www.mathcentre.ac.uk/.../Completing%20the%20Square%20-%20By%20Inspection.mp4

Going to have a look at a very
simple process. It's called

completing the square.

In order to get to it and to
show its potential use, I want

to start with a simple equation.
X squared equals 9.

In order to find out what taxes
we would take the square root of

both sides, so the square root
of X squared is just X and the

square root of 9 is plus three
or minus three.

Be'cause minus three all squared
is also 9.

So that was relatively
straightforward. Both of these

two numbers were square numbers,

complete squares. What if we

got? X squared is equal to

5. Do the same again, so
we take the square root of

each side X equals. Now 5
is not a square number, it

does not have an exact
square root, but

nevertheless we can write
it as root 5 that is exact

or minus Route 5.

So far so good. The same process
is working each time.

Let's have a look at something
now, like X minus 7.

All squared equals 3.

How can we solve this? Well
again, this side of the

equation. We've got something
called a complete square, so we

can take the square root of each
side. X minus Seven is equal to

the square root of 3 or minus
the square root of 3, and so we

can now add 7 to each of these.
So X is equal to 7 Plus Route

3 or 7.

Minus Route 3.

I just have a look at another
One X plus three, all squared is

equal to 5.

Again, this is a complete
square, so again we can take the

square root X +3 is equal to
Route 5 or minus Route 5. Now

to get X on its own, we
need to take three away from

each side, so we have X equals
minus 3 + 5 or minus 3

minus Route 5.

What if we got
X squared plus six

X equals 4?

Problem is this X squared plus
6X is not a complete square, so

we can't just take the square

root. So in terms of handling
something like this, we've got

to have a way of getting a

complete square. So the
process that we're going to

be looking at it's called
completing the square.

The sorts of expressions that we
have before will like this X

plus a all squared.

Or X minus a all squared, so
that's what a complete square

looks like. One of these two.

So let's multiply this out and
see what we get. So this is X

plus a times by X plus A.

And we do X times by X. That
gives us X squared.

And we do a Times by X, which
gives us a X.

And then with X times by a,
which again gives us a X and

then at the end a Times by a,
which gives us a squared. And so

we've X squared plus 2X plus A

squared. I can do the same
with this One X minus a Times by

X minus A and it's going to give
very similar results X times by

X will give me X squared X times
by minus A minus 8X minus 8

times by X minus 8X minus 8
times by minus A plus a squared.

So tidying up the two middle

bits. Minus X minus X, minus
2X and then plus A squared.

So this is what complete
squares look like. They look

like one of these two.

Well. Can I make this
look like a complete square in

some way shape or form? If I
compare this with this, what is

it that I see?

Well, perhaps one of things I
might like to have is this

written as just a function X
squared plus 6X minus four? And

let's not worry too much about
solving an equation. What we

want to concentrate on is this
process of completing the

square, so I'm going to take
this quadratic function X

squared plus 6X minus four, and
I'm going to compare it with the

complete square. X squared
plus 2X plus

A squared. Now The
X squared so the same.

6X2A X I've got to have these
two terms the same. They've got

to match. They've got to be
exactly the same, and that means

that the six has to be equal

to 2A. And of course,
that tells us that the A

is equal to 3.

So if I make a equal to
three, then I've got plus A

squared on the end +9.

So. I can look at this first bit
and I can make it equal to that.

So let's write this down
X squared plus 6X minus

4 equals. X plus
three all squared. Remember this

is X plus three all
squared. Now I'm replacing the

A by three.

Now what more of I got? Well,
I've added on a squared, so I've

added on 9. So I've got some how
to get rid of that. Well, let's

just take it away, minus 3

squared. And then I can keep
this minus four at the end as it

was. So now I've X plus
three all squared minus 9 -

4 gives me X plus three
all squared minus 30.

So I have completed the square.
I made this bit.

Part of a complete square.

And I've done it by comparing
the coefficient of X.

With one of the two standard
forms and I saw that what I had

to do was take half the
coefficient of X.

Let's have a look then at
another example, X squared minus

8X plus Seven and I want
to write this so it's got

a complete square in it.

Well, one of the standard
forms for the complete square

that we had was X squared
minus 2X plus A squared.

And I want to make these two

terms. The same.

So again, we can see that the A.

Has got to be 4 because minus 8
is minus 2 times by 4.

So we've got X squared
minus 8X plus Seven is

equal to. X minus
four all squared.

So I've ensured I've got the X
squared. I've ensured that I've

got the minus 8X, but I've also
added on a squared, so I've got

a squared too much, so I must
take away 4 squared and then

I've got the 7:00 that I need to
add on to keep the equal sign.

And so this is now
X minus four all

squared minus 16 +
7 X minus four all

squared minus 9.

Let's take one more example.

X

squared Plus 5X
plus three and let's see if we

can follow this one through
without having to write down the

comparison. In other words, by
doing it by inspection.

What do we need? We need a
complete square, so we need X

and we look at this number here.
The coefficient of the X turn on

this left hand side.

We want half that coefficient,
so we want five over 2 and we've

got a plus sign, so it's got to
be X +5 over 2.

If we were to multiply out this
bracket, we would be adding on

an additional A squared where
five over 2 is the a. So we've

got to take that away. Takeaway
5 over 2 squared and then at the

end we've got plus 3.

So this gives us
X +5 over 2

or squared minus 25
over 4 + 3.

And of course, we'd like to
combine these numbers at this

end X +5 over 2 all squared
minus 25 over 4 plus. Now we

need to convert these to
quarters as well, so three is 12

quarters. Now we can combine
these, since they're both in

terms of quarters X +5 over
2, all squared minus 13 over

4. And we'd leave you like that.

This process now seems to be
working quite well.

But of course, we haven't
dealt with every kind of

quadratic expression we could
have, because so far we've

only had a unit coefficient
here in front of the X

squared. We haven't had
another number like two or

three or whatever, so let's
have a look at what we would

do in that case.

So we have three X
squared.

Minus 9X. Plus
50, what do we need to do

to begin with? Well, we know how
to do this if we've got a

unit coefficient with the X
squared, so let's make it a unit

coefficient by taking out the
three as a common factor. So

that's three brackets X squared,
minus 3X. Now 50. What are we

going to do with this? Well?

We divide it by three in order
that when we do the

multiplication 3 * 50 over three
will just give us back the 50.

Now we look at this thing here
in the bracket because this is

now exactly the same sort of
expression that we've had

before. Equals 3.

Let's have a big Curly
bracket. I'm going to make

this going to complete the
square around this, so this is

going to be X minus. Look at
the coefficient of X and take

half of it 3 over 2 all
squared.

By doing that, we've added on.

An additional A squared, so we
need to take that off 3

over 2 squared and then finally
plus 50 over 3 and close

the big bracket.

3. X minus
three over 2 all squared minus

nine over 4 + 50 over
3 and closed the big bracket.

Now all we need to do now is put

these together. And to do that
we need a common denominator and

the common denominator. Four and
three is going to be 12.

3 the Big Curly Bracket X minus
three over 2 all squared minus

over 12. We need to change the
nine over 4 into 12.

3/4 gave us 12 so 3
nines give us 27.

Plus we need to change the 50
over 3 into twelfths.

4 * 3 is 12, so 4
* 50 is 200.

And now I've got a little bit of
arithmetic to do. Let's just

write back bracket down again.

Equals 3. The Curly

bracket. X minus three
over 2 all squared.

Minus 27 over 12.

Plus 200 over 12 and that's
the calculation that we need to

do here to simplify 3 Curly
bracket X minus three over 2

or squared. Los all over 12
and we're going to take the 27

away from the 200 is going to
give us 173 and then we can

close the bracket off.

So despite the fact that the
numbers were quite fearsome

there, we've still ended up with
a complete square, and we've

automated the process so that
what we're doing is looking at

the coefficient of X. First of
all, we check that what we've

got the coefficient of X squared
is one. If it's not, we take out

the coefficient of X squared as

a factor. Next we check the
coefficient of X and we take a

half of it, and that's the
number that's going to go here

inside the bracket.

Then we must remember that we've
got take off the square of that

number is effectively we've
added it back on, and then the

rest is just arithmetic.

So now we've developed this
technique of completing the

square. Let's use it to solve
our original problem. If you

remember we had X squared plus
6X is equal to four and we chose

to write that as X squared plus
6X minus 4 equals 0.

So first we need to check has it
got a unit coefficient.

And it has, so we don't need to
take out a common factor. Now we

look at the coefficient of X and
it's 6 and it's half that

coefficient that we want. So we
need 3. So this is going to be

X. Plus three all
squared.

In doing that, we have added on
3 squared, so we need to take

off that 3 squared.

And now we need to include the

minus 4. So that we can
maintain the quality of

this xpression with this
one and it's equal to 0.

So X plus three all squared.

Minus 3 squared. That's minus
nine and minus 4 equals 0,

so we can combine these X
plus three all squared minus 13

equals not. At the 13 to each
side X plus three, all squared

equals 13, and now we're in a
position to take the square root

of both sides. Because here on
the left hand side we have a

complete square. And so this is
X +3 equals. Now 13 isn't a

complete square. It's not a
square number, so we have to

write it as square root of 13.
Or remembering when we take a

square root of a number it's
plus or minus Route 30.

Now we take the three away from
each side and we end up with our

two roots. So we take the three
away we have minus 3 + 13.

Or minus 3 minus Route 30
and so that process of

completing the square can be
used to help us solve a

quadratic equation. But
that's not the real issue

here. You can see another
video on solving quadratic

equations. The point is to
master this technique of

completing the square.