Main content

### Course: Get ready for Geometry > Unit 3

Lesson 4: Completing the square- Completing the square
- Solving quadratics by completing the square
- Worked example: Completing the square (intro)
- Completing the square (intro)
- Worked example: Rewriting expressions by completing the square
- Worked example: Rewriting & solving equations by completing the square
- Completing the square (intermediate)

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Completing the square

Some quadratic expressions can be factored as perfect squares. For example, x²+6x+9=(x+3)². However, even if an expression isn't a perfect square, we can turn it into one by adding a constant number. For example, x²+6x+5 isn't a perfect square, but if we add 4 we get (x+3)². This, in essence, is the method of *completing the square*. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

- That wasn't very clear, may you please do another video about that?(72 votes)
- You don´t need another video because I´m about to explain it to you! Say you have the equation 3x^2-6x+8=23. To complete the square, first, you want to get the constant (c) on one side of the equation, and the variable(s) on the other side. To do this, you will subtract 8 from both sides to get 3x^2-6x=15. Next, you want to get rid of the coefficient before x^2 (a) because it won´t always be a perfect square. Because there is a 3 in front of x^2, you will divide both sides by 3 to get x^2-2x=5. Next, you want to add a value to the variable side so that when you factor that side, you will have a perfect square. In this case, you will add 1 because it perfectly factors out into (x-1)^2. Because you´re taking this value away from the constant, you will add it to the other side of the equation (this might not make sense at first, but if the constant were on the variable side, you would be subtracting). This will all give you the equation (x-1)^2=16. Next, you want to take the square root from both sides so that x-1 is equal to the positive or negative square root of 16 (positive or negative 4). Finally, you add 1 to both sides, taking into account that 4 could be positive OR negative. Therefore, x = -3 or 5. Situations could vary, but this is the basic idea behind the procedure. I hope this helps! :)(276 votes)

- Can someone please post the link to the "Last video" of which Sal speaks?(23 votes)
- The link to
*"Last Video"*that Sal mentions at0:31, is here:

https://www.khanacademy.org/math/algebra/polynomial-factorization/factoring-quadratics-perfect-squares/v/perfect-square-factorization-intro

Hope this helps!(48 votes)

- Between2:04to2:10, May i understand why a=-2. i thought it would be 2 instead since were equating -4x to -2ax.(16 votes)
- I think Sal made a mistake. But 2^2 and (-2)^2 are both 4, so the result was correct.(5 votes)

- When would this be more advantageous than using the quadratic formula?(10 votes)
- After studying I have found that completing the square is useful for finding vertex form whereas quadratic formula is useful for finding roots.(8 votes)

- honestly even if this is useful in real life I hope I don't encounter a circumstance where I need this math(13 votes)
- At1:37, I am confused about where the (x-a)^2 came from and why that is equal to x^2-4x=5

Could you explain this?(5 votes)- I hope these following steps help you:

Now x^2 - 4x = 5

x^2 - 4x + (something) = 5 + something

I want to factorise the left side of the equal sign, so I have to find a value for (something) which would allow me to factorise the left-hand side of the equation.

If (something)=4

x^2 - 4x + 4 = 5 + 4

Notice that now I could factorise the left hand side into (x - 2)^2

(x-2)^2 = 9

x-2 = root of 9 = + or - 3

x = +3+2 or -3+2

x = +5 or -1.

(x-a)^2 is basically the format of the answer you want receive.

You want x^2 - 4x + (something) to be equal to (x-a)^2.

where a is half the coefficient of x(the number before the x), as long a there is no coefficient of x^2. Here a = 2.

A shorter way to do this is:

x^2 - 4x = 5

(x - (4/2))^2 - (4/2)^2 = 5

(x-2)^2 - 4 = 5 and so on. But remember, you only halve the coefficient of x and put it into the brackets only if there is no number before x^2 (coefficient of x^2). If it is there then you have to divide the whole equation first by the coefficient and then halve the coefficient of x and put it into the brackets.

Sal uses (x - a)^2 simply to tell this is the format of the factorised form of :

x^2 -4x +? = (x - a)^2, where a = 2 (4 divided by two).

Sorry for the large number of words used to answer your question.(13 votes)

- Why did he say that we needed a a number times 2 to equal -3 (it was around8:30)?(7 votes)
- It was probably a mistake.(I might be wrong)(1 vote)

- 1:51

isn't a perfect square trinomial

(ax-b)^2 = (ax)^2+2abx+b^2?

I get a bit confused as to why, when using the completing the square to derive the quadratic formula we only divide by 2, whit out also dividing by a.

but we do, right? we do it at the beginning.

ax^2+bx+c = 0

1/a*(ax^2+bx+c) = (0)1/a

x^2+bx/a + (b/2a)^2 = -x/a - (bx/2a)^2

In that step, we divide the second term by 2a to isolate b, and raise it to a second power.

I'm just trying to confirm things, not sure if I’m wrong(4 votes)- Just a quick correction.

(ax-b)^2 = (ax)^2 - 2abx + b^2

(ax+b)^2 = (ax)^2 + 2abx + b^2

And I'm not sure where x^2+bx/a + (b/2a)^2 = -x/a - (bx/2a)^2 came from. Step by step though:

Start with ax^2 + bx + c = 0

Factor out a

a(x^2 + (b/a)x + c/a) = 0

Now we complete the square using the term (b/a)/2 or b/(2a), adding and subtracting it to the one side so we don't change the value. Or we could add it to both sides, but then you would have to take into account the factored out a. Let me know if that doesn't make sense.

a(x^2 + (b/a)x + (b/(2a))^2 - (b/(2a))^2 + c/a)

x^2 + (b/a)x + (b/(2a))^2 = (x + b/(2a))^2 by being a perfect square trinomial [a = 1, b = b/(2a),hope that's not too confusing.]

a((x + b/2a)^2 - (b/(2a))^2 + c/a) = 0

And this will get you into vertex form.

From here you can derive the quadratic formula by solving for x. If you'd like me to walk through that let me know.(4 votes)

- Around6:40, Sal divides the quadratic equation by 5. This process makes the coefficient of x^2 equal to 1. My question is does the coefficient of x^2 need to be 1 to complete the square.(5 votes)
- no it doesn't have to.

for example; 2x^2+18x+16

one can factor this by..

(x+8)(2x+2)

but if you divide everthing by 2,

you can make 2x^2+18x+16 to x^2+9x+8 then you can factor this to

(x+8)(x+1)

you see, this is the same as (x+8)(2x+1) but simpler.

so to answer your question; it doesn't matter, but is's the matter of which one is simpler

hope this helps :)(3 votes)

- Why is
**bx**from**ax^2+bx+c=0**omitted in final answer, by this I mean, how is**ax^2+bx+c***factorised*into a binomial when*completing the square*?

I'm trying to figure out my own equation`x^2+4x+1`

x^2+4x=-1

then add *(b/2)^2* to both sides

x^2+4x+**2^2**=-1+**2^2**

So, then, how is it factorised from here to become

(x+2)^2=3

Please explain to me like I'm five:)(3 votes)- When you factor anything, you don't see the original value in the factors. For example, factors of 144 = 12^2. Where are the 4's? You only see them after multiplying.

The same is true with x^2+4x+4. In factored form it is (x+2)^2. You can confirm these factors are correct by multiplying the 2 binomials.

(x+2)(x+2) = x^2+2x+2x+4

Combine the 2 middle terms and you get x^2+4x+4

Hope this helps.(5 votes)

## Video transcript

In this video, I'm going to show
you a technique called completing the square. And what's neat about this is
that this will work for any quadratic equation, and it's
actually the basis for the quadratic formula. And in the next video or the
video after that I'll prove the quadratic formula using
completing the square. But before we do that, we
need to understand even what it's all about. And it really just builds off
of what we did in the last video, where we solved
quadratics using perfect squares. So let's say I have the
quadratic equation x squared minus 4x is equal to 5. And I put this big space
here for a reason. In the last video, we saw
that these can be pretty straightforward to solve if
the left-hand side is a perfect square. You see, completing the square
is all about making the quadratic equation into a
perfect square, engineering it, adding and subtracting from
both sides so it becomes a perfect square. So how can we do that? Well, in order for this
left-hand side to be a perfect square, there has to be
some number here. There has to be some number here
that if I have my number squared I get that number, and
then if I have two times my number I get negative 4. Remember that, and I
think it'll become clear with a few examples. I want x squared minus 4x plus
something to be equal to x minus a squared. We don't know what a
is just yet, but we know a couple of things. When I square things-- so this
is going to be x squared minus 2a plus a squared. So if you look at this pattern
right here, that has to be-- sorry, x squared minus 2ax--
this right here has to be 2ax. And this right here would
have to be a squared. So this number, a is going to
be half of negative 4, a has to be negative 2, right? Because 2 times a is going
to be negative 4. a is negative 2, and if a is
negative 2, what is a squared? Well, then a squared is going
to be positive 4. And this might look all
complicated to you right now, but I'm showing you
the rationale. You literally just look at this
coefficient right here, and you say, OK, well what's
half of that coefficient? Well, half of that coefficient
is negative 2. So we could say a is equal to
negative 2-- same idea there-- and then you square it. You square a, you
get positive 4. So we add positive 4 here. Add a 4. Now, from the very first
equation we ever did, you should know that you can never
do something to just one side of the equation. You can't add 4 to just one
side of the equation. If x squared minus 4x was equal
to 5, then when I add 4 it's not going to be
equal to 5 anymore. It's going to be equal
to 5 plus 4. We added 4 on the left-hand side
because we wanted this to be a perfect square. But if you add something to the
left-hand side, you've got to add it to the right-hand
side. And now, we've gotten ourselves
to a problem that's just like the problems we
did in the last video. What is this left-hand side? Let me rewrite the
whole thing. We have x squared minus 4x
plus 4 is equal to 9 now. All we did is add 4 to both
sides of the equation. But we added 4 on purpose so
that this left-hand side becomes a perfect square. Now what is this? What number when I multiply it
by itself is equal to 4 and when I add it to itself I'm
equal to negative 2? Well, we already answered
that question. It's negative 2. So we get x minus 2 times
x minus 2 is equal to 9. Or we could have skipped this
step and written x minus 2 squared is equal to 9. And then you take the square
root of both sides, you get x minus 2 is equal to
plus or minus 3. Add 2 to both sides, you get x
is equal to 2 plus or minus 3. That tells us that x could be
equal to 2 plus 3, which is 5. Or x could be equal to 2 minus
3, which is negative 1. And we are done. Now I want to be very clear. You could have done this without
completing the square. We could've started off
with x squared minus 4x is equal to 5. We could have subtracted 5 from
both sides and gotten x squared minus 4x minus
5 is equal to 0. And you could say, hey, if I
have a negative 5 times a positive 1, then their product
is negative 5 and their sum is negative 4. So I could say this is x
minus 5 times x plus 1 is equal to 0. And then we would say that x is
equal to 5 or x is equal to negative 1. And in this case, this actually
probably would have been a faster way to
do the problem. But the neat thing about the
completing the square is it will always work. It'll always work no matter what
the coefficients are or no matter how crazy
the problem is. And let me prove it to you. Let's do one that traditionally
would have been a pretty painful problem if
we just tried to do it by factoring, especially if we
did it using grouping or something like that. Let's say we had 10x squared
minus 30x minus 8 is equal to 0. Now, right from the get-go, you
could say, hey look, we could maybe divide
both sides by 2. That does simplify
a little bit. Let's divide both sides by 2. So if you divide everything
by 2, what do you get? We get 5x squared minus 15x
minus 4 is equal to 0. But once again, now we have this
crazy 5 in front of this coefficent and we would have to
solve it by grouping which is a reasonably painful
process. But we can now go straight to
completing the square, and to do that I'm now going to divide
by 5 to get a 1 leading coefficient here. And you're going to see why this
is different than what we've traditionally done. So if I divide this whole thing
by 5, I could have just divided by 10 from the get-go
but I wanted to go to this the step first just to show
you that this really didn't give us much. Let's divide everything by 5. So if you divide everything by
5, you get x squared minus 3x minus 4/5 is equal to 0. So, you might say, hey, why did
we ever do that factoring by grouping? If we can just always divide by
this leading coefficient, we can get rid of that. We can always turn this into a 1
or a negative 1 if we divide by the right number. But notice, by doing that we
got this crazy 4/5 here. So this is super hard to do
just using factoring. You'd have to say, what two
numbers when I take the product is equal to
negative 4/5? It's a fraction and when I take
their sum, is equal to negative 3? This is a hard problem
with factoring. This is hard using factoring. So, the best thing to do is to
use completing the square. So let's think a little bit
about how we can turn this into a perfect square. What I like to do-- and you'll
see this done some ways and I'll show you both ways because
you'll see teachers do it both ways-- I like to get
the 4/5 on the other side. So let's add 4/5 to both
sides of this equation. You don't have to do it this
way, but I like to get the 4/5 out of the way. And then what do we get
if we add 4/5 to both sides of this equation? The left-hand hand side of the
equation just becomes x squared minus 3x,
no 4/5 there. I'm going to leave a little
bit of space. And that's going to
be equal to 4/5. Now, just like the last problem,
we want to turn this left-hand side into the perfect
square of a binomial. How do we do that? Well, we say, well, what number
times 2 is equal to negative 3? So some number times
2 is negative 3. Or we essentially just take
negative 3 and divide it by 2, which is negative 3/2. And then we square
negative 3/2. So in the example, we'll
say a is negative 3/2. And if we square negative
3/2, what do we get? We get positive 9/4. I just took half of this
coefficient, squared it, got positive 9/4. The whole purpose of doing that
is to turn this left-hand side into a perfect square. Now, anything you do to one side
of the equation, you've got to do to the other side. So we added a 9/4 here, let's
add a 9/4 over there. And what does our
equation become? We get x squared minus 3x plus
9/4 is equal to-- let's see if we can get a common
denominator. So, 4/5 is the same
thing as 16/20. Just multiply the numerator
and denominator by 4. Plus over 20. 9/4 is the same thing
if you multiply the numerator by 5 as 45/20. And so what is 16 plus 45? You see, this is kind of getting
kind of hairy, but that's the fun, I guess, of completing the square sometimes. 16 plus 45. See that's 55, 61. So this is equal to 61/20. So let me just rewrite it. x squared minus 3x plus
9/4 is equal to 61/20. Crazy number. Now this, at least on
the left hand side, is a perfect square. This is the same thing as
x minus 3/2 squared. And it was by design. Negative 3/2 times negative
3/2 is positive 9/4. Negative 3/2 plus negative 3/2
is equal to negative 3. So this squared is
equal to 61/20. We can take the square root of
both sides and we get x minus 3/2 is equal to the positive
or the negative square root of 61/20. And now, we can add 3/2 to both
sides of this equation and you get x is equal to
positive 3/2 plus or minus the square root of 61/20. And this is a crazy number and
it's hopefully obvious you would not have been able to-- at
least I would not have been able to-- get to this number
just by factoring. And if you want their actual
values, you can get your calculator out. And then let me clear
all of this. And 3/2-- let's do the plus
version first. So we want to do 3 divided by 2 plus the
second square root. We want to pick that little
yellow square root. So the square root of 61 divided
by 20, which is 3.24. This crazy 3.2464, I'll
just write 3.246. So this is approximately equal
to 3.246, and that was just the positive version. Let's do the subtraction
version. So we can actually put our
entry-- if you do second and then entry, that we want that
little yellow entry, that's why I pressed the
second button. So I press enter, it puts in
what we just put, we can just change the positive or the
addition to a subtraction and you get negative 0.246. So you get negative 0.246. And you can actually verify
that these satisfy our original equation. Our original equation
was up here. Let me just verify
for one of them. So the second answer on your
graphing calculator is the last answer you use. So if you use a variable answer,
that's this number right here. So if I have my answer squared--
I'm using answer represents negative 0.24. Answer squared minus 3 times
answer minus 4/5-- 4 divided by 5-- it equals--. And this just a little
bit of explanation. This doesn't store the entire
number, it goes up to some level of precision. It stores some number
of digits. So when it calculated it using
this stored number right here, it got 1 times 10 to
the negative 14. So that is 0.0000. So that's 13 zeroes
and then a 1. A decimal, then 13
zeroes and a 1. So this is pretty much 0. Or actually, if you got the
exact answer right here, if you went through an infinite
level of precision here, or maybe if you kept it in this
radical form, you would get that it is indeed equal to 0. So hopefully you found that
helpful, this whole notion of completing the square. Now we're going to extend it
to the actual quadratic formula that we can use, we
can essentially just plug things into to solve any
quadratic equation.