# Video: Improving Your Proficiency in Rearranging Quadratic Expressions to the Completing the Square Form

Learn how to quickly rearrange quadratic expressions into the completing the square form.

16:20

### Video Transcript

In this video we’ll be looking at how to quickly rearrange quadratic expressions into the completing the square form. We have an introductory video on completing the square which gives you a bit of background on why it’s useful and how it works. And we have other videos which use the technique to find the roots of the quadratic equations and help you practice skills on trickier quadratic equations. For now though let’s look at how to get quicker at rearranging quadratic expressions into the completing the square form. So we are not looking at quadratic equations for now, just the quadratic expressions themselves.

So here’s our first example: 𝑥 squared plus ten 𝑥 plus seven. Now we want to rearrange this into a format where we’ve got something in brackets squared plus or minus another constant. Well this bracket’s going to be squared. So the first term is gonna be multiplied by itself to generate this 𝑥 squared term. So the first term in the bracket is gonna be 𝑥. And then we’ve got the number which is gonna multiply by itself to give something. But when we multiply it out, whatever number that is, we’re gonna be adding it to itself and that’s the number of 𝑥s that we’re gonna get. So we want to get an answer of ten 𝑥; so that number should be five because five plus five is ten.

So our standard technique then is we look at the number of 𝑥s and we halve that coefficient and put that in the brackets there. So in multiplying out 𝑥 plus five all squared, we’ve generated an 𝑥 squared term which is great; that’s what we wanted. We’ve generated a plus ten 𝑥, which is what we wanted. But we’ve also generated this plus five squared term, which isn’t what we wanted. We’ve got plus twenty-five. We wanted plus seven on the end. So what we’re gonna do is we’re gonna take away that five squared term first, which would leave us with plus zero. And then the seven is the thing we’re interested in, so we’re going to add that on. So if I just evaluate those numbers on the end, take away five squared, so that’s take away twenty-five plus seven gives us a total of negative eighteen. So this is our completing the square format. Now let’s just quickly check that answer. Well if I multiply that out, I’ve got 𝑥 times 𝑥 is 𝑥 squared plus so that’s that one then plus five 𝑥 plus another five 𝑥 plus twenty-five and then take away eighteen. And simplifying that, I’ve got five 𝑥 and five 𝑥 makes ten 𝑥 and twenty-five take away eighteen is positive seven. So that is in fact the answer that we were looking for, so there we go the completing the square format.

So let’s look at another example then. 𝑥 squared plus six 𝑥 minus two. We’re looking for this format something in brackets all squared plus or minus some other number. And we’re gonna take half of the 𝑥-coefficient. So and because this is 𝑥 squared, we’re gonna put an 𝑥 here and a plus three here. And when we multiply out 𝑥 plus three all squared, we get an 𝑥 squared term, which is great; that’s what we’re looking for. We get three 𝑥 and three 𝑥 make six 𝑥, so that’s good. But then we’re getting plus three squared on the end. So if I just take that off, that gives me plus zero on the end. And what I’m looking for is negative two. So if I take the negative two away, then that’s gonna give me my correct answer. So minus three squared, so that’s minus nine take away another two is minus eleven. And that’s my completing the square form. And if I was just to check that, I’ve got 𝑥 plus three squared minus eleven. Well we’ve ready multiplied out 𝑥 plus three times 𝑥 plus three here. And now we’re saying we’re taking away another eleven from that. And when I do nine take away eleven, I get negative two, which is in fact the version that I was looking for. So this format here is equivalent to this expression up here and this is the completing the square format.

Right let’s look at one more subtly different question here. So I’ve got 𝑥 squared take away twelve 𝑥 take away five. So let’s put the basic format: something squared plus or minus something. We’re taking half of the 𝑥-coefficient and we’ve got 𝑥 squared here. So this is gonna be 𝑥 minus six all squared. And to work out what I need to add on to the end, let’s just quickly multiply out the 𝑥 minus six squared. And I’ve got 𝑥 squared minus six 𝑥 minus six 𝑥 and then I’m adding negative six times negative six. So I’ve got negative six times negative six as my last term and I’m adding that on. So I’ve got my 𝑥 squared, I’ve got my minus twelve 𝑥, and I’ve also got this negative six squared that I added on. So we need to take that away. So that wipes out this term here that was generated by multiplying out this expression. And now we need to take away another five because that was the number that we’re looking for for the constant on the end. So now let’s evaluate this. Well negative six times negative six is plus thirty-six, positive thirty-six. So we’re taking away positive thirty-six as negative thirty-six and then we’re taking away another five; so that’s negative forty-one. And there’s our completing the square form of this original expression.

So what have we learnt so far? Well in order to create our completing the square format, we were halving the 𝑥-coefficient here to create the term in the bracket. And we were then realising that if we multiply that bracket by itself, it generates some sort of term on the end. And we need to take that away and then we need to make the adjustment to get the number that we’re actually looking for and that then leaves us with our completing the square form. So let’s look at the general form of this using letters instead of numbers and then see if we can come up with any conclusions.

So for now we’re still sticking to quadratic expressions, which have only got one 𝑥 squared. But we’re going back to the general form of that. We’ve got plus something times 𝑥 — we don’t know what that number is — plus another number on the end. So we said we need to halve the coefficient of 𝑥 and add that to 𝑥 and that’s our squared term. And when we multiply out that bracket, so multiplying 𝑥 plus 𝑏 over 𝑐 two by itself, we get 𝑥 squared; we get 𝑏 over two 𝑥 plus 𝑏 over two 𝑥. Well a half of 𝑏𝑥 plus another half of 𝑏𝑥 is a whole 𝑏𝑥. And these are in fact the two terms that we were looking for, but we’ve got this plus 𝑏 over two all squared on the end. So in order to get rid of that because we didn’t really want it — we’ve got plus 𝑏 over two all squared — so we’re gonna subtract that from this term here. So if I evaluated the whole of this, I would just end up with 𝑥 squared plus 𝑏𝑥 cause I’ve just got rid of this term on the end here. But of course I want plus 𝑐, so I need to add 𝑐 onto the end.

So that whole process boils down to three things then. So first of all, we have to halve the coefficient of 𝑥 to work out what’s gonna go in the brackets there. We then subtract that number squared from this whole expression. And then we add in the original constant that we wanted to get the original quadratic expression. So let’s practice that a little bit. So firstly, express 𝑥 squared minus twenty 𝑥 plus eight in completing the square form. So that’s gonna be something in brackets squared plus or minus some other number. We know we’ve got one 𝑥 squared here. So we’re just gonna put an 𝑥. And minus twenty is the coefficient of 𝑥. So we’re gonna halve that to make minus ten. And so by now hopefully we realised that when we multiply this out, that’s gonna generate 𝑥 squared minus twenty 𝑥, but it’s also going to have a plus negative ten squared as well. So we need to take that away. So let’s take away negative ten squared. And then we need to add on the constant that we wanted at the beginning; so that’s plus eight. Well negative ten times negative ten is a hundred, positive a hundred. So we’re taking away a hundred and then we’re adding on the plus eight. And negative a hundred plus eight is negative ninety-two. So that’s our completing the square form of that expression. So once you spot this technique — halving the 𝑥-coefficient there, taking away the square of that term, and then adding on the original number that you were looking for — this process actually becomes quite quick and easy to do.

So let’s look at another question then. This one they’ve tried to make it a little bit more difficult by making it more complicated form of wording. Express 𝑥 squared plus six 𝑥 minus nineteen in the form 𝑥 plus 𝑎 all squared plus 𝑏. Well that is the completing the square format: something squared plus or minus another number. 𝑏 could be a negative number, so where 𝑎 and 𝑏 are constants. So 𝑎 is just a number and 𝑏 is just a number. So we’re trying to organise it into the format something in brackets squared plus or minus another number. We’re gonna halve the coefficient of 𝑥, so that becomes 𝑥 plus three in the brackets. We’re then gonna subtract that positive three squared. And then we’re going to add on the original constant term that we were looking for. And if I add negative nineteen, that’s just the same as taking away nineteen. Now I need to evaluate this second part here. Well three squared is nine. So minus nine minus another nineteen is minus twenty-eight. So 𝑥 plus three all squared minus twenty-eight; that’s the completing the square form that we’re looking for. Now in an exam question, it might go on to ask you for the values of 𝑎 and 𝑏. Well in this case, 𝑥 plus 𝑎 squared. So we had 𝑥 plus three squared. 𝑎 is three and-and now we’re adding 𝑏 to the end of that. But our completing the square term is negative twenty-eight. So 𝑏 must have been negative twenty-eight. Well it’s negative twenty-eight that we were adding to that square term. So 𝑏 is negative twenty-eight.

So let’s look at just slightly more complicated question. Express 𝑥 squared minus ten 𝑥 in the form 𝑥 plus 𝑎 all squared plus 𝑏, where 𝑎 and 𝑏 are constants. So this is the completing the square format. But the problem is we’ve got 𝑥 squared minus ten 𝑥. Well we haven’t got a number; we’ve not got a constant term on the end. Now this throws quite a lot of people. And it’s very tempting to look at 𝑥 squared minus ten 𝑥 and think “oh! I can factor that.” So 𝑥 brackets 𝑥 minus ten; that’s a factored form of 𝑥 squared minus ten 𝑥 yeah. But it’s not in the form 𝑥 plus 𝑎 all squared plus 𝑏. So it’s wrong. I mean that’s probably the most common mistake that people make when they’re answering questions like this. So just be aware of that. So just writing the question out. If I had written plus zero on the end cause we’re not adding anything, we would have been able to just follow our little formula that we used before: we’re gonna halve the 𝑥-coefficient to get that square term, we’re then going to have to take away this squared, minus five squared, and then we want to add on the constant that we’re actually looking for. So we’re gonna add the zero to the end. So that’s 𝑥 minus five all squared. And then we’re taking away — negative five times negative five is positive twenty-five. So we’re taking away twenty-five. Negative twenty-five plus nothing is just negative twenty-five. And here it is. That’s our answer. If you multiplied this out and evaluated it, took away twenty-five, this is the expression that you would get at the top of the page.

So we’ve used that standard technique and we have seen some slightly different wordings of questions — some slightly different examples there. Let’s have another look at one; there’s one more little twist I can put in here which throw some people. So express 𝑥 squared minus five 𝑥 plus nine in completing the square form. Well the trick here is that the 𝑏 term, the multiple the coefficient of 𝑥, is an odd number. So we’re not gonna have a nice integer to work with in our-in our brackets. So we’re gonna have 𝑥 and half of negative five. Well you could say that’s negative two and a half, but generally speaking I would advise just writing as a top heavy fraction. So minus half of five is just five over two. So just leave it in that format. We then need to subtract the square of that and then add the number that we were originally looking for, the plus nine onto the end. Now I need to evaluate this a little bit over here. Well negative five over two times negative five over two is gonna be positive. five times five is twenty-five. Two times two is four. So we’re subtracting twenty-five over four. And then we’ve got to add nine to the end, so this confuses some people. We now got to go back to primary school and do some-some fraction or addition. So we need to find an equivalent version of nine that is a fraction with a common denominator of four. So that is thirty-six over four. If I do thirty-six divided by four, I get nine. So I’ve just written nine as an equivalent top heavy fraction with the same denominator as the other one. So I’ve got negative twenty-five over four plus thirty-six over four. Well that’s the same as thirty-six over four take away twenty-five over four, which is positive eleven over four. And there it is. That’s the completing the square form of 𝑥 squared minus five 𝑥 plus nine. Now if you wanted to convert those top heavy fractions to mixed numbers or even into decimals, then they’re perfectly acceptable forms of the answer. But you just don’t need to do it. This version here we’ve got there first; that’s the-that’s the perfectly good version of the answer.

So here is a general form of our question here. Express 𝑥 squared plus 𝑏𝑥 plus 𝑐 in completing the square form. So that means something in brackets all squared plus or minus some other number. We’re gonna halve the 𝑥-coefficient. So that’s gonna be 𝑥 plus 𝑏 over two, half of 𝑏. Then we need to subtract that 𝑏 over two all squared. And then we need to add back our original 𝑐. And that is our completing the square form of the quadratic expression.

So things to watch out for in these questions. 𝑥 plus 𝑎 all squared plus 𝑏 means the completing the square form. You need to be very careful with negative signs in these expressions. Sometimes you’d better off writing out the expression with a plus zero on the end, so you are clear about what you’re doing. And it only works this way when the coefficient of 𝑥 squared is one. We’ll talk about when you’ve got two 𝑥 squared or five 𝑥 squared or something like that in another video another day. So hopefully now you’re feeling a bit more proficient with completing the square form. It’s a really nice little piece of algebraic manipulation that comes in really handy. So it’s well worth going away and practicing that now and becoming really really good at it. Okay good luck. Thanks.