www.mathcentre.ac.uk/.../Completing%20the%20Square_maxima_minima.mp4

Completing the square is a
process that we make use of in a

number of ways. First, we can
make use of it to find maximum

and minimum values of quadratic
functions, second we can make

use of it to simplify or change
algebraic expressions in order

to be able to calculate the
value that they have. Third, we

can use it for solving quadratic
equations. In this particular

video, we're going to have a
look at it for finding max- and

min-imum values of functions,
quadratic functions.

Let's begin by looking at a very specific
example.

Supposing we've got x squared,

plus 5x,

minus 2. Now.

x squared, it's positive, so one
of the things that we do know is

that if we were to sketch the
graph of this function.

It would look something perhaps

like that. Question is where's
this point down here?

Where's the minimum value of
this function? What value of x

does it have? Does it actually
come below the x-axis as I've

drawn it, or does it come up
here somewhere? At what value of X

does that minimum value occur?
We could use calculus if we knew

calculus, but sometimes we don't
know calculus. We might not have

reached it yet.

At other occasions it might be
rather like using a sledgehammer

to crack or not, so let's have a
look at how we can deal with

this kind of function.

What we're going to do is
a process known as

"completing ... the ... square"

OK, "completing the square",
what does that mean?

Well, let's have a look at something
that is a "complete square".

That is, an exact square.

So that's a complete and exact
square. If we multiply out the

brackets, x plus a times by x
plus a, what we end up with

is x squared...

that's x times by x...

a times by x, and of course
x times by a, so that gives

us 2ax, and then finally a
times by a...

and that gives us a squared.

So this expression is
a complete square, a complete

and exact square. Because it's "x
plus a" all squared.

Similarly, we can have "x minus a"
all squared.

And if we
multiply out, these brackets

we will end up with the same
result, except, we will have

minus 2ax plus a squared. And
again this is a complete

square an exact square
because it's equal to x minus a...

all squared.

So,... we go back to this.

Expression here x squared, plus
5x, minus two and what we're

going to do is complete the
square. In other words we're going to

try and make it look like this.
We're going to try and complete it.

Make it up so it's a full
square. In order to do that,

what we're going to do is
compare that expression directly

with that one.

And we've chosen this expression
here because that's a plus sign

plus 5x, and that's a plus sign
there plus 2ax.

So.

x squared, plus
5x, minus 2.

And we have x squared
plus 2ax plus a squared

These two match up

Somehow we've got
to match these two up.

Well,... the x's are the same.

So the 5 and the 2a have got
to be the same and that would

suggest to us that a has got to
be 5 / 2.

So that x squared plus
5x minus 2...

becomes x squared plus 5x...

now... plus a squared and

now we decided that 5 was
to be equal to 2a

and so a was equal
to 5 over 2.

So to complete the square, we've
got to add on 5 over 2...

and square it.

But that isn't equal to that.

It's equal, this is equal
to that, but not to that

well. Clearly we need to
put the minus two on.

But then it's still not equal,
because here we've added on

something extra 5 over 2 [squared]. So
we've got to take off that five

over 2 all squared. We've got to
take that away.

Now let's look at this bit.

This is an exact square. It's
that expression there.

No, this began life as x
plus a all squared, so this

bit has got to be the
same, x plus (5 over 2) all

squared. And now we can
play with this. We've got minus 2

minus 25 over 4. We
can combine that so we have

x plus (5 over 2) all squared...

minus...

Now we're taking away two, so
in terms of quarters, that's

8 quarters were taking
away, and we're taking away 25

quarters as well, so
altogether, that's 33

quarters that we're taking
away.

Now let's have a look at this
expression... x squared plus 5x minus 2.

Remember what we were
asking was "what's its minimum value?"

Its graph looked like that.
We were interested in...

"where's this point?"

"where is the lowest point?"

"what's the x-coordinate?"

"and what's the y-coordinate?"

Let's have a look at this
expression here.

This is a square.
A square is always

positive unless it's equal to 0,
so its lowest value

that this expression [can take] is 0.

So the lowest value of
the whole expression...

is that "minus 33 over 4".

So therefore we can say
that the minimum value...

of x squared, plus 5x, minus
2 equals... minus 33 over 4.

And we need to be able
to say when

"what's the x-value there?"

well, it occurs when this bracket
is at its lowest value.

When this bracket is at
0. In other words, when x equals...

minus 5 over 2.

So we found the minimum
value and exactly when it happens.

Let's take a second example. Our
quadratic function this time, f of x,

is x squared, minus 6
x, minus 12.

We've got a minus sign in here, so
let's line this up with the

complete square: x squared, minus
2ax, plus a squared.

The x squared terms are the same,
and we want these two to be the

same as well. That clearly means
that 2a has got to be the same

as 6, so a has got to be 3.

So f(x) is equal
to x squared, minus 6x,

plus...

a squared (which is 3 squared),
minus 12, and now we added on

3 squared. So we've got to take
the 3 squared away in order

to make it equal. To keep the
value of the original expression

that we started with.

We can now identify this as
being (x minus 3) all squared.

And these numbers at the end...

minus 12 minus 9, altogether
gives us minus 21.

Again, we can say does it have a
maximum value or a minimum value?

Well what we know that we began
with a positive x squared term,

so the shape of the graph
has got to be like that. So we

know that we're looking for a

minimum value. We know that that
minimum value will occur when

this bit is 0 because it's a
square, it's least value is

going to be 0, so therefore we
can say the minimum value.

of our quadratic function f of x

is minus 21, [occurring] when...

this bit is 0. In other words,
when x equals 3.

The two examples we've taken so far
have both had a positive x squared

and a unit coefficient
of x squared, in other words 1 x squared.

We'll now look at an example
where we've got a number here

in front of the x squared.

So the example
that will take.

f of x equals 2x squared,

minus 6x, plus one.

Our first step is to take out
that 2 as a factor.

2, brackets x squared,

minus 3x,...

we've got to take
the 2 out of this as well, so

that's a half.

And now we do the same as we've
done before with this bracket here.

We line this one up with x squared,
minus 2ax, plus a squared.

When making these two terms the
same 3 has to be the same as 2a,

and so 3 over 2 has to
be equal to a.

So our function f of x is going
to be equal to 2 times...

x squared, minus 3x,...

now we want plus a squared,
so that's plus (3 over 2) all squared

Plus the half that
was there originally and now

we've added on this, so we've
got to take it away,...

(3 over 2) all squared. And finally we
opened a bracket, so we must

close it at the end.

Equals... 2, bracket,... now this
is going to be our complete square

(x minus 3 over 2) all squared.

And then here we've got some
calculation to do. We've plus a half,

take away (3 over 2) squared,
so that's plus 1/2

take away 9 over 4.

The front bit is going to stay the same

And now we can juggle with
these fractions. At the end,

we've got plus 1/2 take away 9
quarters or 1/2 is 2 quarters,

so if we're taking
away, nine quarters must be

ultimately taking away 7 quarters.

So again, what's the
minimum value of this function?

It had a positive 2 in front
of the x squared, so again, it

looks like that. And again,
we're asking the question,

"what's this point down here?"
What's the lowest point and that

lowest point must occur when
this is 0.

So the min ...

value of f of x must be
equal to... now that's going to be 0

But we're still multiplying
by the 2, so it's 2 times

minus 7 over 4. That's minus
14 over 4, which reduces to

minus 7 over 2. When?

And that will happen when this
is zero. In other words, when x

equals 3 over 2.

So a minimum value of minus
7 over 2 when x equals 3 over 2.

Let's take one final
example and this time when the

coefficient of x squared is

actually negative.

So for this will take our quadratic function
to be f of x

equals... 3 plus

8x minus
2(x squared).

We operate in just the same way
as we did before. We take out

the factor that is multiplying
the x squared and on this

occasion it's minus 2.
The "- 2" comes out.

Times by x squared, we
take a minus 2 out of the 8x,

that leaves us minus 4x and the
minus 2 out of the 3 is a

factor which gives us minus
three over 2.

We line this one up with X
squared minus 2X plus A squared.

Those two are the same. We want
these two to be the same. 2A is

equal to four, so a has got to
be equal to two.

So our F of X is going
to be minus 2.

X squared minus 4X plus A
squared, so that's +2 squared

minus the original 3 over 2,
but we've added on a 2

squared, so we need to take it
away again to keep the balance

to keep the equality.

Minus two, this is now our
complete square, so that's X

minus two all squared. And here
we've got minus three over 2

- 4. Well, let's have it all
over too. So minus four is minus

8 over 2, so altogether we've
got minus 11 over 2.

And we can look at this. We
can see that when this is

zero, we've got. In this case
a maximum value, because this

is a negative X squared term.
So we know that we're looking

for a graph like that. So it's
this point that we're looking

for the maximum point, and so
therefore maximum value.

Solve F of X.

Will occur when this square term
is equal to 0 'cause the square

term can never be less than 0.

And so we have minus two times.

Minus 11. And altogether that
gives us plus 11.

And it will occur when this.

Is equal to 0. In other
words, when X equals 2.