CIE A Level Maths Paper 1: 1.3 Quadratics: Inequalities and Equations

Hey guys, welcome back to another CIE A level maths video for pure paper one again. This time we're looking at quadratic equations and equalities. I've just combined it together because solving quadratic equations is part of solving quadratic inequalities anyway. So, may as well kill two birds with one stone. Just going to do a few of the exam questions. Shouldn't be too uh tricky. By the way, if you have a look at the link in the description or in the pinned comment, there is links to free CIE A level maths tuition classes that I'm running. Only for the next few weeks though. Anyways, solving the inequality. So first things first, let's make this an actual quadratic by minusing 20 from both sides. Now what is this actually asking me? It's asking me here's my function. Give me all of the values of x which will give me a positive number. So for example, if I sub x= 1 here, I get 1^ 2 which is 1 + 8 which is 9 - 20 which is -1. That would not be bigger than zero. But let's say if x was 10. Well, 10 squar is 100 + 80 180 - 20 160. Okay, that is bigger than zero. So I want all of the values of x for which I get a positive number out. In my opinion, the best way to do this is to sketch the graph because if I sketch the graph, as you know with quadratics a lot of the time they have roots where they intersect the x-axis. So they might intersect at two points here and here. Everything below that graph is the values of x where it's negative. So we're literally just going to sketch this function. How do you sketch functions? Well, the first thing you need to do is work out what their roots are. So we're going to work out where these values are equal to zero. Again, this is let's keep going. I was going to moan. So that's going to be 10 and minus 2. So my two solutions here, my two roots are - 10 and 2. Now technically you should get the y intercept as well but I don't actually care about the y intercept in a quadratic inequality question. So minus 10 will be somewhere over here and two is going to be somewhere over here. And then we get a big old quadratic. So let's just draw something. Cool. Now if I have a look at this function when I sub x= like 10 for example that's on the right of two. That's somewhere over here. And as you can see it gives me a positive value. So I can see from here all of these values are going to be positive, right? So if I sub in x= 2.1, I should get a positive number out of here. And you can check that down a calculator if you want to. Likewise, everything on the left over here. So if I subbed in -1, I actually end up getting a positive number. So - 11^ squar is 121. 8 * -1 is - 88 - 20 gives me um - 108. So 121 - 108 is positive. It's 13. Cool. So I'm going to get two regions in my inequality. One which is to the right of two. So for every x value bigger than two, this function over here is bigger than zero. Likewise, for all of the values less than minus 10, this function is also bigger than zero. So in short, my two solutions are this. that technically you can also write the or sign if you really want to but those are my two solutions. So again if you draw the graph it's super super easy. Let's have a look at another one. So here find the set of values of x for which this quadratic here is less than 32. There's a few ways to do it. If you if you leave it as is you could draw the graph of this and less than and the line 32 and just see where it's below. But the problem with that is it's very hard for you unless you draw it accurately. it's very hard for you to pick those values off. And because I don't want to draw it accurately, I'm actually going to expand it, rearrange it, and then factoriize it again, which sounds long, but I promise you it's better than accurately plotting this graph. So that's less than 32. So I'm going minus 32 from both sides. We get x^2 + 6x - 27. 27 + 5 is 32. That should be fine. And again, what we're going to do, we're going to try and sketch the graph. And when we sketch it, we'll be able to visually see when this whole function is negative. So again, it's not less than 32 anymore. We've changed it to be less than zero, which is completely legal. That's completely fine. So we're going to look for where this graph here, this function, this graph is below the xaxis. So once again, we need to work out its roots, which is when it equals zero. So two numbers that times to - 27, but add to six. That's going to be 9 and -3 equals 0. So our two roots are - 9 and positive3. So - 9 let's say is over here. Pos3 let's say is over here somewhere. And again if we draw this quadratic like that we're looking for it below zero now. So this is where we have to be careful. Yeah, I just realized you can definitely hear that helicopter that went over just then. Uh anyways um so we're looking for the values when it's below zero which if I highlight that region is when it's below the graph right and what this graph is telling me what this function is telling me is if I sub in a value bigger than -9 so say minus 8 I should get a negative number which is what I want but it's not just all values bigger than minus 9 because eventually I get to three if I sub in three I get zero if I sub in four I get a positive number so instead there's a lower bound and an upper bound that we have to be careful of. So the way we write those, we're at - 9 less than x less than 3. So we're saying it has to be bigger than - 9 and less than three, right? So bigger than - 9 less than three. That will work. And that's how you describe an inequality like that. If we have a look at one last one, again, we have this. Once again, I personally think the easiest way to do this is to expand the brackets and just resolve the whole thing. um unless you want to draw a really accurate graph which I can't imagine why you'd want to because that's painful and you can probably imagine exactly what we're going to do now right now again the tricky part here is the less than equals to bit I'm going to show you that now uh once again we're going to sketch and now we can start skipping a few steps right so for example if I factoriize that we get x + 4 xus 3 which means our solutions are minus4 and plus three so minus4 over here see how quick we can make this now look at that Beautiful. And again, if you noticed in all the other inequalities, I've never used the equal to sign because this said less than zero. It didn't count zero. That's why we don't put three because it has to be less than three. Even though that's 2.9999999999, it's still less than three. However, here I want the values where it equals zero as well. Once again, this is going to be between the two numbers. So now I am going to do -4 less than= x less than equal to 3. If you don't believe me, you can sub it in. If I sub in minus4, I do get a value less than equal to zero. I get zero, which is less than equal to zero, right? And that's it. That's it for quadratic inequalities. It's not too bad. But once again, this is something we're going to be solving a lot of.