# Video: Completing the Square for Quadratic Expressions with the 𝑥² Coefficient Not Equal to 1

Tim Burnham

Using a series of examples, we explain how to use the technique of completing the square to rearrange quadratic expressions in which the coefficient of 𝑥² ≠ 1.

14:47

### Video Transcript

When you’re rearranging quadratic expressions into the completing the square format, you really need the 𝑥 squared coefficient to be one. If it isn’t, then first we’re gonna need to factor the expression and then we can use our tried and tested method of rearranging into completing the square format. Let’s have a look at some examples in this video.

So let’s say we wanted to rearrange this into the completing the square format. Well as we said, here we’ve got two lots of 𝑥 squared, so this idea of just creating something plus or minus something all squared minus something isn’t quite gonna work. So first of all, let’s factor out that two. So if we factor out the two, the first term becomes 𝑥 squared because two times 𝑥 squared is two 𝑥 squared.

The next term must be minus four 𝑥 because two times minus four 𝑥 is minus eight 𝑥. And the last term will be negative five because two lots of negative five are negative ten. So those two lines are equal, but now we’ve got an expression inside the brackets here which has got a coefficient of 𝑥 squared of one. This means that we can easily use our tried and tested method for completing the square.

So the trick is as we go through the rest of the calculation, we just leave that two and the big brackets all around the outside and sort of ignore it whilst we work on the inside of the bracket. So remember, to complete the square, to get 𝑥 squared minus four 𝑥, we’re going to need 𝑥 to be our first term and then we half the coefficient of 𝑥 to make negative two and that’s gonna be 𝑥 minus two all squared. But when we multiply out 𝑥 minus two all squared, we get the 𝑥 squared minus four that we want. But we’re also left with this negative two times negative two which is four, and we need to get rid of that to create an equivalent expression.

So we need to subtract that second term, negative two all squared, and we’ve still got the negative five on the end, the take away five, so let’s just tidy that up a bit. And negative two times negative two is four, so what we’re doing here is taking away four and then we’ve still got the minus five on the end. So 𝑥 squared- sorry 𝑥 minus two all squared take away four take away five; that’s take away nine.

So we’re nearly there. We just need to remember the we’ve got two lots of that first term and two lots of the second term so let’s evaluate that. So the completing the square format of this expression is two lots of 𝑥 minus two all squared minus eighteen.

So sometimes the numbers aren’t quite so easy as that one, so let’s look at this slightly more difficult example: three 𝑥 squared plus six 𝑥 minus one. So we need to have a single 𝑥 squared, so we’re gonna need to factor out three. So that first term is gonna become three lots of just 𝑥 squared. And then to get the plus six, we’re gonna need to make that two 𝑥 because three lots of two 𝑥 are six 𝑥. But the last term, what do I need to multiply by three to get negative one? It’s minus a third because three lots of minus a third would give us negative one. So this is an equivalent format, but it’s got an 𝑥 squared term of one. So now let’s concentrate inside the brackets and ignore what’s outside of the brackets for now.

Well the completing the square bit is gonna be 𝑥 plus one all squared, but of course I’m gonna to need to take away that one squared, so this second term here all squared, in order to make this term here- oops this term here equivalent to this term here. And then I’ve got my minus a third on the end. So obviously one squared is just one so that’s equivalent to three lots of 𝑥 plus one all squared minus one minus another third. So that’s minus one and a third, but I’m gonna leave it as a top heavy fraction. So minus one is equivalent to three thirds minus another third would make that minus four thirds.

And that’s really a top tip for these sorts of questions, tend to work as with top heavy fractions like this because it’s gonna make the maths a little bit easier as we go through and do the questions. So now we need to multiply the three back in we need three lots of 𝑥 plus one all squared and then three lots of negative four over three.

So the first term is three lots of 𝑥 plus one all squared, and three lots of negative four over three is negative four. So there we have it: these two things here are completely equivalent to each other, so let’s just check that just for the sake of argument.

So 𝑥 plus one all squared is 𝑥 plus one times 𝑥 plus one, so let’s multiply that out. And that gives us inside the bracket three lots of 𝑥 squared plus 𝑥 plus another 𝑥 plus one.

And of course 𝑥 plus 𝑥 is two 𝑥, so now I’ve got to do three lots of 𝑥 squared, three lots of two 𝑥, and three lots of one. So I’ve got my three 𝑥 squared that I was looking for. I’ve got my six 𝑥 that I was looking for. And three take away four is negative one, so I’ve got the negative one I was looking for. Yep, they are correct.

Okay let’s look at one more example like that. Take two 𝑥 squared plus nine 𝑥 plus five, so we need to factor out the two. So the first term is just gonna be 𝑥 squared because two lots of 𝑥 squared is two 𝑥 squared. Now we need to generate nine 𝑥, so two times what is nine 𝑥?

Well obviously four and a half. But again as we said before, it’s good idea to leave these as top heavy fractions so two lots of nine over two would give us nine, so it’s nine over two. And then the last term is gonna be five over two because two lots of five over two is five. So we’ve got our expression now in the brackets, and we’re going to complete the square with this expression, which has just got one 𝑥 squared.

So we’ve got umm 𝑥 squared plus nine over two 𝑥, so we’ve got 𝑥 as the first term and half of the coefficient of 𝑥 half of nine over two is nine over four. So that’s 𝑥 plus nine over four all squared, but of course we need to take away the square of the nine over four. Otherwise we’ll have a too high a number, too high a constant value on the end, and then we’ve still got our plus five over two on the end there as well.

So just evaluating that term there nine over four all squared is eighty-one over sixteen. So looking at this expression here, I’ve got minus eighty-one over sixteen plus five over two. Well in order to add fractions, we need to have a common denominator, so I need to find an equivalent version of five over two that is over sixteen. So if I multiply the top by eight and the bottom by eight, I get forty over sixteen. So five over two, forty over sixteen, just the same fraction expressed in a different way. And now it’s easy to do that; minus eighty-one over sixteen plus forty over sixteen is negative forty-one over sixteen.

So now I can multiply the two back in, so the first term multiplied by two and the second term multiplied by two. And forty-one over sixteen times two is gonna be forty-one over eight.

So that’s our answer; this is the completing the square format of this. So the process then is use the coefficient of 𝑥 squared as a factor to create an expression that we’re gonna do the completing the square of which has only got one lot of 𝑥 squared. Now the top tip is to leave your fractions in top heavy format because that just makes things a little bit easier to calculate as we go through. Then we ignore this bracketed out this factored out two all the way through; we just work inside the brackets completing the square. And then right at the very end, we need to multiply it back in again to get our final answer.

Okay just to round off with another couple of examples, if you’re using the completing the square format to find the roots of the quadratic equation with an 𝑥 squared coefficient isn’t one, then you can use the same method we’ve just seen. But rather than factoring at the start to you know maintain an equivalent expression, then you can divide through by the coefficient of 𝑥 squared because we’ve got an equation. We can do the same things to both sides of that equation, which just makes things a little bit easier to look at. So let’s see this one: solve five 𝑥 squared minus ten 𝑥 minus fifteen equals zero using completing the square. So we’ve got five 𝑥 squared, so I’m gonna divide both sides of the equation by five.

So going through term by term, five 𝑥 squared divided by five just gives me 𝑥 squared, minus ten 𝑥 divided by five gives me minus two 𝑥, minus fifteen divided by five gives me minus three, and zero divided by five just gives me zero. So now I’ve got a nice easy example to work with.

I’m just gonna start off by adding three to both sides to give me a nice simple expression over here to complete the square for. So taking half of the 𝑥-coefficient, that would give me negative one. So I’ve got 𝑥 minus one all squared, but of course I need to take away the square of that minus one. And then I’ve got nothing else to adjust over here, and that whole thing is equal to three. Just evaluating, negative one times negative one is just one so we’re taking away one on the left-hand side.

And to try to simplify the left-hand side, I’m gonna add one to both sides of my equations, which makes the left-hand side just equal to 𝑥 minus one all squared and the-right hand side three add one is four. Now I’m gonna take the square root of both sides. And the square root of 𝑥 minus one all squared is just 𝑥 minus one, and the square root of four is two. But don’t forget, I’ve got positive or negative versions of that because negative two times negative two is also four

So 𝑥 minus one is equal to positive or negative two. So if I add one to both sides of that equation, I’ve got 𝑥 is equal to one plus or minus two. So there’re two possible versions: I can have 𝑥 is equal to one plus two or 𝑥 is equal to one minus two.

So that’s 𝑥 is either equal to three or negative one. So really what we were looking at there was just showing you this-this point that if we’re solving an equation, because we’ve got an equation with a left-hand side and the right-hand side, we can do the same thing to both sides. So we can we can create an equivalent expression, which we can then solve for 𝑥 that does have an 𝑥 squared coefficient of one. We don’t need to sort of factor out the five and then work with more tricky numbers all the way through the calculation and then multiply things back in. We just be treat it as an equation.

Right. Okay, let’s do one final example when maybe the numbers don’t quite work out so nicely. So we’ve got three 𝑥 squared plus seven 𝑥 minus eight is equal to zero, and we’re gonna solve this using the completing the square method. So as we were just saying, we’ve got an equation so it’s perfectly acceptable to divide through everything by three, both sides of the equation divided by three. Those two things will still be equal.

Now when doing that, I’ve now got seven over three 𝑥 minus eight over three. So the numbers aren’t particularly nice, but we can still work with this. So what I’m gonna do is add eight over three to both sides to give me a simpler expression here to complete the square for. So we’ve got 𝑥 squared plus seven over three 𝑥 is equal to eight over three, so what I need to do is find a bracket that I can multiply out, that I can square. So it’s gonna be 𝑥 as the first term and then half of seven over three is seven over six.

So that’s the bracket that I’m squaring. But if I multiply that bracket, I would end up with a plus seven over six all squared term on the end. But I don’t want that because I’ve got nothing else on that side here, so I’m gonna subtract that seven over six all squared to create my equivalent expression. Now that is still equal to eight over three so now we can evaluate seven over six all squared; that’s forty-nine over thirty-six.

And then because I only want this term on the left-hand side, I’m going to add this term to both sides. And now over on the right hand side, I’ve got eight over three plus forty-nine over thirty-six. So what I’m really looking for is an equivalent version of eight over three that’s got a denominator of thirty-six to match the forty-nine over six.

And if I multiply the bottom and top by twelve, I’ve got ninety-six over thirty-six. So adding ninety-six over thirty-six to forty-nine over thirty-six, I get a hundred and forty-five over thirty-six. so now we’ve got 𝑥 plus seven over six all squared is equal to a hundred and forty-five over thirty-six. Well I wanna know what 𝑥 is, so I’m gonna take the square root of both sides of this equation.

And that gives me 𝑥 plus seven over six on the left-hand side, while on the right-hand side I’ve got positive or negative the square root of a hundred and forty-five over thirty-six. And of course the square root of thirty-six is six. So finally to get 𝑥 on its own, we’re gonna subtract seven over six from both sides, which leaves us with minus seven over six plus or minus the square root of a hundred and forty-five over six on the right-hand side. So that’s two possible answers.

Either 𝑥 is minus seven over six plus that or minus seven over six minus that. And if I’d been asked to round them to one decimal place, that would give me 𝑥 is nought point eight or negative three point two to one decimal place.

So that about wraps it up for completing the square for quadratic expressions when the 𝑥 squared coefficient isn’t equal to one. If we’ve got an equation, then it’s perfectly okay to divide both sides of the equation by the-the coefficient of 𝑥 squared, which gives us a nice easy expression or hopefully a slightly easier expression to work with. And it’s still true; it’s still a true expression because we’ve done the same thing to both sides of the equation, so both sides are still equal. But if we just started off with an expression that wasn’t an equation, then you have to factor out the coefficient of 𝑥. You have to keep it in there now. Otherwise if you just divide through by that number then you wouldn’t have an equivalent expression to what you started off with. So subtle difference there, if you got an expression, you have to factor; if you got an equation, then it’s okay to divide through by the coefficient of 𝑥 squared.