In this video we’re gonna use the completing the square technique to find the roots of quadratic equations. We’re gonna look at a range of examples and talk a little bit about some of the benefits of answering questions this way. You probably already know how to solve equations using factoring or the quadratic formula and they’re great methods. But for now, we’ll just concentrate on looking at completing the square method.
So here’s our first example. Solve 𝑥 squared plus two 𝑥 minus three equals zero. Now that means find the values of 𝑥 for which this equation is true. And it’s sometimes called “finding the roots of the equation.” Now actually this question would be easier to solve by factoring. So we can see that we can factor that as 𝑥 plus three times 𝑥 minus one. So if we use the factoring method, it would be pretty straightforward. We have ended up with two brackets multiplied together to make zero. So either the first bracket is zero or the second bracket is zero. In other words, 𝑥 would be either to equal to negative three or equal to one. So let’s go ahead and look at the completing the square method instead now.
Right well I got a couple of choices: I could either complete the square for the whole expression here or I could just complete the square for that bit of the expression there, 𝑥 squared plus two 𝑥. And that turns out to be a little bit easier; so that’s in fact what I’m gonna do. So let’s add three to both sides of that equation. And that gives us 𝑥 squared plus two 𝑥 is equal to three. So looking for a bracket that we can square on the left-hand side to come up with an expression which is equivalent to 𝑥 squared plus two 𝑥, we’ve basically gotta have something times itself gives us 𝑥 squared; so that’s gonna be 𝑥. And the other term we need it to basically add to itself to give two; so we’re gonna take half of the coefficient of 𝑥 and that’s gonna be our other term. But remember if I multiply it out 𝑥 plus one all squared, I would get 𝑥 squared plus two 𝑥. But I’d have this plus one which comes from multiplying one times one, one squared on the end. So if I want this expression here to be equal to the expression on the line above, I’m gonna need to subtract that one that I’ve got here. So the 𝑥 plus one all squared gives us this. If I take away the one squared, I’m getting rid of that and I’ve got the expression that I was looking for.
So just evaluating one squared — clearly that’s just one — now we’re gonna add one to both sides of the equation just to leave ourselves with 𝑥 plus one all squared on the left-hand side, which is equal to four. Now I want to know what 𝑥 is equal to. So I need to take square roots of both sides of my equation. And the square root of 𝑥 plus one all squared is just 𝑥 plus one and the square root of four — well we know that’s two. But in fact there are two answers: I can multiply two by two to get four or negative two by negative two to get four. So I need to put in this plus or minus sign here. So I’ve got 𝑥 plus one is equal to plus or minus two. In fact because I’ve got brackets around the whole left-hand side here, I don’t actually need those anymore; I can remove them. So 𝑥 plus one is equal to positive or negative two. Now I want to get the 𝑥 on its own. So the opposite of adding one is taking away one. So I’m gonna take away one from both sides of the equation. Now the order in which I do this on the right-hand side doesn’t actually make a difference. I could say I’m starting off with positive or negative two and then I’m taking away one or I could stay [say] that I’m starting off with negative one and then I’m either adding two or taking away two. And in fact I find this second version easier to work with, but it’s up to you which-which of those you use.
So the plus or minus sign there means we’ve got two different calculations to do. So 𝑥 could equal negative one plus two, which is obviously one, or 𝑥 could be negative one take away two, which would be negative three. And let’s put our answer in a nice big box to make it lovely and clear. Luckily the answers agree with the answers we got earlier using the factoring method. So that’s good news. Now obviously this took a little bit more working out when I’ve probably been very generous with the amount of working out I’ve written down. And you wouldn’t necessarily right every step of that along the way. So there are a few shortcuts that you can take. But clearly this method was longer and more torturous than just factoring. So if your quadratic equations do factor nicely, then that’s probably gonna be a quicker easier method to use. But as we’ll see a bit later on, there are some other advantages of approaching your question in this particular way, but we’ll look at that in another example.
Okay let’s move on to our next question then. Again this one we’ll factor, but let’s forget about that for now. Well just the reason I’m using examples of factors because they got nice easy numbers to work with, which is gonna make demonstrating the method a little bit easier. So solve 𝑥 squared plus seven 𝑥 plus ten equals zero. So first of all I want to isolate the 𝑥 squared plus seven 𝑥 just to make my life a little bit easier. So I’m gonna subtract ten from both sides to give me x squared plus seven 𝑥 is equal to negative ten. Now I’m gonna try to complete the square on this expression. So that’s 𝑥 plus seven over two. Half of seven, I could write that as three point five. But it’s generally easier to leave these top heavy fractions as we’ll see as we go through the rest of the examples. Now if I might- if I do square that out, I’ll find that I’ve got a plus seven over two all squared on the end, which I want to get rid of. So to make this expression equivalent to the expression above, I need to put this term in here take away the seven over two all squared. And that’s equal to negative ten. So now I’m going to add seven over two all squared to both sides.
And of course seven over two all squared is forty-nine over four. So on the left-hand side, I’m left with 𝑥 plus seven over two all squared is equal to negative ten plus forty-nine over four. So to evaluate the right-hand side there, I need to convert ten into a top heavy fraction that’s got a common denominator of four. And of course that’ll be forty over four because forty over four is ten. So now I need to evaluate that. So negative forty over four plus forty-nine over four gives me nine over four. So I’ve got this expression all squared is equal to nine over four. So if I take square roots of both sides, then I start to move towards a situation where I’ve just got 𝑥 on its own.
And the square root of the left-hand side is just 𝑥 plus seven over two and the square root of nine over four is the square root of nine over four. But don’t forget I need to have the positive version and the negative version of that and now I can evaluate that the square root of nine is three and the square root of four is two. And again I’ve bracketed the whole of the left-hand side. I get rid of those brackets, just to leave me with 𝑥 plus seven over two. Now I want to get 𝑥 on its own. So I’m gonna have to subtract this seven over two from both sides. And when I do that on the left-hand side, I’m left with just 𝑥.
And on the right-hand side, I’m left with negative seven over two plus or minus three over two. Again I could do this either way around. I just tend to find it a bit easier to do the calculation in this order. And as luck would have it, we’ve got common denominators already here. So the first possibility is that 𝑥 is negative seven over two plus three over two, which is equal to negative four over two. And negative four over two is negative two. And the second possibility is that 𝑥 is equal to negative seven over two take away another three over two, which is negative ten over two and of course that is negative five. So the answer to the question is that there are two values that 𝑥 can take which would satisfy that equation at the very beginning: 𝑥 could be negative two or 𝑥 could be negative five.
So it was in two examples now where you’re probably thinking why is he telling me this complicated method because factoring would have come up with the same answers a lot more quickly. And you know that’s a really really good point, but very often there’s a bit more going on with the question. So for example, let me just give you an example of the sorts of situations where you might use completing the square. So let’s say we’ve been asked to sketch the curve 𝑦 equals 𝑥 squared plus seven 𝑥 plus ten and find the coordinates of the lowest point on the curve. So the sort of reason that we were doing this kind of calculation here solving 𝑥 squared plus seven 𝑥 plus ten equals zero is in order to sketch that curve we need to find out where does the curve cut through the 𝑥-axis. So if that’s all we’re doing, then yeah factoring will be a great method. But if we’re trying to do the sketching in the overall analysis and finding the coordinates of the lowest point on the curve, then putting the equation into the completing the square format makes life a little bit easier.
So what we’re doing here we are setting our 𝑦-value to be zero. So if we rearrange this line here to be 𝑥 plus seven over two all squared minus nine over four equals zero. Remember that the 𝑦-coordinate was zero, so we can replace that with 𝑦. This is the completing the square format of our equation and now we can very easily find the minimum point on the curve because in order to make this as small as possible we’ve gotta find the 𝑥-coordinate which-which when we plug it into this equation we get the lowest possible value. But remember 𝑥 only appears in this squared term here. So whether 𝑥 is very large and positive or very large and negative, it doesn’t matter how big a negative number we’ve got after adding seven over two to it. We’re gonna square that value. So that’s gonna come up with a positive value. So this term here the smallest value that can take is zero and that would happen when 𝑥 is equal to negative seven over two. And when 𝑥 is negative seven over two and this bit becomes zero, the 𝑦-coordinate corresponding to that would be just zero minus nine over four. So now we’ve got the coordinates of the lowest point, we’ve got the coordinates where it cuts the 𝑥-axis, we’ve got the coordinate where it cuts the 𝑦-axis, and we know that this was a positive value of 𝑥 squared here. So we know it’s kind of a happy smiley curve. So we know it goes like that; we can very easily sketch that curve. So completing the square comes into its own when we’ve got you know a bit more going on in the question.
Well let’s have a look at another question then. So slightly different format, solve for 𝑥: 𝑥 squared minus five 𝑥 plus twenty-four equals twenty. So we’ve now got numbers on both sides of the equal sign. So we’re just gonna try subtract twenty-four from both sides to leave ourselves with just 𝑥 squared minus five 𝑥 on the left-hand side that we can use for the completing the square method on that. And now we’re right back where we were before — same kind of question that we’ve just done many times. So we’re gonna complete the square on the left-hand side. So remember we’ve got a single 𝑥 squared. So that gives us our 𝑥 as our first term and then we halve the coefficient of 𝑥 to give us our second term. And then we’ve got to take away that second term squared. So take away minus five over two all squared. And now the two lines, this term here, is completely equivalent to this term here. And that is equal to negative four. So evaluating negative five over two all squared, we get twenty-five over four. So we’re taking away twenty-five over four. So I’m gonna try get rid of that by adding twenty-five over four to both sides. And keeping out numbers in top heavy fraction format and doing fraction, addition, and subtraction believe it or not is probably about the hardest part of this. This is what most people find difficult on these completing the square questions. But once you get used to it, obviously it’s quite easy to do. So negative four plus twenty-five over four, I need to convert negative four into an equivalent fraction — top heavy fraction — with a common denominator of four. So that’ll be negative sixteen over four. So on the right-hand side, I’ve got negative sixteen over four plus twenty-five over four, which is nine over four.
Now I need to take square roots of both sides so that I can simplify the left-hand side. And the square root of the left-hand side is just 𝑥 minus five over two. And on the right-hand side, I’ve got a square root nine; I’ve got a square root four. But remember I do need the positive and negative version of that. So now to get 𝑥 on its own — find out what 𝑥 is equal to. I need to add five over two to both sides. So there are two possible versions of 𝑥: 𝑥 could be five over two plus three over two or it could be five over two minus three over two. And when I evaluate those, I’ve got either 𝑥 is equal to four or 𝑥 is equal to one. So that’s that one now. The only bit that was different really was this at the beginning; we had numbers on both sides of the equations, but you can very quickly rearrange it to something you’re more familiar with.
Right let’s do another one then. Solve 𝑥 squared plus three 𝑥 minus six equals zero. So exactly the same technique; it’s just that towards the end of this question the numbers are gonna get a little bit trickier. But never mind we’ll work out how to do that when we get there. So we’ll clear off a little space on the left-hand side by adding six to both sides and then we’ll complete the square on the left-hand side. So we’ve got 𝑥 and we take half of the 𝑥 coefficient, three over two; we’re squaring that. To make that equivalent to the expression above it, we’re gonna need to take away that three over two term all squared. So that’s equal to six. So I’m just gonna evaluate the three over two all squared. So I’ve got 𝑥 plus three over two squared minus nine over four is equal to six and I’m gonna add the nine over four to both sides. So I’ve got six plus nine over four on the right-hand side. So I need to find an equivalent version of six, which is actually a top heavy fraction over four. So I’ve got my common denominators. And that’s obviously twenty-four over four because twenty-four over four is the same as six. Now twenty-four over four plus nine over four equals thirty-three over four. So I’ve got 𝑥 plus three over two squared is equal to thirty-three over four. Now I’m gonna take square roots of both sides. And the square root of the left-hand side is just 𝑥 plus three over two and we got two possible values for the right-hand side: it’s either the positive version of the square root of thirty-three over four or it’s the negative version. Now obviously that’s the same as root thirty-three over root four. So that’s the root of four is two.
So this is slightly different to the previous questions because always before when we were doing this square rooting stage, we ended up with nice simple numbers. But the square root of thirty-three isn’t anything nice and easy. So we just leave it as root thirty-three. And in the next step to get 𝑥 on its own, on the left-hand side here I’m subtracting three over two from both sides. And so the right hand becomes negative three over two plus or minus root thirty-three over two. So as luck would have it here, I’ve already got my common denominator. So I’ve now got two possible values for 𝑥: it could be negative three over two plus the root of thirty-three over two or it could be negative three over two subtract this root of thirty-three over two. Now I could gather those terms together but because these numbers are not particularly nice, I have to leave them in-in that sort of format. Now the question might-might have asked me to give my answers rounded correct to one decimal place or it might have asked me to leave them in this exact format. And just for the sake of argument, I’m gonna pretend they asked me to leave it in the exact format in this case. So sometimes the answers work out really simply and sometimes the numbers are a little bit more complicated, but the method is just the same.
So let’s look at another question. In this case, we’ve got a negative one of 𝑥 squared and we don’t know how to complete the square when we’ve got a coefficient of 𝑥 squared that isn’t equal to one. So what I’m gonna do here is just adjust the questions slightly. I’m gonna multiply everything by negative one on both sides of my equation and then I’ll have positive 𝑥 squared that I can work with. So multiplying each of those terms by negative one, I’ve got negative five take away two 𝑥 plus 𝑥 squared equals zero. In fact I’m just gonna rearrange the left-hand side a bit. Putting the terms in order from the highest power of 𝑥 going downwards and now we got a question just like the ones we’ve just seen. So very quickly we add five to both sides. Then we complete the square on the left-hand side, which gives us 𝑥 minus one because half of two is one all squared. Take away that negative one all squared, which is 𝑥 minus one all squared minus one is equal to five. Now let’s add that one to both sides, giving us 𝑥 minus one all squared equals six. Take square roots of both sides. Again we need to have the positive or negative version of the square root of six. And again root six doesn’t come out to be a nice simple number for us, so we’re gonna add one to both sides. And that gives us two possible values for 𝑥: one plus root six or one minus root six. And of course we could round those to one decimal place if that’s what it wants to do in the question.
And just finally one more quick variation on the theme. This time we’ve got two 𝑥 squared. And all we need to do here is to divide everything through by two. Because we’ve got an equation, we can do the same things to both sides of the equation to keep this the same. So if I divide everything by two, I’ve got an entirely equivalent equation. And you can see here that I’ve now just got one lot of 𝑥 squared, so I can do the method as I did before. And so the rest of the technique is just the same and I end up with my answer 𝑥 is equal to three or 𝑥 is equal to negative seven.
So let’s summarise then. Completing the square bit more tricky than factoring or using the quadratic equation, but it gives you a little bit more information about the curve. Sometimes you have to do a bit of manipulation at the beginning. I like to do this little step here, where we are getting a simpler expression to complete the square with. And you must always remember that when you’re taking square roots of both sides, you’re gonna end up with two possible different answers to work with.