Solving quadratic simultaneous equations graphically

Key Stage 3

Maths

Quadratic equations (9.10)

﻿Hello, and welcome to today's lesson on solving quadratic equations graphically. For today's lesson all you need is a pen and paper or something to write on and with. If you could please take a moment to clear away any distractions, including turning off any notifications that would be brilliant. And if you can't please try and find a quiet space to work where you won't be disturbed. Okay, when you're ready, let's begin. So it's time for the try this task. I'd let you to use the table of values to plot the quadratic. So pause the video and have a go, pause in three, two, one. Okay, welcome back so let me quickly work through some of this. So we have a negative one. Well, we've got to find out this Y value, our X equal to negative one. So when you're negative one squared, minus three times negative one plus one. So negative one squared and notice I've put it in brackets. Now, the reason I've put in brackets is because we're squaring the negative as well as the one. And if you don't put in brackets on a calculator for instance, you would get the wrong answer. So negative one times negative one. Give me positive one minus three times negative one, which is negative three. So I'm going to do minus negative three, plus one. So one plus three plus one, which equals five. Okay, zero is a nice one. Zero is a nice easy one because zero squared is zero, minus three times zero, zero, plus one, one. Okay, Y is it for one? One squared, one minus three times one, three plus one. So we have one minus three plus one, which gives us negative one. For two, I'll do two up here. So we've got for two, we've got four, two squared minus three times two, three times two is six, minus six, plus one. So four minus six, plus one, which gives me an answer of negative one. For three, three squared is nine, nine minus three times three, also nine plus one just gives me one. And finally for four. Four squared, 16, 16 minus three times four, 12 plus one, 16 minus 12 , four, four plus one five. Do you notice this? Well, this is symmetrical. And in fact, all quadratic graphs are symmetrical. Now we may not always see the symmetry in the table of values, but all quadratic graphs are symmetrical. And that is a very important point for quadratic graphs. Okay, so where you going to plot these points? Negative one, five, zero, one, one negative one two negative one. Three, one and four, five. Okay now what you don't want to do is you don't want to join them up with straight lines. You do not want to use a ruler here. You want to, and this is going to be very bad, but you want to try and do a free hand curve. So if I can manage to do it with this pen, you can definitely manage to do it with a pencil on paper. Okay so it should... And that doesn't look very symmetrical, but it should look something like that. Going through those points. Now notice that it doesn't flat out there. It goes down a bit. Okay, it's always just going to be one coordinate, that's the lowest and that's actually called the turning point. And so that would be a line of symmetry here. Remember I said, they've all got lines of symmetry. So your quadratic should look something like this. Or if you made one earlier, should look something like that, which is much, much nicer. But you see you see how this is curved. We need to make sure that when we're drawing them, we do it as curve and it doesn't have to be perfect. Mine wasn't perfect. We just, we have to, we just try, we try and make it nice. Find approximate solutions to the following simultaneous equations. Well, why are they simultaneous equations? Because I've got two of them. If I just have one of them, I would have infinitely many solutions, but because I've got two simultaneous equations, I'll get, well, you can sometimes have simultaneous equations. Like the infinite solutions, but for this, we'll have less than an infinite number of solutions. Well, let's see how many solutions we'll get. So we've got Y equals X squared plus one. So I need to plot that. So let's make a table of values. I'm going to do it and from negative three. That's a lovely three, isn't? It's lovely two, to improve all the way out, we're getting better there. So, I want to start off with zero, 'cause it's an easy one. Zero squared plus one, one, zero squared, sorry, one squared, one plus one two. Two squared, four plus one, three squared, I'll write this one, just so you've got something to remember it by, three squared plus one, nine, nine plus one 10. Okay, well this one, well, this one is negative one in brackets squared, plus one. Negative one squared is one, one plus one is two. Maybe we're starting to see our symmetry here, but let's just do them to check. So negative two squared is four, four plus one, five. And finally negative three squared, so remember we put our negative three in brackets. Negative three squared is nine, nine plus one is 10. And yet we have a nice symmetrical pattern there. So for this bottom point is there, and then it kind of goes out like this and it increases quite quickly. More kind of steep, awesome. I'm quite happy with this, go on, awesome. All right, so there is, there is my curve. Now it doesn't have to perfect, this is because we are finding approximate solutions to the simultaneous equations. So it doesn't have to be perfect. So now we need to find out when this line, Y equals X squared plus one, when they have the same value. So they have the same value when they meet. So we can say that X squared plus one, we can think of it like that. When is X squared plus one equal to three? And the graph drawing this line in, will tell us. So line Y equals three, the line where every Y coordinate, has a Y coordinate of three. So the line where every point, every coordinate has a Y quadrant of three, every single Y coordinate on that line is three. That's why it's Y equals three. And people get that confused, I want you to try and remember that. Y equals three is horizontal because every single point has a Y coordinate of three. So now we've got our one graph, Y equals X squared, plus one, notice I've labelled them. That's a really good idea and it's really important that you remember to do that. And now I can see where they meet. So my approximate solution, well is so something like that, that's going to be straight there. Y ends up there, so my X coordinate of this point is, I'll write here is about negative 1.6. Can you see how it's negative 1.6 because there's five of them in one, and then there's not point two, not 1.2, 1.4, 1.6, negative 1.6. Y coordinate well, that's pretty obvious that's going to be three. So that is one solution, X equals negative 1.6 and Y equals three. And our other solution is, X equals 1.6. Sorry, I should have I don't why I wrote X there. X equals 1.6 and Y equals three. Now in this graph, they asked symmetrical because the minus, the negative 1.6 and the 1.6, because the line of symmetry is this line here that does not always happen is just in this situation, that's happened. There are not always the same, all of the negative and positive version of each other that is just in this situation. So here's the accurate drawing and you can see that, well that was 1.4, negative 1.5. And that was 1.4. My solution was reasonably accurate, and we'll look in future lessons, how to actually get an accurate solution to these. So, here, I've got and I've drawn this graph already, 'cause this is the one we did earlier. This was in the try this task. So now I've just got to draw in my line, Y equals a half X plus two. So this tells me the Y intercept, the Y intercept is two. And you could draw a table of values, drawing a table of values is very sensible, but we do not need to. So I know it goes past that point. And then my gradient, which we write as an M. My gradient tells me for every one I go across, I got half off, or we could think about that as for every two, I go across, I go one, up. I prefer to think of it like that. Just 'cause it'll make drawing it slightly easier for me. Let's say I'm ever going backwards one down, two across, so I kind of get this line here. And we want to connect them up, and this is when we'd actually use a ruler for. So yeah, there's my line Y equals a half X plus two and I can get approximate solutions. So what is that coordinate there? Well, that is about, I would say not negative, not point two, and 1.8. I think that's reasonably accurate. And here we've got 3.8 and three point, no, yeah. 3.8 and 3.8 outset. So you can see from this one that we don't have that same symmetry. And that's because the line of symmetry is this line here and not the Y axis. So here is the accurate version and you can see that my line was fairly accurate and my solutions would have been very similar. So just before the independent task, I just want to have a go at this because people... This is really common mistake and I don't want you to make it. So I want you to try and find what is the value of Y when X equals five. I'll give you a five second count down, well you may need to pause the video if you need a bit longer. Five, four, three, two, one. Here is my answer. Well, why is that? Because X squared, well X is five. So we do the five squared first. So five squared is 25, three times 25, 75. We do the squared first. So it's three times five squared. Three times 25, do the squared first before we do the times by three. What is the value of Y when X equals negative five? So five, pause the video if you need to. Four, three, two, one. 75, again, why is it 75 again? We'll let us check. So we are doing negative five squared, a negative five squared is 25. So we'd get the same answer because five squared is the same as negative five squared. Now, if you did that on a calculator and you did not put your negative in brackets, your calculator would have give you this answer, which is incorrect. So you need to be really careful and include brackets when you do in this. So now it's time for the independent tasks, There are one, two, three or five, questions. So I'd like you to pause the video to complete your task and resume once you're finished. Okay, welcome back, now, here are my answers. Now, I've actually drawn the graphs. So your graphs, aren't going to look exactly like mine. They're not going to be perfect or really well done if they are. So you don't need to get exactly the same solutions as me, but your graph needs to look similar and your solutions need to be similar. If, for instance, you got at negative four, negative six. Well, I don't think that's an acceptable answer, but if you got something around that and something around that, then, then that is sensible, okay? So again, these solutions are approximate and your solutions are approximate, so they might not be exactly the same. But a good thing to check is your table of values. Okay you may have to pause the video, if you need more time to mark. So on this one, this one's quite nice, only crosses once and it just touches it there and that's a tangent and it can do it in lots of places. So it can just cross once on a diagonal like that and just touch it and it's called a tangent. And that actually is useful, 'cause it helps us work out a gradient of a curve. These ones, these don't intersect. So thinking about a number of solutions for this and this, we'd say that there's no solutions. So now it's time for the explore of task. So this is the graph of Y equals six X squared plus three. And I want you to give an equation of a line such that there are two solutions, one solution, zero solutions. So you'll give another line like Y equals something and you can make it an easy one, you can make it hard, try and push yourself okay? So I'd like you to pause the video to complete the task and resume once you've finished. So here are some possible answers, now I could have gone for this line here. I did that quite, I'm quite happy with that. Y equals negative six. And that would have been one on one solutions or I'm not happy with that though, or we could have gone for, no, no, it's not going to be right. Something maybe like that, which just touches it. Just touches the edge. And that just touches that point there and doesn't touch any other point. And then you'd have to find the equation of that line. So I've kind of drawn two of them, it's kind of hard to work out, which one. So I can't fully work out my gradient, say it was that one there. We'll say we'll draw a triangle to work out a gradient of that line. That goes from 1.8 to 3.8. So that's a change of two that goes from negative eight to negative 5.6. 'Cause each one of them is, each one of them is not point four So negative eight, 5.6. So that's approximately, 5.6 plus point four, approximate 2.4. So my gradient is approximately, I'm going to times them both by 10, 24 over 20, which simplifies to 12 over five as what 12 over 10, so which simplifies to six over five. And then I would find my Y intercept. So it does get quite tricky, this task, is not an easy task, but if you want, you can go for some of the fancy ones like this, then you could have had, you could have had a horizontal line for two solutions or you could have had maybe gone for, I'll go for a nice, easy one this time, 'cause that was about, well that was meant to be Y equals X, but that is not going through all the corners that does not quite have a gradient of one, but that would have had two solutions. Y equals X, Y equals four, and anything with no solutions is a nice kind of flat line down here that won't touch this curve. So hopefully you had an explorer, hopefully you, you were very creative and came up with some solutions to this. 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.

Solving quadratic simultaneous equations graphically

Solving quadratic simultaneous equations graphically

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UnitMaths / Quadratic equations (9.10)

Maths