# Video: Using Completing the Square to Explore Quadratic Equations

Learn about using the technique of completing the square to find the coordinates of roots of quadratic equations and the highest or lowest points on the curves. We also show how to use this information to sketch the curves.

16:22

### Video Transcript

In this video, we’re gonna look at the completing the square technique and use it to help us to sketch quadratic curves, and find the coordinates of the points where the curves cut the 𝑥-axis. We’ll also look at how to find out the coordinates of the highest or lowest point on the curve using this technique as well.

Okay. Let’s start off with the curve 𝑦 equals 𝑥 squared plus two 𝑥 minus seven. Well there’s nothing in front of the 𝑥 squared and-and we all know that actually that means it’s one lot of 𝑥 squared. And more than that, it’s in fact, it’s positive one. So the fact that this is a positive value, and if you’re feeling positive you’re nice and happy, that means it’s a smiley parabola. All quadratic curves are parabolas. They’re symmetric curves that look either sort of like a happy smile, or-or a sad face. And in this case, because it’s positive, we know that it’s gonna be a symmetric curve that looks like that. This value on the end here, okay, minus seven, that tells us where it cuts the 𝑦-axis. So in other words, the- on the 𝑦-axis, the 𝑥-coordinate is always zero. If we put an 𝑥-value of zero in here, that’ll be zero squared, will be zero. An 𝑥-value of zero in here, two lots of zero is zero. So the corresponding 𝑦-coordinate will be zero plus zero minus seven, would be minus seven. So it cuts the 𝑦-axis at zero, minus seven.

So without doing any completing the square or anything, we’ve already gleaned a bit of information about what this curve is gonna look like on our graph, as part of our sketch. So let’s go ahead and complete the square now. And remember, to complete the square, we’ve got to have an expression like this all squared, and then we’re going to be taking something away, in order to make everything match up again at the end. So we’ve taken half of the 𝑥-coefficient, so that’s one. And then to generate one 𝑥 squared, it’s just 𝑥. So 𝑥 times 𝑥 would give us 𝑥 squared. But of course, if I multiplied out that bracket there, 𝑥 plus one all squared, I would get my 𝑥 squared plus two 𝑥 which I’m looking for. But I would also get this one, plus one on the end, which comes from multiplying the one by one, the one squared. So that’s what I need to take away from this expression here, to make it equivalent to the bit up here, the first two terms of that expression above. So this bit here is equivalent to this bit here, and I’ve still got to take away seven from this. So there we’ve got an equivalent version of that equation. So let’s just tidy that up. One squared is just one, so we’ll take away one. Then we’ll take away another seven.

So we’ve now got this in its completing the square format, 𝑦 equals 𝑥 plus one all squared minus eight. Now when we look at that, if I think about all the different 𝑥-values that I might put in my 𝑥-coordinates, I’m taking that, I’m adding one, and I’m squaring that. So it doesn’t matter whether 𝑥 is hugely negative, and I add one, I’ve still got a negative number. When I square a negative number, I’m gonna get a positive answer. And then I’m gonna subtract eight from it. If I start up with a very large positive 𝑥 number, I’m gonna have a positive number plus a positive number. I’m gonna have a really big positive number squared, which makes a positive number, taking away eight from it. So this term here is always gonna be positive, or is it, we- it could be zero. If 𝑥 was equal to minus one, then the smallest I could make this term equal to, is zero because it’ll be 𝑥- negative one plus one, would be zero. And zero squared is zero. It can’t be any smaller than that. So the smallest value we can get for 𝑦, in other words, the lowest point on this curve, as low down as that curve goes, is when the 𝑥-coordinate is equal to negative one. And when the 𝑥-coordinate’s equal to negative one, this bit becomes zero. So the 𝑦-coordinate is gonna be zero take away eight; the corresponding 𝑦-coordinate is gonna be negative eight.

So by rearranging into the completing the square format, we’ve managed to work out the coordinates of the lowest point in the curve. We can also use that to work out where it cuts the 𝑥-axis. Now on the 𝑥-axis, the 𝑦-coordinates are zero. So if I put 𝑦 equal to zero, then I’ve got 𝑥 minus one all squared minus eight is equal to zero. So if I add eight to both sides of that equation, I’ve got 𝑥 minus one all squared is equal to eight. So now if I take square roots of both side, well the square root of 𝑥 minus one all squared is just 𝑥 minus one and the square root of eight, well there can actually be two posit- posi- possible answers for that, they could be the positive version or it could be the negative version, so positive eight or negative eight. And if I add one to both sides, 𝑥 could either be equal to one plus root eight or one take away root eight, which to one decimal place gives me the one point eight or negative three point eight. So the curve cuts the 𝑥-axis in two different places. So now I’ve got all the information that I need to be able to do my sketch.

So first of all, let’s draw out our 𝑥- and 𝑦-axes and there we can mark some things on. We know that it cuts the 𝑥-axis here and here. We know that it cuts the 𝑦-axis here. And we know that the lowest point on the curve, the 𝑥-coordinate is negative one and the 𝑦-coordinate is negative eight. So we can draw in our sort of line of symmetry here, for the middle of the curve, and mark on the lowest point of the curve there. We know that this is a symmetrical parabola, and in fact my drawing is not gonna be entirely accurate but let’s go with it. And it is only a sketch, so we’re not plotting it. But it has to look roughly right. We’ve marked on, on the 𝑥- and the 𝑦-axes where it cuts those. And we’ve got the line of symmetry, we’ve got the-the lowest point on that curve. So we’ve got a pretty good idea of what that curve would look like. So the completing the square technique is a great way of getting all the information we need, to visualise the curve related to that equation that we were given in the question.

Right. Let’s look at another question then 𝑦 equals 𝑥 squared minus three 𝑥 plus five. Well we can see that it cuts the 𝑦-axis at positive five. And being a positive one 𝑥 squared, it tells us it’s a positive and happy curve. So we’ve got some basic information there. Let’s now go on and do our completing the square. And we’ve got one 𝑥 squared, so it’s gonna be 𝑥 as our first term. And half of the 𝑥-coefficient would be negative three over two, so negative three over two all squared. And now we’ve got to subtract that negative three over two all squared in order to make it equivalent to the first few two terms of that expression. Now we’ve still got the plus five that we need on the end to make those two lines completely equivalent. So that’s our completing the square version. And we know just need to tidy up so negative three over two times negative three over two is nine over four. So that equation becomes 𝑥 minus three over two all squared minus nine over four plus five. And if we just change five into a version of five which is a top heavy fraction that’s got a denominator of four, that’s gonna make our adding up a little bit easier. And that will be twenty over four. So we’ve got negative nine over four plus twenty over four, which gives us eleven over four. So this is another version of our equation for our curve.

And again, we’ve got a term here that is squared involving 𝑥. So however we vary 𝑥 when we-when we do stuff with it, whether if the answer comes out positive or negative, we’re gonna square that; we’re gonna get a positive answer. So this time here, the smallest value it could take would be zero. And that would happen when 𝑥 is equal to three over two. And when the first term is zero, that means that the 𝑦-coordinate will be zero plus eleven over four. So the 𝑦-coordinate would be eleven over four. So to make the 𝑦-coordinate as small as possible, we make this term equal to zero. And at that point, 𝑥 is equal to three over two and 𝑦 is equal to eleven over four. So they’re the coordinates of the lowest point on that curve.

Okay. Let’s try and find when it cuts the 𝑥-axis then. It cuts the 𝑥-axis when the 𝑦-coordinate is equal to zero. So 𝑥 minus three over two all squared plus eleven over four would be equal to zero. So if we take eleven over four away from both sides, we get 𝑥 minus three over two all squared is equal to negative eleven over four. And now we take square roots of both sides. We get 𝑥 minus three over two is positive or negative the square root of negative eleven over four.

Well now we’ve got a problem. How do you find the square root of a negative number? Well you can’t. Because if you’ve got a negative number and you multiply it by itself, you get a positive answer. If you’ve got a positive number and you multiply it by itself, you’ve got a positive answer. You can’t take a real number and multiply it by itself and get a negative answer. So there’re no real solutions to this. So that means that it doesn’t cut the 𝑥-axis. There ar- there are no values of 𝑥 which are gonna get- generate a 𝑦-coordinate of zero when you’re using that equation. So this is a curve that comes down and doesn’t cut the 𝑥-axis. And actually, if we’d have looked at the lowest point quite carefully, we could’ve spotted that a little bit earlier. When 𝑥 is three over two, the lowest point on that curve has got a 𝑦-coordinate of eleven over four; that’s positive. It never comes down as low as zero, certainly never goes negative.

So now we can go ahead and sketch our curve. We know that it cuts the 𝑦-axis at positive five. And we know that the lowest point on the curve is three over two, eleven over four, which is one and a half and two and three quarters. So that basically makes it about here somewhere. So using that symmetry, we can now sketch in the curve, looks something like this. It comes down to the bottom here, and then comes back up exactly the same way the other side. So that’s what our curve looks like.

So this is where the completing the square method comes in really handy. Let’s just think back to the previous example for a moment. So we had 𝑦 equals 𝑥 squared plus two 𝑥 minus seven and it looked like this. It cut the 𝑥-axis in two places. And because it cut the 𝑥-axis in two places, and because it’s am-a symmetrical curve, if we’d have used the quadratic formula, or we’d have been maybe been able to factorise that to calculate what these 𝑥-coordinates are when 𝑦 is equal to zero, then we would have been able to take the midpoint of those two to work out what the 𝑥-coordinate of this minimum point down here would be. And once we knew the 𝑥-coordinate, we can plug that back into this equation to work out the corresponding 𝑦-coordinate. So we had a method, even if we didn’t know about completing the square, to work out the value of this minimum point. Now with this question, because it never cut the 𝑥-axis, we didn’t have two points that we could find the midpoint for to be- to help us to find this lower point here. Now there are other ways of working out that minimum point, but completing the square is one of them and it’s a very useful method of doing that.

So for example, we knew this point here when 𝑦 was equal to five, we knew that 𝑥 is equal to zero. So if we set our equation equal to five, we could find the coordinates of this point over here as well, and we could find the midpoint of that. So I’m not saying there aren’t other way of doing it, but the completing the square method is a nice way of doing this.

Okay. Let’s look at one more example with slightly more tricky numbers. 𝑦 is equal to negative a third of 𝑥 squared plus five 𝑥 minus twenty. So we can see straightaway that it cuts the 𝑦-axis at negative twenty and it’s a negative, so it’s a sad curve. So it’s a negative sad curve. Now to complete the square on this, we’re gonna have to factor out a negative a third to leave us with one 𝑥 squared, which we can do the completing the square with. So that’s gonna be negative a third on the outside of the bracket. So then negative a third times 𝑥 squared gives us negative a third 𝑥 squared. What do we need to multiply negative a third by in order to give us five 𝑥? Well that’s gonna be negative fifteen 𝑥. So negative a third times negative fifteen would be positive five, so we need negative fifteen 𝑥. And then likewise, what do we need to multiply negative a third by to give us negative twenty? The answer is positive sixty.

So we’ve rearranged this expression as negative a third times the whole of 𝑥 squared minus fifteen 𝑥 plus sixty. So if we concentrate inside the bracket and use our completing the square technique in there, I’ve got 𝑥 for my first term in the square bracket and a half of the 𝑥-coefficient, so minus fifteen over two is the second term, that’s all squared. But of course, I need to subtract negative fifteen over two all squared. And now this expression here is the same as this first two expressions here. So to get those two lines to be completely equivalent, I still need to add on my positive sixty. And now I can evaluate negative fifteen over two all squared is two hundred and twenty-five over four. So the equation becomes 𝑦 equals minus a third of the whole of 𝑥 minus fifteen over two all squared minus two hundred and twenty-five over four plus sixty. So if I rewrite sixty as two hundred and forty over four, then I’ve got two terms here which have got the common denom- denominator of four, and I can add them together. And negative two hundred and twenty-five fourths plus two hundred and forty fourths is plus fifteen over four. And now I need to multiply back in my negative a third times that and negative a third times that, which gives me 𝑦 is equal to negative a third of 𝑥 minus fifteen over two all squared minus five over four.

Now looking at just the bracket here 𝑥 minus fifteen over two all squared, that has always gotta be positive. The smallest we can make that is zero because whether 𝑥 minus fifteen over two gives me a positive or a negative answer, I’m gonna multiply it by itself and get a positive answer. So the smallest I can make that little bit there is zero. And given that, generally speaking, that’s gonna be positive. And then multiplying that by negative a third, this whole term here is always going to be negative, and- unless it’s zero. So it’s negative or zero. So the biggest it can be is zero. So the biggest I can make this first complete term here is zero, and that’s when 𝑥 is equal to fifteen over two. And this bit here in the bracket becomes zero. And when 𝑥 equals fifteen over two, the corresponding 𝑦-coordinate will be zero take away five over four.

So the highest point on the curve is when 𝑥 is fifteen over two and 𝑦 is negative five over four. Now we-we learned in the last example to look for this kind of thing. So if the highest point on the curve has a negative 𝑦-coordinate, that means it’s never getting up as far as the 𝑥-axis. So I don’t need to go on and solve this to try and find when it cuts the 𝑥-axis because it doesn’t. The highest point has a negative 𝑦-coordinate. It never reaches as high as zero, so it’s not gonna cut the 𝑥-axis. So I can draw my axes. I know that it cuts the 𝑦-axis at negative twenty and I know that the highest point is at fifteen over two, so that’s seven and a half minus five over four, which is minus one and a quarter. And I can sketch in my curve. So it goes up to the maximum point here. It comes down here. And that is a line of symmetry going through my curve here. And that’s the highest point on the curve of fifteen over two, minus five over four. So that’s what my sketch would look like.

So just to summarise, completing the square can help you to find the roots of a quadratic, if they exist. And it can help you find the coordinates of the highest or lowest point on a quadratic curve, whether or not that curve cuts the 𝑥-axis.

So we’ve got other videos about completing the square that explain how to do it, explain how it works, and enable you to kind of practice and get better at doing it. But hopefully, this has brought it all together and just showing you how you can use the technique to investigate quadratic equations.