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Video: Introducing the Completing the Square Format of a Quadratic Equation

Tim Burnham

Explaining the purpose and structure of the completing the square format for quadratic equations and starting to apply it to quadratics with 1𝑥².

17:49

Video Transcript

In this video, we’re gonna look at the completing the square format of quadratic equations. We’re gonna look at how to organise quadratic expressions into that format and we’re gonna look why that’s such a useful thing to do. We’re gonna see how to use that in practice.

Here’s a quadratic equation. Well let’s think, “What do we know about quadratic equations?” Well, we know that they’re curves; in fact, we know that they’re symmetrical parabolas. We also know that they can either be what we might call “happy curves” or “sad curves.” They can either face downwards or they can face upwards. And we know that they cut the 𝑦-axis somewhere and sometimes they cut the 𝑥-axis.

So let’s try and do a sketch of this curve. Well before we do that, let’s look at this 𝑥 squared term. There’s nothing in front of it; there’s no coefficient in front of there. And that really means that actually we’ve got one times 𝑥 squared. In fact, it’s positive one times 𝑥 squared. So the fact that this is a positive number — If you think you’re positive, you’re happy; if you’re negative, you’re feeling sad. So positive means it’s a happy curve. The curve- the parabola is gonna look something like that. And this term here on the end, the constant term on the end, tells us about where the curve cuts the 𝑦-axis. And we call that the 𝑦-intercept. So now we’ve got some basic information. We can start to sketch the curve.

So here’s a pair of coordinate axes and we can sort of plot the point here, negative six. So that’s where it cuts the 𝑦-axis. Now we know it’s happy curve. But we don’t know exactly how it’s aligned, left or right? So it could be the case that this curve comes down here. This is the minimum point at-at six- minus six down here, and then it goes up like that. It could be that it’s slightly to the right of that; so in fact the curve comes down here through this point here, keeps going down, and then goes further up to the right. Or maybe it’s further to the left, and maybe it comes down here, goes down through here, and then back up through here over this way.

So to work out which of these three scenarios we’re talking about, what we really need to do now is work out where does that curve cut the 𝑥-axis. We know where it cuts the 𝑦-axis; we need to know where it cuts the 𝑥-axis.

Of course what we know is that the points on the 𝑥-axis have got a 𝑦-coordinate of zero. So we could replace this 𝑦 with zero, and we could try to solve that equation and work out what the solutions are.

So I’ve written that equation out: 𝑥 squared plus 𝑥 minus six equals zero. And one of the obvious ways of doing this is we can factorise the expression 𝑥 squared plus 𝑥 minus six. So that gives us 𝑥 minus two in one bracket and 𝑥 plus three in the other bracket. Of course we can check that: 𝑥 times 𝑥 is 𝑥 squared, 𝑥 times three is three 𝑥, negative two times 𝑥 is negative two 𝑥, and three 𝑥 add negative two 𝑥 is just plus one 𝑥, and then negative two times three is negative six. So yea that indeed does multiply out to give us the same as we had on the line before.

Now the thing is we’ve got something times something is equal to zero. So one of those things must be zero. So either the first thing is zero, so in other words either 𝑥 minus two is equal to zero, or the second thing is zero, so 𝑥 plus three is equal to zero. If the first thing is true, then 𝑥 must be equal to two. And if the second thing is true, then 𝑥 must be equal to negative three. So we’ve got the two values of 𝑥, which give us a 𝑦-coordinate of zero. And those points, well the first one is two to the right of the 𝑦- axis and the second one is three to the left of the 𝑦-axis. So we’re looking at this point and this point. So this curve here was the curve that we were looking at.

Now we know that quadratic curves, these parabolas, are symmetrical. So if we drew a line of symmetry down the middle here and because that whole curve is just slightly to the left of the 𝑦-axis there, we know that the-that the minimum point of the curve, this-this lowest point on the curve here, and the 𝑥-coordinate must be midway between negative three — so this distance here is the same as this distance here — and positive two. So I see the difference between negative three and two is five and half of five is two and a half. So if we come back two and a half from two or we go forward two and a half from negative three, we end up at negative nought point five.

So this minimum point on the curve here has got an 𝑥-coordinate of zero point five- negative zero point five. We can work out what the 𝑦-coordinate is by just putting that value back into our original equation. And when we do that, we see that the corresponding 𝑦-coordinate is negative six point two five.

So really we’ve found out everything we need to know about this curve by doing the following steps. We can first of all factor the quadratic. That enabled us to find out where it cut the 𝑥-axis. Then we were able to find the 𝑥-coordinate of the midpoint between those two points. So it’s pretty easy to sketch that curve knowing where it cut the 𝑥-axis, where it cut the 𝑦-axis, and what the minimum point on the curve was.

Okay let’s try a different quadratic equation. So this time we’ve got 𝑦 equals 𝑥 squared plus two 𝑥 plus seven. Now again, nothing in front of the 𝑥 squared, so that means it’s a one and that’s a positive number. And a positive number in front of the 𝑥 squared, the positive coefficient, means it’s a happy curve. So again that parabola is gonna be facing up like that. We’ve also got the fact that the 𝑦-intercept is at positive seven.

So we can start trying to sketch the curve. So first of all, let’s mark on the point that it cuts the 𝑦-axis. And now we’ve got again several different options for how that curve might look. It could be the case that the curve comes down and that minimum point there is on the 𝑦-axis at the point zero, seven. It might come down through the point zero, seven and then down to the left or it could come down to the right. Or it could come down to the left and just touch the 𝑥-axis. Or it could come down to the right and just touch the 𝑥-axis. Or for that matter, it could come down and to the left and even further, and it could cut the 𝑥-axis in a couple of places or down further to the right cutting the 𝑥-axis in a couple of places. So to finish off sketching this curve, again we’ve got to work out where does it cut the 𝑥-axis.

Well when it does cut the 𝑥-axis, we know that the 𝑦-coordinate here will be zero. So we can write 𝑥 squared plus two 𝑥 plus seven is equal to zero and try to evaluate that. Well let’s try factorising that. Well I can’t think of two factors which multiply together to give positive seven and add together to give positive two. So it doesn’t look like that factorises. So we’re gonna have to give up on that. So we might next try the quadratic equation.

And if you remember, it’s 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. And in our case, 𝑎 is one, 𝑏 is two, and 𝑐 is seven from our original equations. So let’s plug those values in. And this is what we get. So let’s have a look in the- at the discriminant here, inside that square root, and that gives us two squared is four and we’re taking away four times seven is twenty-eight. Well that’s negative twenty-four. So what we’re trying to do in here is find the square root of negative twenty-four. Well, that’s impossible. There aren’t any real values which when you multiply them by themselves you’re gonna get a negative number; you’re certainly not gonna get negative twenty-four.

All of this is telling us is that there isn’t any value that we can make up for 𝑥 which is going to generate a 𝑦-coordinate of zero. That equation isn’t solvable. In other words, you know, it’s not the green scenario; it doesn’t come down and touch the 𝑥-axis. And it’s not the red scenario; it doesn’t come down and go through and back up again. So it’s gotta be one of those blue scenarios.

So we’re in one of those blue situations, but we don’t know which one. And we can’t use the symmetry of the curve for working out the minimum point because we-we-we haven’t got any points at which it cuts through the 𝑥-axis. So what are we gonna do? So for those who have been patiently waiting for over eight minutes for this moment now, this is where the completing the square format of a quadratic expression comes in handy. What it amounts to is rearranging our expression here in such a way that it becomes obvious which of these three scenarios we’re talking about for our graph. And it also will help us to find the coordinates of the minimum point, wherever that happens to be. So the completing of the square format is essentially we find an expression that we square and then we complete the square by adding or subtracting something from or to that.

So we’re gonna have an initial guess of what this might be. But the first thing is, we know we’ve got to generate 𝑥 squared and we’re squaring some brackets out here. So if I made that first term 𝑥, I know that 𝑥 times 𝑥 is 𝑥 squared. Now I’ve gotta think what can I put here, adding or subtracting a number, which is going to generate the two 𝑥 term here. Well if I put plus one, if I just half that coefficient, that’s a pretty good guess as to how it’s gonna be. And the reason I’ve halved that coefficient there is because if we look at the brackets here, and I’ll try them out: 𝑥 times 𝑥 is giving me my 𝑥 squared and I’ve got 𝑥 times one and then I’ve got one times 𝑥 here. And one times 𝑥 plus one times 𝑥 is gonna give me my two 𝑥. And then we’re also gonna get this term in one times one which is plus one. So we’ve got the 𝑥 squared; we’ve got the two 𝑥, but we wanted positive seven. What that expression 𝑥 plus one all squared is generated is a plus one. So we’re a bit short: we’ve got plus one; we wanted plus seven. So I need to add another six onto the end of that.

So I’ve generated a different version of that equation which if I multiplied it all out, I would still get the original equation. But this form of the equation has a distinct advantage. So remember, an-an-an equation is generally you put in an 𝑥-coordinate, you then do the math — you do the calculation here — and the answer that you get when you substitute in the particular value of 𝑥 is your 𝑦-coordinate. So this expression is generating our 𝑦-coordinates from our 𝑥-coordinates.

And what we do is we take our 𝑥-coordinate. We add one and that gives us a number. We then square that number. And that’s interesting because by squaring that number, we’re always gonna generate a positive answer. Or if the number that we squared was zero, we could get zero squared is zero. So what we’re saying is that this part of our expression can never be negative. The smallest it could possibly be is zero. So the 𝑦-coordinate that we’re generating is the smallest it could possibly be is zero plus the six. We’re always gonna be adding six to that value. So what we’re seeing is that the 𝑦-coordinate of the minimum point must be zero plus six is six.

So we still don’t know whether we’re in the-the left-hand scenario or the right-hand scenario. But now we’ve ruled out the middle one because that would have a minimum point for the 𝑦-coordinate of seven. So we’re in one of these cases. We’ve now got to work out are we going to the left or are we going to the right.

Well to work that out, we need to know which value of 𝑥 is going to generate this minimum point which is gonna generate the contents of the brackets to be equal to zero. Well 𝑥 would have to be negative one because negative one plus one is zero. So when 𝑥 is negative one, this bracket goes to zero and we add six to that and we get our minimum point; so this must be the minimum point here.

Now that’s all looking a bit messy after all that crossing-out. So I’m just gonna draw that out quickly again. So we’re gonna have a look at another couple of examples and — well this seems quite painful so far. Once you get the hang of it, it becomes very very quick to generate a sketch of your curve to work out where the minimum or the maximum point on the curve is and to work out where that curve is cutting the 𝑦-axis.

So now we know where the whole completing the square thing comes from. Let’s look at a couple of examples where we just go straight in and use it as we would normally use it. So 𝑦 equals 𝑥 squared minus two 𝑥 plus nine. So we want to rearrange that, where we’ve got something squared plus something else — or in fact sometimes you have to take something away. So yeah I won’t put the sign in them just yet. We’ll work that out later. Now we’re looking at 𝑥 squared. So we’re gonna put an 𝑥 in here. And negative two is the coefficient. So we’re gonna halve that to make it negative one.

Now if I multiply out 𝑥 minus one all squared, I get 𝑥 squared minus two 𝑥 plus one. But remember, I was looking for 𝑥 squared minus two 𝑥 plus nine. So I’ve got the 𝑥 squared; great. I’ve got the negative two 𝑥; great. But this particular expression here has generated plus one on the end. But I wanted plus nine. So I’ve got to add another eight. So that second line there is the completing the square format of the first quadratic equation.

So that’s all there is to it. Now let’s use that information to try to complete a sketch of what that curve would look like. Well from the original equation, we know that there was one 𝑥 squared. It was positive value. So positive value of 𝑥 squared gives us a happy curve. And the number on the end, the constant on the end, was positive nine. That tells us where it cuts the 𝑦-axis.

Now we can look at the second version here. The smallest value that this expression here can take is zero because it’s something squared. And it would have to have an 𝑥-value of one in order to make that zero. So the minimum 𝑦-coordinate that we can possibly generate is gonna be zero plus eight. So the minimum 𝑦-coordinate is going to be eight. That’s gonna be just below the nine. And that is when the 𝑥-coordinate is one. So we’re looking at this situation here. So one, eight is the minimum coordinate. So we know we’ve got a curve that comes down through the 𝑦-axis and slightly to the right and then up. So that’s what a sketch of our curve would look like. We’ve marked all the key points where it cuts the 𝑦-axis. In this particular case because the minimum point is at one, eight, it can’t possibly cut the 𝑥-axis. But the minimum point here’s at one, eight.

Let’s look at this one then. 𝑦 equals 𝑥 squared plus fourteen 𝑥 minus two. So our completing the square format is gonna be something squared plus something or minus something. We’ve got 𝑥 squared here. So when we multiply these brackets out, we’ve gotta have an 𝑥 in the first position of each of those brackets, so 𝑥 times 𝑥 is gonna generate that 𝑥 squared. We’re gonna halve the coefficient of 𝑥 here, so plus seven. And when we multiply out 𝑥 plus seven all squared, we get 𝑥 squared plus fourteen 𝑥 plus forty-nine. So the 𝑥 squared term yeah we’ve got that, the fourteen 𝑥 yeah we’ve got that. But we wanted negative two, but we’ve ended up with positive forty-nine. So — how — what do I gotta do to forty-nine to turn it into negative two? I’ve gotta take away fifty-one.

So there’s the completing the square version of that equation. So we can use both of those lines there, both those versions of the equation, to do a little rough sketch of the- what the curve is gonna look like. Nothing in front of the 𝑥 squared means it’s one 𝑥 squared, which is a positive happy curve. This value over here tells us where it cuts the 𝑦-axis. So we can put those two things onto the graph. So we kn- as we say, we know it’s gonna be a happy curve, but we don’t quite know wher — so it’s gonna come up and cut the 𝑥-axis somewhere, we’re not quite sure where. But this other version for jus- for calculating the 𝑦-coordinates, remember the minimum value that this expression, this term here, could be is zero. And when that’s zero, this is always fifty- negative fifty-one. So the minimum 𝑦-coordinate is gonna be minus fifty-one. What’s the 𝑥-coordinate that’s gonna generate that 𝑦-coordinate of negative fifty-one? Or to make that bracket zero, 𝑥 must be negative seven because negative seven plus seven is zero. And if we were to go on and work out where it cut the 𝑥-axis using the quadratic formula, it’s about minus fourteen point one four and positive nought point one four.

So the basic method then is to halve the coefficient of 𝑥 and use that in the squared term here and then work out what we need to add or subtract to get the constant that we’re actually looking for in the first place.