Solving adfected quadratic equations (Part 1)

Key Stage 3

Maths

Quadratic equations (9.10)

﻿Hello, I'm Mr. Coward and welcome today's lesson on solving adfected quadratic equations. For today's lesson, all you'll need is a pen and paper or something to write on and with. And if you could please take a moment to clear away any distractions, including turning off any notifications, that would be great. And if you can't, please try and find a quiet space to work where you won't be disturbed. When you're ready, let's begin. So time for the try this task. Find as many possible pairs of side lengths for each rectangle. So here we've got a rectangle with area six units squared, and here we've got a rectangle with area of zero units squared. And obviously I don't want you to find every single possible combination, but have a go, see if you can find three, four, maybe even five different pairs that get you that area. And what do you notice? So pause the video and have a go. Pause in three, two, one. Welcome back. So hopefully you found something. Now we could have had for this one, we could have had six and one. We could have had three, in fact, I'll write it somewhere else. So we could have had one and six and three and two. They're the only ones I can think of, but then I could have had 1/2 and 12. I could have had, 2/3 times nine. So there's loads of different ones and you could have had improper fractions, loads of different decimals. There's infinitely many ways to find the area for this. Now this one. Well, I could have had zero and six. I could have had zero and five, zero and 3.71, zero and 92.8, 471 times zero. 1413.56 times zero. All of these ones. What is different about them? What is special about them? Well, they've all got a zero in. So two things times together to get zero. One of those things must be zero or both of them. We could have also had zero times zero. But just remember that. If two things times together to get zero, at least one of those things must be zero. Internalise that in your head. Think about that. You need to remember that. It's really important for today's lesson. If two things multiply to give zero, what do we know? Go on, say it. That one of them or that at least one of them, must be zero. That's really important. If two things multiply to give zero, what do we know, that at least one of them is equal to zero. So you might want to pause the video and have a little think about this. So pause in three, two, one. Welcome back. So hopefully you've said for this, A must be equal to zero. Which is fine. And this one, A must be equal to two. This one, A can actually be anything. So there's lots of different values that A could take here. Because no matter what that is, A times zero will give us zero. What do we know here? Well, we know either, A or B or both is, so we know that either A or B is equal to zero. Now that's given us a constraint on A or B. So we know now something about A or B. Whereas if we compare this to the next one, well, we can't say either A or B is two or four or A or B is one or A or two numbers that multiply to give A. Because A and B might be fractions, they might be decimals. So with this, we can't really say anything more about A and B. We don't have any extra information about A or B. Whereas this statement here, this statement tells us that either A or B is equal to zero. So this statement gives us more information than the second statement. So we're going to use this principle now to solve a quadratic equation, and this is different to the quadratic equations that we've seen. And I don't want you to worry too much about the differences for now. We'll look at that at a future time. But generally speaking, we're going to use different methods to solve this. Now, why is this quadratic? Well, because we get an X times an X, so it'd be an X squared. So how am I going to solve this? Well, if that, and that means times when we have two brackets next to each other. So that times that equals zero. So what do we know if we have two things multiplied together to get zero, what do we know? Well, we know that one of these things must be zero and I don't actually need to write with the brackets anymore when I'm writing this statement. So we know that this bracket, X plus three is equal to zero. Or this bracket is equal to zero. If two things multiply to get zero, one of those things must be zero. So now we need to find what value of X would make that equal to zero, this bracket, and what value of X would make this bracket equal to zero. And when we do that, we solve our quadratic equation. So what does it mean by solve? It means find the value of X that makes this statement true. So when we find the value of X that makes one of the brackets equal to zero, that is a value of X that makes the statement true. And because it can be either bracket or both, but in this case, it's not going to be both. But because it can be either bracket, we're going to have two answers. The value that makes one bracket zero or the value that makes the other bracket zero. So let's finish this off then. And I would solve this just like a normal linear equation. So we get that, X equals negative three. And we'll solve this like a normal linear equation. Add five, add five. Divide by two, and I'll leave it as a fraction. So they are my two different values for X. So what I'd like you to do now is I would like you to have a go. So pause the video and have a go. Pause in three, two, one. Welcome back. Now, either this bracket, is equal to zero or this bracket is equal to zero. So solve this. So we'd add on three to both sides. I'm going to get that. And when we solve this. Divide by three. So now we have our two solutions. Either this solution that makes the first bracket equal to zero or this solution that makes the second bracket equal to zero. And if you substitute them in, if you do three minus three, you get zero there. If you do three times negative 10 over three. Three times negative 10 over three gives you a negative 10. Negative 10 plus 10 that gives you that bracket as zero. So those are just the values that give you a zero in one of those brackets. So they are our solutions. What about this one? How's this one different? We've got an X, an X times X minus five. That means that either X itself is equal to zero, because this is two things that times together to get zero. So either this X is equal to zero or X minus five is equal to zero. So that's one solution already. We don't have to do anything to that one. Or we can add five to both sides and find out that X could also be five. So here again, we've got two solutions. So just try this one, pause the video and have a go. Pause in three, two, one. Welcome back. Hopefully you got X is equal to zero, which is one solution or X plus 10 equals zero. And then you take 10. So you get X equals negative 10. Really well done if you've got those correct. Awesome. Another trickier now. Well, we've got a squared. So here we've got just one bracket that's going to be equal to zero. That bracket there must be equal to zero. So that means we can solve this equation. Take three from both sides, multiply by two, so that means X would be equal to negative six. And there is just one solution here. Because there's only one bracket that can equal zero. Because it's that bracket squared. So it's X plus two. X divided by two plus three times X divided by two plus three. So it's either one of those brackets equal to zero, but because it's the same bracket, we just solve the one equation. So we'll just get the one solution. So I'll let you to pause the video and have a go at this one. Pause in three, two, one. Welcome back. Now hopefully you've recognised that the thing in the bracket must be equal to zero. So then we rearrange or solve to find the value of X that makes that statement true. And we get that X equals negative 30. So our one solution in this case. We're going to decide if the following equations are ready to be solved or need expanding first. And we'll look more what happens if they do need expanding first. But for now, I just want you to think about what was the key principle. And the key principle was that, our quadratic expression needed to be equal to zero. So is this one ready to be solved or does it need expanding? I'll give it a tick if it's ready to be solved. I'll give it a cross if it needs expanding. Yes, that is ready to be solved. Why? Because it's equal to zero. Is this ready to be solved? Why? Because it's equal to zero. And I just want to point out here, we'd actually get the same solutions to this one. We'd get X equals negative five and X equals negative three. And we'd get the same for that one. So it doesn't matter if your brackets swapped around because multiplication is commutative. So the order doesn't matter. So we'd get the same solutions. Is this ready to be solved? What's changed? A lot has changed. But does it affect it? Nope, because it's still equal to zero. Is this ready to be solved? No, it is not equal to zero, it is not ready to be solved. We need to expand these brackets first. Is this ready to be solved? No, it's not equal to zero. It's not ready to be solved. Is this ready to be solved? Yeah. Is this ready to be solved? No. Why? Because it's not equal to zero. Is this ready to be solved? Nope. Is this ready to be solved? Yes. Why? Because it's equal to zero. Is this ready to be solved? No. Why not? Because it's not equal to zero. What about this one? Nope, not equal to zero. This one? Yes, this one is ready to be solved. This one? Yes, this one is ready to be solved. Why? Because it is equal to zero. So it doesn't matter if we only have one bracket, that is still ready to be solved. This one. Is that ready to be solved? No it is not. Why is it not ready to be solved? Because it's not equal to zero. So hopefully because one of the things that people always make a mistake on is they'll get something that is not equal to zero and they'll try and solve it. So I just wanted this exercise was just to get you thinking about that. And just to keep you alert to when things are equal to zero, and when things are not equal to zero. And we'll deal with how to solve things like this in a future lesson. So now it's time for the independent task. Three questions. So I would like you to pause the video to complete your task and resume once you're finished. Welcome back. So here are my answers. You may need to pause the video to mark your work. Number two. What did you do wrong or what did he or she do wrong or did they do wrong? They tried to solve it when it was not equal to zero. That's equal to three. So here's your answers to three. I threw a quintic in there and we solve it in exactly the same way. And for C, any equation that contains a square bracket. So if we've had X minus two squared, instead of this bracket there, then that would give us false solutions. And because we would have had almost essentially X minus two, twice. So we would have got X equals two, twice. And there's something else. There's another way that we could have written that. And we'll discuss that in a second. So Zaki says that the brackets will have two unique solutions because the brackets are different. Show Zaki is incorrect. Can you explain why the brackets give the same solution? Can you find another bracket that would have worked to go with two X minus six? Can you create your own question like this? So what I'd like you to do is I would like you to pause the video and have a go. So pause the video to complete your task and resume once you've finished. So here are my answers for these spots and still I'll just kind of talk through it. So here we have two X minus six and here we have three X minus nine. Now what I've done here is I factorised them both. Can you see what's the same or what's different? Well, they've both got an X minus three when factorised and they both give the solution of three. So what I'd say is, dividing the X term and the constant term by the same number doesn't affect it, or neither does multiplying. So some of the things that we could have had is we could have had this, the answer to this, five X minus 15. Why could we have had that? Well because, if we had this, five X minus 15, we said that that was equal to zero. If we divide by five, on both sides, we get X minus three is equal to zero. And because we have zero divided by five, that still stays zero. So that's why they are both solutions. So we can actually have any multiple of X minus three. So 17X minus 51 times the X by 17 and the three by 17. You could even have 1/2 X minus three over two which times them both by 1/2. So there's loads and loads of different solutions that you could have had there. Can you create your own question like this? Well, yes, what's my con factor here. If I take out a factor of three, I've got three X plus five. If I take out a factor of nine here, I've got nine X plus 45. So we can even think about it like this. Here I've got this, I've taken out my, sorry, that should be five. I've taken out my factor of nine and my factor three. Now, because multiplication is commutative, I can swap this order around. So I'm going to do three times nine and get 27. So I've got 27. So I can actually take out factors from the brackets and I can just divide by 27. I can just divide by 27 because it's equal to zero. Divide by 27 on that side and divide by 27 on that side, I'll just get zero on that side and X plus five times X plus five on that side. So hopefully you enjoyed that and that is all for this lesson. Thank you very much for all your hard work. And I look forward to seeing you next time. Thank you.

Solving adfected quadratic equations (Part 1)

Solving adfected quadratic equations (Part 1)

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UnitMaths / Quadratic equations (9.10)

Maths