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
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
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
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
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
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
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
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.