Lesson video

Welcome to today's lesson.

My name is Miss 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 will help you out in any way I can as we work our way through.

Let's get started then.

Welcome to today's lesson on solving quadratics by factorising.

We're gonna put together all your skills on quadratics and actually be able to solve some now.

By the end of the lesson, you'll be able to use this method of factorising to solve quadratic equations.

With that in mind, you need to be able to factorise in order to be able to solve by factorising.

So make sure you are happy with the concept of factorising a quadratic before you carry on with this lesson.

We're also gonna talk today about solutions.

If you're not sure what a solution is, there's the definition on your screen.

We're also gonna talk today about quadratic graphs.

I'm gonna use this word, parabola.

If you haven't come across it before, a parabola is a curve which has a special property.

The property is that any point on the curve is an equal distance from a fixed point, which is called the focus, you can see it on our diagram, and a fixed straight line called the directrix.

The key point is that they have a line of symmetry that goes through that focus and makes rectangles up to the directrix.

That might sound quite complicated, but essentially we're looking at special curves which have this line of symmetry down the middle, and the important thing being that quadratic graphs are parabolas.

So we're gonna start by looking at solutions from a graph.

We can use graphs to solve equations.

I've drawn the graph of y equals 18 subtract two thirds x.

So how can I find a solution to the equation 18 minus two thirds x equals 12.

What do you think? So you might have said that we could solve this equation using inverse operations, or because we have the graph, we could use the graph to find the solution.

If I want the solution where this is equal to 12, I can plot the line y equals 12, and then I can find where that intersects with my graph of 18 minus two thirds x.

We can see that when y is 12, x is 9.

So there's a solution to the equation 18 minus two thirds x equals 12, when x is 9.

Are there any other solutions to this equation? How do you know? So in this case, there aren't any other solutions.

This is a linear graph, there will only be one intersection between those two straight lines.

So now we're gonna find solutions to quadratic equations using their graphs.

We can do this in the same way.

So we are gonna solve x squared plus 2x minus 3 equals 5.

We can use a table of values to help us plot our graph.

So if we do -2 squared plus two lots of -2 minus 3, we get -3.

We can substitute -1 in instead and we get -4.

And we can carry on substituting values.

So Lucas says, "I don't need to draw the graph now, "I can see the solution from the table." Do you agree? Okay, so yes, we can see the solution from the table.

We can see that x squared plus 2x minus 3 equals 5 when x equals 2.

However, by drawing the graph we can see if there are any other solutions.

So let's have a look.

So there are the points that I have plotted and I'm gonna use technology to add some more points and draw the parabola.

Now I can find where y equals 5.

And you can see that there's actually two solutions.

There's the one that Lucas found before at 2, but also this parabola intercepts with the line y equals 5 when x equals -4 as well.

We didn't have that in our table 'cause we didn't plot those values in our table.

When drawing graphs by hand, it is sometimes not possible to find the exact solutions.

Instead, we can approximate solutions to equations.

Even when using technology, sometimes it's hard to see the exact solutions if they're not integer values.

So let's try x squared plus 2x minus 2 equals 5.

So I've plotted the graph of x squared plus 2x minus 2 and I've plotted the line y equals 5.

You can see that the solutions are not on integer coordinates.

Drawing a vertical line can help you make sensible estimates.

So let's draw a vertical line from our intersection points down to my x-axis and then we're gonna make a sensible estimate, so I've gone a -3.

8 and 1.

8.

A quadratic equation can have two solutions as we've seen.

They could also have one solution or no solutions.

So let's have a look at some examples.

This time I've got x squared minus 6x plus 5.

How many solutions are there to the equation x squared minus 6x plus 5 equals 5? Use the graph to help you.

So let's find where that expression is equal to 5.

And you can see that we have two solutions, one when x is zero and one when x is 6.

How many solutions are there to the equation x squared minus 6x plus 5 equals 1? Hopefully you looked at that line there, and again you can see that we've got two solutions.

They are approximate solutions because they're non-integers.

I think they're roughly 0.

8 and 5.

2.

Even though they're approximate solutions, they are still solutions and there are two of them.

I wonder if you can see this.

Where on the graph is there only one solution? Right, at that very, very bottom point of our graph, the lowest the graph gets to before it starts increasing again, there's only gonna be one solution.

If you follow that solution to our x-axis, we can see it's when x equals 3.

When x equals 3, that has a value of -4.

So what we're saying is the equation x squared minus 6x plus 5 equals -4 only has one solution.

I'm wondering if you can think of an equation which wouldn't have any solutions.

Of course, it's gonna be anything below that line.

So if x squared minus 6x plus 5 equaled -5, we would not have a solution because the graph does not reach values less than -4.

In fact, you could have that expression equal to any value less than -4 and it would not have a solution.

Andeep is looking at the same expression and has said the graph of y equals x squared minus 6x plus 5 crosses the x-axis at 1 and 5.

What do those two points represent? What equation are these solutions to? Well Lucas has pointed out that the x-axis is when y equals zero.

You've possibly used that fact before when you've found x-intercepts.

So x equals 1 and x equals 5 are solutions to the equation x squared minus 6x plus 5 equals zero.

Bear that thought in mind because we're gonna come back to that when we're looking at solving by factorising.

Okay, I would like you to use this graph to match the equations to their solution type.

So the diagram shows the graph of x squared plus 4x plus 4, and then I've written four equations, x squared plus 4x plus 4 equals 10, then equals 4, then equals zero, and equals -2.

Which of those has no solutions, one solution, two integer solutions, and two non-integer solutions? Match them up and then we'll check your answers.

Let's have a look.

So equals 10 has two non-integer solutions, equals 4 has two integer solutions, -4 and zero.

Equals zero has one solution, when x is -2.

And equals -2 will have no solutions 'cause the graph does not cross the line y equals -2.

Time for a practise then.

I've drawn the graphs for you.

All we need to do is find the solution or solutions to these equations.

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

Well done, and a few more for you to have a try this time.

This time you might have your solutions as non-integers, therefore you can only make an estimate.

Make sure your estimate is as accurate as it can be.

And when you are happy with those, come back for the final question.

And finally, I've drawn the graph of x squared minus 4x plus 4 and I'd like you to write an equation with each of these following features.

So first, an equation with two integer solutions, then two non-integer solutions, one solution, and no solutions.

When you are happy, come back for the answers.

So the first one, you should have x equals -5 and x equals -1, that's where that curve intersects with the line y equals 3.

For b you should have one single solution which is when x is -3.

For c you've got two solutions, when x is -3 and when x is zero.

x equals zero is a valid solution so that should be written as a solution.

And then we've got x equals -5 and x equals 2, and that's where our graph crosses the x-axis.

Then for the third one we've got x equals 2 and x equals 7, then x equals 3 and x equals 6.

Again, that's where that crosses the x-axis 'cause that's when y equals zero.

For the second set, if you didn't get exactly the same answers as me this time, that's absolutely fine as long as you are in the right region.

So for a, something like -5.

5 and 1.

5.

As long as you've got -5 point something and positive 1 point something, then you are close enough.

It's always good to be as accurate as possible though, so well done if you did get within one decimal place.

For b, -4.

2 and 0.

2.

For c, -0.

4 and 5.

4.

d actually had no solution, so well done if you wrote that.

And finally we've got 1.

8 and 7.

2 approximately.

And then f was actually exactly 1 and 8, so I can use the equals sign there rather than my approximately equal.

For 3, you've got some options.

So for a, if you're just using values on the graph, you'd have x squared minus 4x plus 4 equals 1, or equals 4 or equals 9.

Interestingly you might have spotted that those are the square numbers.

For b, you want x squared minus 4x plus 4 equal to, and then any value greater than zero, but not 1, 4, 9, or any of those square numbers that we talked about having integer solutions.

So you might have gone with x squared minus 4x plus 4 equals 2 or equals 3.

Didn't have to be an integer, you could have that expression equal to 4.

8 or 6.

2.

There's only one answer for c, an equation with one solution is x squared minus 4x plus 4 equals zero.

And then for d, equal to anything less than zero.

So for example, x squared minus 4x plus 4 equals -1.

Now we're gonna have a look at finding solutions from factors.

We're gonna look at some products which are equal to zero.

I'd like you to have a look at these and think about what could the value of a be in these equations.

Pause the video and have a think about what's gonna happen here.

Right, well if 5 multiplied by something is zero, that something has to be zero, 5 times zero is zero.

Doesn't matter how big that value is, 2,461 times something is gonna equal zero, that something has to be zero.

But this one, 4 times something is zero, well that's something has to be zero, so 2a is zero.

Well that means a has to be zero.

I might have caught you out with this one, I don't know if you spotted it.

If it's 2 times a minus one is zero, then a minus 1 has to be zero.

In order for a minus 1 to be zero, a has to be 1.

Well done if you spotted that already 'cause that's what we're going to use in this lesson.

b times a equals zero, well actually in this case, a could be anything if b was zero.

So if b was zero, a could be 2, -3, 1,264, anything as long as b is zero.

But if b wasn't zero, then a would have to be zero.

What we're saying is either a or b must be zero if they multiply to zero, and that's the exact fact we're gonna use for this part of the lesson.

When two values have a product of zero, one of them must be zero.

So if I had axe equals zero, either a is zero or x is zero.

This is actually a really useful feature.

So Sam's trying to get their head around this, "But surely this would work for lots of numbers." If I said axe equals 5, then a equals 1 or x equals 1 and the other number is 5.

Can you explain to Sam why they are incorrect? So Sam is actually only thinking about integer values and the fact that 5 is prime.

So yes, 1 multiplied by 5 does give you 5, but there are other things that multiply to 5, 0.

5 times 10, 2.

5 times 2, -5 times -1.

So there's infinite pairs of values with a product of 5.

Okay, Sam's still thinking and they say, "Okay, but there are infinite pairs of values "which have a product of 0 too.

"1 times zero, 2.

5 times zero, zero times 15, "zero times anything." Jun's got this figured out.

Jun says "Yes, but the point is that one of the values "has to be zero." That is the fact when two things multiply to zero, one of them has to be zero, even though the other one could be anything.

It's really important that you come to this understanding that Sam has that you know why this is a fact.

So take time if you need to to check you are now happy with this fact.

Let's see how that is useful.

If we can write an equation as two or more expressions with a product of zero, then we know that one of the expressions has to be zero.

So if I had x multiplied by x plus 4 and I know that it's equal to zero, then either x equals zero or x plus 4 equals zero.

That means we can find values that satisfy this equation.

Well we just said x equals zero would work.

Let's check it.

Zero multiplied by zero plus 4 is zero multiplied by 4, which is zero.

Now there's another number that works as well.

If x was -4, then that bracket becomes -4 plus 4, so we've got -4 lots of -4 plus 4, or -4 lots of zero, <v ->4 times zero is also zero.

</v> So we found two solutions to that equation.

No other values is gonna work.

You can't multiply any other things to get zero unless one of them is zero.

So Sam's had a revelation, "I see now.

"So if I had x multiplied by x minus 3 equals 5, "I couldn't work out x straight away, "but if I had x multiplied by x minus 3 equals zero "I could." Well let's check their theory.

If I have x multiplied by x minus 3 equals zero, I know that either x has to be zero or x minus 3 has to be zero.

So one solution is when x equals zero and the other solution is when x minus 3 equals zero.

Where if I solve that equation by adding 3 to both sides, that means x has to be 3.

So there's another solution when x equals 3.

You could substitute these into the original equation and check they worked.

This method works when the product of two binomials equals zero as well.

So if we have x plus 5 multiplied by x minus 3, that means either x plus 5 is zero or x minus 3 is zero, 'cause their product has to be zero.

We've got two linear equations now then.

If x plus 5 is zero, x is -5.

If x minus 3 is zero, x is positive 3.

So those are our two solutions.

True or false then, if the product of two binomials is zero, one of the solutions will be zero.

What do you think? Well done if you saw that was false? If you're not sure why, have a look at these justifications.

Which one's the correct justification? Right, if the product of two binomials is zero, it's the fact that one of the binomials has to equal zero.

That doesn't mean x is zero, it means the binomial is zero.

Here's an example that might help you see.

Which of these is a solution to x plus 3 times x minus 7 equals zero.

Well done if you saw both solutions.

There's one when x is positive 7 and one when x is -3, 'cause what we need is those binomials to be equal to zero.

How about this one? Which of these is solution to x plus 5 multiplied by x plus 5 equals zero? Right, we've just got one solution this time.

Well done if you saw that it was x equals -5.

So we can put our skills together now and solve by factorising.

So if I wanted to solve x squared plus 2x minus 3 equals zero, I could factor it.

So I've written it as two expressions with a product of zero.

I can use an area model to help me factorise if I wish.

And in this case I get x plus 3 x minus 1 equals zero.

That means either x plus 3 is zero or x minus 1 is zero.

That means x equals -3 or x equals 1.

So I've draw the graph of y equals x squared plus 2x minus 3.

What do you notice about our two solutions? Right, they're the values when y equals zero, which means they're the values where the parabola crosses the x-axis.

If we're looking for when an expression equals zero we are looking for where it crosses that x-axis.

Let's try another one, x squared minus 8x plus 16 equals zero.

Factorise it any way you wish, I've done it by splitting the x term into -4x and -4x.

Then I can factorised the first two terms, so x lots of x minus 4.

And the second two terms to -4 lots of x minus 4.

That factorises to x minus 4 multiply by x minus 4 equals zero.

This time I've got a solution when x equals 4, notice both my brackets are the same.

What we can see when we draw our graph is that there's only one solution because the parabola touches the x-axis at one point, but it doesn't actually cross the x-axis twice.

So that means there's gonna be one solution when x equals 4.

So a quick check, which of these statements are true? Quadratic equations always have two solutions.

To solve a quadratic equation you should expand the brackets first.

A quadratic equation in the form x plus a, all squared, equals zero will have one solution.

The solution to a quadratic of the form x squared plus bx plus c equals zero will be where the parabola crosses the x-axis.

Which of those are true? So, top one is false.

We've seen examples where there's one solution or no solutions.

Definitely false this second one, we don't want to expand the brackets if we're solving a quadratic equation.

We want the factorised form equal to zero.

This is not to say that there won't be equations where you'll have to expand the brackets and rearrange things first if you're looking at some trickier quadratics, but we want them in factorised form equal to zero.

c is true and d is true.

Time for a practise then, I'd like you to write down the solution or solutions to the following equations.

Off you go, come back for the next bit.

Good start, now you need to put a few of your skills together.

I'd like you to write down solution or solutions to these equations.

You are gonna need to factorise first, give it a go.

And finally, Jun wants to solve the equation x squared minus 10x minus 24 equals zero.

He's shown his working out really well, so have a read through his working out and he's got the solutions x equals 6 and x equals 4.

Jun says, "I've drawn the graph "of y equals x squared minus 10x minus 24 "and I think my answer must be wrong." What I want to know is why does Jun think his answer must be wrong and can you spot a mistake, which means his answer is wrong.

Give that a go, come back when you're ready for the answers.

Fantastic, I would like you to take your time to mark your answers to these, checking you've got both solutions where there were two solutions.

I'm gonna draw your attention to a couple of them.

To start with c, if 2x plus 6 is the bracket, so 2x plus 6 is zero, that means 2x has to equal -6 and x has to equal -3, there was an extra step in that one.

Notice also h and i both had one solution because it had a repeated bracket, which meant that either x plus 9 is zero or x plus 9 is zero, both of which give the solution of x equals -9.

The same for i, but with x equals positive 9.

Pause the video if you need to check the rest of your answers.

And secondly, so you're gonna need to watch out for your factorization and then your final solutions.

Please don't skip that step of writing the factorised form first.

If you do skip that step, generally people get in a muddle as to whether their solutions are positive or negative.

So again, check you've got the correct factorization, then check your final answer.

I'm gonna draw your attention to e, 'cause e when it was factorised only gave one solution, and to i, well done if you spotted you could factorise that as the difference of two squares, x squared minus 49 is x plus 7 x minus 7.

You'll get two solutions that time, one's -7, one's positive 7.

Pause the video and check the remainder of your answers.

And Jun, so Jun thinks he made a mistake because the parabola crosses the x-axis at -2 and 12, but his solutions were 4 and 6.

So he realised that something must be wrong, either his graph or his solutions.

Now if he drew his graph with graphing software, he's probably expecting his solutions to be wrong.

Right, the mistake came where he decomposed his x terms. He thought his factors were gonna be -4 and -6.

That's not gonna work 'cause -4 multiplied by -6 is positive 24, he needs a constant of -24.

The correct factorization would've been x minus 12 and x plus 2.

And that would give those solutions as seen from the graph.

Fantastic reasoning in that last bit, you are really showing that you've understood how this structure of factorising works and it's link to the graph.

So today we've seen how quadratic graphs can be used to find solutions or approximate solutions where they exist 'cause we've also seen that there could be no solutions, one solution or two solutions.

If a quadratic can be written as the product of two expressions equal to zero, then one of the expressions must be zero.

Setting those expressions equal to zero allows us to find the solutions.

We brought that together at the end to show where the solutions to the equation, when it's equal to zero, is where the graph crosses the x-axis.

Read that back through if you want to consolidate what you've learned today.

Thank you for joining us and I look forward to seeing you again.