Lesson video

Hello, Mr. Robson here.

Welcome to maths.

Lovely to see you again.

It's especially lovely for you today because we're solving quadratic inequalities.

This is an awesome bit of maths, so let's go and take a look.

Our learning outcome is that we'll be able to solve a quadratic inequality algebraically.

Some key words we're going to hear throughout the lesson.

An inequality is used to show that one expression may not be equal to another.

A quadratic is an equation, graph, or sequence where the highest exponent of the variable is 2.

The general form for a quadratic is axe squared plus bx plus c.

We've got two parts to our learning today.

We're going to begin by solving quadratic equations.

We're not gonna be able to solve quadratic inequalities without the skills that we use when solving quadratic equations.

So, we're gonna spend the first half of today's lesson reminding ourselves how to solve quadratic equations.

We can solve quadratic equations in a variety of ways.

For example, x squared plus 4x minus 5 equals 0.

A very efficient way to solve this one is to factorise.

We can factorise it to x plus 5, x minus 1.

If x plus 5 multiplied by x minus 1 equals 0, then the solutions are x equals negative 5 and x equals negative 1.

Those are the two values that we'll substitute into our quadratic in order to make it equal to 0.

When we look at the graph of the equation, we can see why we have two solutions.

Our two solutions are what we call the roots of the quadratic.

Quick check that you can factorise.

I'd like you to match these quadratics to their factorised form.

Five quadratics on the left-hand side, the same five on the right-hand side in factorised form.

Pause and pair these up.

Welcome back.

Hopefully, you matched them like so.

I didn't do that deliberately, but what a pretty pattern the solution has made.

These are the same five quadratic equations.

Underneath each, I've written them in factorised form.

All that remains to do is to define the solutions.

I'd like you to do this.

Pause and find the solutions to these five quadratics now.

Welcome back.

In the first case, x plus 5 multiplied by x plus 1 equals 0.

The two values that are going to make that quadratic 0 are x equals negative 5 and x equals negative 1.

For x squared plus 7x plus 10 equals 0, our solutions are x equals negative 5 and x equals negative 2.

Those are the two values that are going to make that quadratic equal to 0.

Our next solution x equals negative 5 and x equals 2.

x equals positive 5 and x equals negative 2.

And for the last example, slightly trickier case, 2x plus 5 multiplied by x plus 1 equals 0.

Solutions there will be x equals negative 2.

5 and x equals negative 1.

You might have written one of those answers as a fraction, negative 5 over 2.

That's perfectly fine to communicate that one as a fraction or a decimal.

Some quadratics only have one solution.

X squared plus 8x plus 16 equals 0.

When we factorise this one, we factorise it to x plus 4 squared.

When I graph that one for you, you'll see why it has only one solution.

X equals negative 4.

That's because this quadratic has one repeated root.

There's only one moment when the curve touches the x-axis.

Hence, our equation only has one solution.

Quick check that you can spot such a case.

I'd like you to factorise to spot which of these quadratics have one solution.

Four to think about there.

Pause.

See you in a moment to go through the answers.

Welcome back.

Hopefully, you said a and d.

a factorised to x plus 5 squared.

That's got one solution: x equals negative 5.

B factorised like so and had two solutions.

C factorised like so and had two solutions.

D factorised to 2x plus 3 squared equals 0.

It's got just the one solution: x equals negative 3 over 2.

Some quadratics will not factorise.

X squared minus 4x plus 2.

If you want to, pause the video and try and factorise that one, but you might be here a while.

But this is not something to worry about.

We can still find a solution.

There's a few ways we can do this.

One of which is to use the quadratic formula.

The quadratic formula is x equals negative b plus or minus the root of b squared minus 4ac all over 2a.

If you say it enough times, you will remember it.

Quadratics come in the general form a squared plus bx plus c equals 0.

So, in the case of our quadratic that we're trying to solve, our a value is that x squared coefficient.

1 in this case.

Our coefficient of x is negative 4.

So, our b value is negative 4.

And our constant c is equal to 2.

We're going to substitute those a, b, and c values into the quadratic formula.

Notice when I'm substituting, I use brackets so anyone reading my work can clearly see what I've done.

From here, I'm going to simplify the bits that easily simplify.

I've now got x equals 4 plus or minus the root of 8 over 2.

This is lovely and simple to type into my Casio classwiz.

Firstly, I'm going to write 4 plus root 8 over 2.

My calculator simplifies that to 2 plus root 2.

I can turn that into a decimal where the answer will be 3.

414 and so on.

I'll just truncate the answer.

I'd like to leave it exact.

That's our positive solution.

I now go back into my calculator display, and change it to read like.

So, this will be our negative solution 2 minus root 2 or 0.

585 and so on.

Quick check, you can do this.

Which of these is the correct use of the quadratic formula to solve the equation 4x squared plus 3x minus 17 equals 0.

The quadratic equation is in the right-hand side at the top of the screen to help you.

Which one's the right substitution in this case? Pause, see if you can spot it.

Welcome back.

Hopefully, you said B.

Why was it not A? Well, the 2A term in the denominator position was incorrect.

It should be two lots of four down there.

For C, I put the A and B values in the wrong positions.

It was option B we were looking for in this case.

It's the same quadratic equation, and we know that's the correct application of the quadratic equation.

What I'd like you to do is enter it into your calculator to solve the equation.

You can enter it like so with everything you see written there, or you can simplify it before entering into your calculator.

The choice is yours.

You might want to do it both ways and just check that you get the same answer using either method.

Pause, give this a go now.

See you in a moment to check your answer.

Welcome back.

Now before I put it into my calculator, I simplified.

So, it reads negative 3 plus or minus root of 281 over 8.

When I substitute that into my calculator, my positive solution is negative 3 plus root of 281 over 8.

That's 1.

720.

I've truncated that answer.

The negative solution looks like that in my calculator display.

It's negative 3 minus the root of 281 over 8.

That's negative 2.

470, and so on.

Another method is to complete the square.

X squared minus 4 x plus 2 equals 0.

You might recognise that quadratic equation.

We've solved it already.

We know the two solutions that we're going to get, so we'll know if we completed the square correctly when we reach them.

We need to rearrange our equation.

My first rearrangement is to leave just x squared minus 4 x on the left-hand side.

I've added negative 2 to both sides.

From here, I'm going to turn the left-hand side of this equation into x minus 2 squared.

What do I write on the right-hand side? Negative 2? No.

It's a really important part about completing the square here.

I want to write the left-hand side as x minus 2 squared, but when you expand x minus 2 squared, we get x squared minus 4 x plus 4.

When we change the left-hand side to this form, we've added 4.

If we've added 4 to the left-hand side, we must do the same to the right-hand side.

Negative 2 add 4 takes us to positive 2.

Be really careful in this step when you're completing the square.

From here we can take the square root of both sides.

I must remember solving a quadratic here.

I have to take the positive and negative roots of 2 on the right-hand side.

A further rearrangement and our solutions will be x equals 2 plus or minus the root of 2.

Our positive solution is 2 plus root 2 3.

414, et cetera.

Our negative solution 2 minus root 2 0.

585, et cetera.

And you'll recognise those two solutions.

They're the exact same two we had earlier when we solved this example using the formula.

Quick check that you can do this skill.

Which of these is the correct rearrangement of the quadratic equation x squared minus 8 x plus 9 equals 0.

If we want that in completing the square form, which of these three options, is it? Pause.

Have a think.

Welcome back.

Hopefully, you said option C.

It wasn't option A or B.

So, when we look at this x squared minus 8 x plus 9 equals 0, we want to rearrange like that.

Then we're going to write that left-hand side as x minus 4 squared.

Are you surprised to see 7 on the right-hand side? Hopefully not.

When I rewrote the left-hand side to read x minus 4 squared, that is the equivalent of x squared minus 8 x plus 16.

I was adding 16 to both sides.

Negative 9 plus 16 is positive 7.

Hence it was C.

If you said B, you fell into my trap.

You didn't add 16 to the right-hand side.

Sorry about that, but it's important you learn this about completing the square.

What I'd like you to do next is solve this equation by completing the square.

We've got it in the correct rearranged form.

We've got it down to x minus 4 squared equals 7.

What do we do from here to find our two solutions? I'll leave you to ponder that.

I'll be back in a moment with the solutions.

Welcome back.

Hopefully the first step was to take the square root of both sides, and I do hope you took both the positive and negative roots of 7.

From here we can rearrange x equals 4 plus or minus the square root of 7.

A positive solution and a negative solution.

Well done.

If we can find the roots of a quadratic, it can help us to sketch the graph.

Sketching graphs is going to be really important later in the lesson when we come to solve quadratic inequalities.

So, let's practise it now while we're solving quadratic equations.

X squared minus 8x plus 12 equals 0.

The left-hand side will factorise giving us roots at positive 2 and positive 6.

This is just a sketch, so I'll just put them approximately there.

How does the graph look from here? Well, we've got a positive x squared coefficient.

An x squared coefficient of positive 1 in this case.

That gives us a positively shaped parabola, i.

e.

a parabola with a minimum point and y values increasing in both x directions from this point.

This graph has some similarities but one key difference.

Can you spot it? Well done.

A negative x squared coefficient.

So how do we sketch this one? There's lots of ways we could do it.

I'm going to factorise negative 1 out on the left-hand side.

Inside that bracket now x squared minus 8x plus 12.

Of course that factorises.

I've still got roots at positive 2 and positive 6, but crucially, the negative x squared coefficient in this case gives a negatively shaped parabola.

A parabola with a maximum point and y values decreasing in both x directions from this point.

Let's check that you can do some sketching or identify some sketches.

I'd like you to match these three quadratics to their respective graphs.

Pause, see if you can pair these up.

Welcome back.

Hopefully, you paired them like so.

x squared minus 2x minus 15 equals 0 for a.

Why is that? Well, that's how it factorises, giving us a root at positive 5 and negative 3.

And it's a positive x squared coefficient, giving us a positively shaped parabola.

For b, x squared minus 8x plus 15 equals 0 because it factorises like so with two positive roots at positive 3, and positive 5 and a positive x squared coefficient.

I hope you can see the difference between a and b there in terms of the roots.

C, negative x squared plus 2x plus 15 factorises like so.

Roots at positive 5 and negative 3 but crucially a negative x squared coefficient, so a parabola was negatively shaped.

Practise time now.

Question one.

I'd like you to solve this one quadratic using three different methods.

If you can use all three methods for solving a quadratic and arrive at the same pair of solutions at each point, you are a very fluent mathematician.

That will be nice to see.

So, pause, give this a go now.

Question two.

I'd like to solve these quadratics by selecting an appropriate method.

Five quadratics to solve, and you can solve them however you choose to.

Pause and try these now.

Feedback time.

Let's see how we got on with question one.

Part a solved by factorising.

It factorised to x plus 9, x minus 2, so solutions are x equals 2 and x equals negative 9.

We should get those same two solutions when we use the formula.

Substituting into the formula for this example would look like so.

That simplifies like so giving us a positive solution 4 over 2, which is 2, and a negative solution negative 18 over 2, which is negative 9.

The exact same two solutions.

For part c, that same problem but completing the square.

Rearrange like so and then turn the left-hand side into x plus 7 over 2 squared.

That means we're gonna add 7 over 2 squared to both sides.

Simplify that right hand side.

Take the positive and negative roots, and then we get a positive solution of 2 and a negative solution of negative 9.

Again, the exact same two solutions.

For question two, I asked you to solve these quadratics, selecting an appropriate method.

You don't have to have copied my method.

You're welcome to use any one you wish, but you should have the same solutions.

For a, I could see that one factorising quite easily giving us solutions x equals 1 and x equals 11.

For b, I couldn't factorise that one, so I used the formula as my positive solution and negative solution.

For c, that one did factorise, giving us the solutions x equals negative 2 and x equals negative one third.

Question two-part d I couldn't factorise that one, so I used the formula giving me these two solutions x equals negative a half and x equals negative two thirds.

For e, I mixed things up.

I completed the square rearranging like so, giving me a solution of 0.

48 and negative 12.

48.

Those two answers are truncated.

On to the second half of our lesson now, where we're going to take a lot of those skills we used in solving quadratic equations and apply them to solving quadratic inequalities.

You'll enjoy the link between these two topics.

We now have to solve quadratic equations.

For example, we could factorise and find those two solutions.

So, how does it differ when we solve a quadratic inequality? A quadratic inequality, for example, x squared plus 4x minus 5 is less than zero.

Can you see the similarity between the two questions? There's a fundamental key difference that we need to appreciate.

The first one, the quadratic equation, asks us when is that curve equal to zero, whereas our inequality is asking us when is the curve below zero.

If you can imagine this every time, you're asked a problem with quadratic inequalities, you'll have conquered this topic.

The graph can be incredibly useful here.

We need to see when that curve is below zero.

Sketching the graph reveals a solution for the inequality.

Solving the quadratic equation told us the roots.

We've got roots at negative 5 and 1, where the curve is equal to zero.

We can now see where the curve is below zero.

This curve is below zero in this region between negative 5 and 1.

Of course, we wouldn't say between negative 5 and 1.

That's lacking technical accuracy.

We'd say the solution to our inequality is x is greater than negative 5 and less than 1.

That's how you communicate the solution to this inequality.

It's important to pay close attention to the sign when solving inequalities.

We've solved x squared plus 4x minus 5 is less than zero.

What if it said x squared plus 4x minus 5 is greater than zero? Same problem.

Different problem.

Same solution.

Different solution.

Of course, it's a different solution.

It's a very different question.

Can you see how the question is different? Pause.

Tell the person next to you.

Have a good think to yourself.

See you in a moment.

Welcome back.

I wonder what you said.

I wonder if you verbalised it like this.

The one we solved earlier was asking us when is the curve below zero.

That's that highlighted region we can see on the graph.

Whereas the one I'm now presenting to you is asking when is the curve above zero.

When is the curve above zero is an entirely different answer altogether.

The curve is above zero in these two regions.

X is less than negative 5 or x is greater than 1.

That's the solution to the inequality x squared plus 4x minus 5 is greater than zero.

Those are the two moments when the curve is above zero.

Quick check, you can do this.

I'm gonna solve two quadratic inequalities and ask you to do two very similar examples.

I'm going to solve x squared minus 7x plus 10 is less than or equal to zero.

It'll factorise.

I've got roots at positive 2 and positive 5.

I'm now thinking to myself, when is that curve below or equal to zero.

It's in this region between 2 and 5, but of course, I don't say between 2 and 5.

I say x is greater than or equal to 2 or less than or equal to 5.

In the second example, it's the same factorization on the left-hand side, but I'm interested in when this curve is above or equal to zero in this problem.

That's those two regions when x is less than or equal to 2 or x is greater than or equal to 5.

We might see that solution written in set notation like so.

The same answer just communicates differently.

Be prepared you might be asked to write some solutions in set notation.

Your turn now.

Two quadratic inequalities to solve.

Pause and enjoy.

Welcome back.

Let's see how we can factorise the left-hand side like so, and we know we'll get roots at negative 7 and positive 2.

When is the curve below or equal to zero? Well, it's that region there.

So, we communicate that solution as x is greater than or equal to negative 7, and less than or equal to 2.

For the second example, factorise is exactly the same, but the question is different.

When is the curve above or equal to zero? Well, that'd be those two regions.

So, our solution x is less than or equal to negative 7 or x is greater than or equal to 2.

You might see that written in set notation.

Without visualising the graph, it's easy to make mistakes.

Sophia's done this math to solve negative x squared minus 4 x plus 5 is less than zero, and she's done it faultlessly.

Well done, Sophia.

However, she now says with roots at negative 5 and 1, I think the solution is x is greater than negative 5 and less than 1.

Sophia has made an error.

What advice would you give her to support her thinking on this problem? Pause and have a think about that.

Welcome back.

What advice you had for Sophia? There's all sorts of things we could have said.

One good piece of advice would have been to advise Sophia to sketch the graph.

Sketching the graph will enable her to see the solution.

Visual representations are hugely powerful in mathematics, especially in this case.

This is a sketch.

Sophia told us about the roots at negative 5 and positive 1, but now we can see the solution.

When is that curve below zero? Sophia says I see the solution now.

It is x is less than negative 5 or x is greater than 1.

Well done, Sophia.

That's right.

Another useful thing to do is to test your solution.

We might have advised Sophia to do this.

Remember, that's the solution she originally proposed for us.

If that is the solution, then when x equals zero, our inequality should be satisfied.

So, let's substitute that value in.

Negative zero squared minus four lots of zero plus five is less than zero.

No, it's not.

So, we know that's not the correct solution.

Alex is trying to solve this inequality.

He does this maths and concludes x is less than three, includes all the values in the region x is less than negative three.

So, the solution is simply x is less than three.

Alex is wrong.

What advice do you think Sophia is about to give him? Pause and see if you can help Alex out.

Welcome back.

You might have said firstly test your solution.

When x equals zero, zero squared minus nine is less than zero.

It is indeed.

But when x equals negative 10, negative 10 squared minus nine is less than zero.

No, it is not.

So, x is less than three cannot be the solution.

Alex concludes I must be wrong.

There's nothing to fear being wrong in maths.

In fact, knowing that we're wrong here just takes us one step closer to knowing what is actually right.

Some additional advice you might have given Alex would be visualise the graph.

If we visualise this example, we can see when x squared minus nine is less than zero.

Alex has got it now.

Alex says, of course, the solution is x is greater than negative three and less than three.

Well done, Alex.

Some examples will not factorise when we're solving quadratic inequalities.

This is not a problem.

In this case, we have other methods to find the roots, such as the quadratic formula.

We'll find a positive root of two plus root two and a negative root of two minus root two.

From there we can sketch with roots at two minus root two and two plus root two.

When is the curve below zero in that region? So, we get a solution x is greater than two minus root two and less than two plus root two.

Quick check, you've got that.

I'd like you to use the quadratic formula to find the exact roots then solve this quadratic inequality.

Pause.

Give it a go.

Welcome back.

Substituting into the formula looks like so.

We get a positive root and a negative root.

Positive x squared coefficient.

So, our curve is above zero outside of these roots.

So, our solution is x is less than 11 minus the root of 161 over four or x is greater than 11 plus root of 161 over four.

Practise time now.

Question one.

I'd like you to solve these quadratic inequalities.

Pause.

Give them a go now.

Question two.

Alex is solving this inequality.

What I'd like you to do is write a sentence explaining what they've done wrong and find the correct solution.

For question three, it's a tricky looking one, but don't fear it.

My hint for question three is rearrange.

Pause and try these problems now.

Feedback time.

Let's see how we did on question one.

You'd have noticed that that quadratic factorised.

There's an x squared coefficient, which is positive, so the curve's below zero between the roots, giving us that solution.

B was an incredibly similar question but crucially different.

The same factorization on the left-hand side, but the curve this time we're looking for when it's above zero.

That's outside of the roots giving us those two solutions, and you might have written this answer using set notation.

C we can factorise out negative one on the left-hand side, then further factorise.

We've got roots again, but we've got a negative x squared coefficient.

The curve is greater than or equal to zero between these roots, giving us that solution.

For D, it won't factorise.

We could complete the square to find the roots.

That would look like this giving us those two roots, therefore that solution to our inequality.

Question two.

Alex was solving this inequality, and we were writing a sentence explaining what they've done wrong and finding the correct solution.

You might have said negative four and negative two are not the roots.

You have to make the right-hand side zero.

Well done for spotting that.

When we make the right-hand side of the inequality zero, factorises like so, giving us that solution.

Question three looked incredibly tricky at first glance, but by rearranging to make the right-hand side equal to zero, it looks a lot easier, especially seeing as it will factorise.

It's a positive x squared coefficient, so our curve is going to be above zero outside of the roots, giving us those solutions.

You might have written this answer using set notation.

We're at the end of the lesson now, sadly, but we have learned that we can solve quadratic inequalities algebraically.

We know the importance of finding the roots and visualising the graph when solving these problems. We know to pay particular attention to the less than and greater than signs to understand when we get a solution of one closed set versus two open sets.

I hope you've enjoyed this bit of maths as much as I enjoy it, and I look forward to seeing you again soon for more mathematics.

Goodbye for now.