x and then adding 18x gives me 8x squared plus 18x minus 5 equals 0.

This is a tricky one to factorise, so I'm gonna look at decomposing that x term.

To help me, I know those two values have got to multiply the same thing as 8 multiplied by negative 5.

So I'm looking for values that sum to 18 but have a product of negative 40.

Those are gonna be 20 and negative 2, so I can split my x term into 20x and negative 2x.

Now I can factor that.

I've got 4x lots of 2x plus 5 minus 1 lots of 2x plus five.

The brackets are the same, so I've got 4x minus 1 lots of 2x plus 5 equals 0 of course.

If 4x minus 1 equals 0, then 4x equals 1 and x is 1/4.

2x plus five is 0, 2x equals negative 5 and x is negative 5 over 2.

There was quite a few steps there, so feel free to read back through my working before you have a go at yours.

Off you go.

Come back when you're ready for the answer.

Okay, so we need to start by adding x squared and then adding negative 3.

We get 6x squared plus 7x minus 3 equals 0.

If you didn't get that rearrangement, then pause the video and have a go at working out the correct factors from here.

Okay, well, 6 multiplied by negative 3 is negative 18.

So if I want to use that method, I'm looking for values that adds to 7 and multiply to negative 18.

Of course that's positive 9 and negative 2.

So I can decompose my x term into plus 9x minus 2x.

And then I can factorise separately, 3x lots of 2x plus 3 minus 1 lot of 2x plus 3.

So 3x minus one modified by 2x plus 3 equals 0.

That means that 3x equals 1 and x is 1/3.

2x equals negative three, so x is negative 3 over 2.

Well done if you got that one right first time.

If you didn't, don't panic, but please do pause the video and read back through it to see where you weren't so sure.

Let's have a go at this one, 4x squared equals 25.

I'm gonna rearrange to equal zero.

The reason I'm doing that is I can now see that this is the difference of two squares, 'cause it's 2x all squared minus 5 squared.

I can write that as 2x plus 5, 2x minus 5 equals 0.

So my solutions are negative 5 over 2 and positive 5 over 2.

Have a go at 9x squared equals 36.

Okay, there's probably two methods you could've used here, so I'll go through one and then I'll explain the other.

Firstly if we rearrange equal to 0, this is the safest method.

You're less likely to miss any solutions this way.

This is the difference of two squares, so you can write it as 3x all squared minus 6 squared, which gives you 3x plus 6, 3x minus 6 equals 0, which gives you the solutions of negative 2 and 2.

However, you may have spotted that both sides of the equation have a factor of nine.

So you could start by dividing 3 by 9 to get x squared equals 4.

Then you could square root both sides of your equation.

Remembering that 4 has two roots, positive 2 and negative 2.

So do make sure you don't miss that second solution if you solved it that way.

So if x squared equals 4, x equals 2 or negative 2.

There's also an option on the second line to divide through by 9, 'cause everything had a factor of 9.

Make sure you check back through it and you're happy with how that came together.

Okay, so sometimes, we can use this factorising method to solve equations in context.

Let's look at this right-angled triangle, which has side lengths expressed in terms of x.

What can we use to form an equation for x? Can you think about anything you know about right-angled triangles that is gonna help us here? Aisha spotted it.

It's Pythagoras' theorem.

The square of the smaller sides sum to the square of the hypotonus.

That's our right-angled triangle theorem that helps us find one length if we know the other two lengths.

So we can form an equation.

We know that 2x squared plus x plus 5 squared equals x plus 7 squared.

Now this time we're gonna need to do a little bit more manipulation until we have the equation equal to 0.

Yes, we've got some elements that have been factorised, but the whole expression hasn't been factorised and is not equal to 0, so it's no use to us at the moment.

So that gives us 4x squared and then we've got x squared plus 10x plus 25 and that's equal to x squared plus 14x plus 49.

We need to do some collecting like terms and some rearranging.

We now have our equation equal to zero and we could factorise.

Equally, you might spot, we have a common factor of 4, so we can make our lives a little bit easier by dividing each term by 4.

That just makes our trinomial a little bit easier to factorise 'cause we haven't got a coefficient of x squared greater than one.

That would factorise to x minus 3, x plus two equals 0, giving us our solutions of x equals 3 and x equals negative 2.

In real-life problems, sometimes a solution for x is not valid, is it gives a negative value for a dimension which can only be positive.

In this case, if x is negative two, then the height of this triangle, 2x, would be negative 4.

That does not work in the context of this question.

So x equals 3 is the only valid solution.

We can substitute the value into the expressions for the side lengths to check that the side lengths make a right-angled triangle.

So if x equals 3, that would give us 6, 8 and 10.

Now if you know your Pythagorean triples, you'll know that they do make up the side lengths of a right-angled triangle.

If not, we can check.

So 6 squared plus 8 squared is 100 and square root gives us 10.

So those three values would make a right-angled triangle.

In questions like this, it's important that you find both your solutions first and then explore whether either of them or both of them are valid.

What I recommend you do then is circle the one that is the correct solution and just write a quick sentence explaining why that is the only correct solution.

Quick check then, which would be the first step to solve this equation? Read the answers, pick the best first step.

Lovely, expanding the brackets is gonna be the best first step.

This is not equal to zero yet, so we're gonna need to do some manipulation first.

Now that we expanded the bracket, what is the next correct step? This one is probably add 4 to both sides.

That will give us an expression equal to zero.

So Aisha has finished solving this equation.

How many solutions are there to Aisha's equation and what does that tell us about the graph of y equals 9x squared plus 12x plus 4? Off you go.

Well, I dunno if you spotted, there is one solution, because the brackets are identical.

That solution is negative 2 over 3.

What that means is the graph will touch the x-axis at one point, but it won't cross the x-axis at 2 points.

There's a picture of our graph so that we can see where that solution is.

Time for you to have a practise.

I'd like you to find the solutions to these quadratic equations.

There's loads of space for your working out, so take your time to present this accurately.

You're gonna make less mistakes this way.

Off you go.

Well done, second set, you've got some more bracketed expressions this time.

See if you can find the solution or solutions to each one.

And finally, the area of these rectangles are the same.

You can see their side lengths are given as expressions.

First, I'd like you to form an equation for the length of x.

Then find the solutions and then use that to tell me the area of the rectangles.

I need you to explain any decisions that you have made.

Give that a go.

Come back when you're ready to look at the answers.

Well done.

Let's have a look.

Make sure you check your working out, especially if you didn't get the answers negative three or 5 over 2 or 2.

5.

For b, you should have 4 over 3 or 3.

Again, check the working if you're not sure where you went wrong.

And C, you get negative 1 or negative 4.

You might have got different factors to me.

Notice 2x plus 2 is the same as 2 lots of x plus 1.

So you might have divided 3 by 2 earlier on or you might have ended up with your second bracket doubled.

If you've got x plus 2, 2x plus 8, that's fine as well.

All of them lead back to the solution of x equals negative 1 or x equals negative 4.

For the second set, so the first one you're gonna want to expand your bracket and rearrange.

Then factorise again.

Your solutions are 1.

5 or negative 2.

5.

If you left them as fractions, that's 3 over 2, or negative 5 over 2.

Now e, if you expanded those brackets first and rearranged to equal 0, I hope you're now feeling a little bit silly.

Because what you should spot is the brackets are the same.

So you could have rearranged this as 4x lots of x minus 2 minus 8 lots of x minus 2, which gives you 4x minus 8 lots of x minus 2.

What you notice is that both brackets give you the same solution.

Just like before, because 4x minus 8 is 4 lots of x minus 2, that factor of 4, you could've divided through by earlier.

Now if you've spotted an even quicker away, well done.

Because those brackets are the same, that means 4x must equal 8.

X therefore is 2.

So just keep your eye out that you're not over-complicating a question and feel proud of yourself if you spotted that.

It's okay if you did it the other way, but you're making extra work for yourself.

So just take a moment to go, oh, yeah, I could've done that a better way and then look out for that in the future.

For f, we've got the difference of two squares, which gives us 2 over 5 or negative 2 over 5.

Again, at this stage, you could have divided by 25 first and got x squared equals 4 over 25 and then square rooted the fraction.

But that does require you to remember that there are two square roots, a positive and a negative.

Take your time to check those and then we'll have a look at the last answer.

Hopefully you've started to see now how these skills can help us solve different problems. An equation for the length of x, the simplest way is to write them as a product of two binomials.

3x plus 8, x plus 5 equals 2x plus 12 multiplied by 1 minus x.

If you expanded those to get your equation, that's absolutely fine.

I'm gonna do that in my next step.

So to get my solutions, I need to expand my brackets.

She gives me 3x squared plus 23x plus 40 equals 2x minus 2x squared plus 12 minus 12x.

You could've simplified that as well, but it was getting a bit complicated, so I put in an extra step.

So now I can add 2x squared to both sides and I can collect like terms on that right-hand side.

So I've got 5x squared plus 23x plus 40 equals 12 minus 10x and do a little bit more rearranging to get my equation equal to 0.

Okay, so I'm looking to factorise this now.

I've got 28 times 5 is 140 and I'm looking at things that multiply to 140 and add to 33.

Luckily, we can see the factors already.

It's gonna be 5 and 28 again, because a and c are 5 and 28, there's a chance that the coefficients of the 2x terms are 5 and 28 as well.

Luckily they are, which saved us a bit of effort.

You can write those either way around and then factorise.

Either way, you'll get x plus 1, 5x plus 28 equal 0.

Solutions then are negative 1 or negative 5.

6, which is negative 28 over 5 if you left it as a fraction.

Now we want the areas of the rectangles.

Those are our options.

Let's see if either of them work.

If 3x plus eight is a side length and x was negative 5.

6, then that would give us negative 8.

8 as that side length.

Well done if you've spotted that that wasn't going to work.

You might have also used the height as x plus 5, because negative 5.

6 plus 5 is gonna give you negative 0.

6.

So x equals negative 5.

6 is not gonna be a valid solution.

Therefore x equals minus 1 may be a correct solution.

Let's substitute it in to check.

We get a side length of 5 and 4 for the top one, which gives us an area of 20.

Let's check with the other rectangle.

This time we've got 10 and 2, which also gives us an area of 20.

Remember we started this whole thing by saying the areas of the rectangles were the same.

Well done, there's lots of steps towards solving that problem.

I'm hoping you can now use those skills to solve more of these in the future.

Thank you for joining us in today's lesson.

We've looked at how to rearrange equations so it equals 0 and the fact that this allows us to solve by factorising.

We've talked about when equations are in context, sometimes one of the solutions, or both of the solutions are not valid.

So we need to check those final solutions to see if they will work in our context.

Well done, and I look forward to seeing you again.