www.mathcentre.ac.uk/.../Solving%20quadratic%20equations.mp4

This video is about how to solve
quadratic equations. Let's begin

by saying exactly what a
quadratic equation is. It's an

equation that must contain a
term that has got X squared in

it, so we can have three X
squared minus five X squared, or

perhaps just X squared on its
own. The equation can also have

terms in. X so it might have
a term 5X or perhaps minus 7X

or perhaps North Point 5X. It
can also have constant terms

just numbers, so perhaps 6.

Minus Seven a half, etc.

It can't have any other terms in
it. It can't have any higher

powers of X in it, so it can't
have X cubed in it. Can't have

things like one over X in it.

So what does our most general
quadratic look like?

AX squared it must have this X
squared term in it and a is just

a constant. It can be one as it
is here, One X squared. It could

be 3. It could be minus five. It
could be any real number.

Plus BX, now this term
may not be there.

Be could be 0.

Plus, CC is the constant term
and again that term doesn't have

to be there. See could be 0.

And then we say equals Nord, so
that's our most general

quadratic. And what we're going
to have a look at is how to

solve all the different
varieties of this kind of

equation that you can meet.

There are four

basic ways. 1st
Way is by Factorizing.

I'm going to assume that you
know how to factorise, but there

is another section of the video
that is to do with Factorizing

quadratic expressions. So if
you're not sure how to factorise

perhaps it would be a good idea
to look at that.

2nd method is by

completing.
The

square. Now again,
I'm going to assume that you

know how to complete the square,
but again, that is another video

that relates to completing the
square, so if you're not sure

how to do that, perhaps you
might look at that video first.

Then we're going to have a look
at how you use the formula.

There's a special formula for
solving quadratic equations.

We're going to be using that.

And finally, how to solve a
quadratic equation using graphs?

Of these four methods, two of
them are by far the most common.

And Whilst one doesn't want to
say you mustn't understand

everything, you can pick a part
of what you do. You really do

need to concentrate on
factorising and the formula.

Those are the two most important

ones. So let's begin by having
a look at how to factorise and

solve a quadratic equation.

I'll begin with
three X squared

equals 27.

There is no extra, but that
still doesn't stop it being a

quadratic equation 'cause it's
got this X squared term in.

So we begin by writing it
equals 0, so I take the

27 of each side.

So we have three X squared
minus 27 equals 0.

That's the first step, and one
that you must do every time.

Next, look for a common factor.
Is there a common factor in the

terms on the left hand side of
this equation, and if there is,

let's take that common factor
out and here we have three X

squared and also 27 which has a
factor of three in it. 27 is 3

times by 9, so we can take out
the common factor.

All three and that will leave us
with X squared minus 9 equals 0.

Inside this bracket, now we have
the difference of two squares.

We have X squared takeaway 9
which is 3 squared and this is

a standard piece of Factorizing.
So that's three brackets, X

minus three X +3 equals 0.

Now when we multiply 2
expressions together and we're

not quite sure of their value,
but we do know what they give

us. In this case zero. We can
make certain deductions about

them, so if I have one number
multiplied by another number

and the answer is 0, then one
of these numbers.

That one or that one or both
of them are zero. So we have

X minus three equals 0.

All X +3 equals 0,
so we end up with

X equals 3 or X
equals minus three.

Notice 2 answers and that will
be a theme that I will

return to again.

Let's just recap there what we

did. First of all, we put
everything equal to 0. Then we

extracted, took out any common
factors. Why should you do that?

Well, let's just run through
this one. Again, 3X squared

equals 27 and let me break a few
of the rules that I've just

said. So let me cancel both
sides by three, since obviously

three goes into three X squared
and leaves us.

X squared and free goes into
27 and leaves us with nine, so

I've got X squared equals 9.
Now let me do the natural

thing which is to take the
square root of both sides.

Square root of X squared is X
and the square root of 9 is 3.

Only got 1 answer if we

look back. 2 answers.

I've lost one by factorizing. I
was able to get two factors and

that gave Me 2 answers.

In doing this step from here to
here and taking the square root,

what I have forgotten is that as
well as 3 squared be equal to

9 - 3 squared is also equal
to 9. So at this point I

should have said or X equals

minus three. Now it's very easy
to make this kind of mistake and

lose one of the answers. So the
best way to tackle these is the

way that I just set out.

Write them as equals. 0 take out
common factors and then look at

a factorization leading to
giving you 2 answers.

Let's take another example. This
one will take us five X squared

plus three X equals 0. It's
already written as equals 0, so

we don't have to worry about
that first step. So the second

step is to look for a common

factor. I've got five X
squared plus 3X, so I've got

a common factor here of X
so I can take that out,

leaving me with five X +3

equals 0. And here again I've
got 2 numbers.

X and 5X plus three. I'm not
sure what they are, but what I

do know is that in multiplying
them together, I've got 0.

This means one of them must be 0
or the other one must be 0, or

they're both 0, so I can write
down X equals 0 or 5X. Plus

three is also equal to 0, so
there's one answer, X equals 0,

and now this is just a simple
linear equation, linear.

Remember because it's just got
an X in it. Now X squared's ex

cubes, no one over X is.

Just an X and this solves quite
easily. 5X equals minus three,

taking three away from both
sides and then dividing both

sides by 5 - 3 over 5.

So there are my two solutions, X
equals 0 and X equals minus

three over 5. What I'd like to
do is just perhaps run through

this one again and show you
where the mistakes can come when

tackling equations like this.
'cause what's special about this

one is there's no constant term
an ex squared term, an ex term,

but no constant term.

So what might be the
mistake that we would

make when doing this?

Let me state that we might very
well make it say, Oh yes, a

common factor of X one there in
the X squared term and one

there, so I'll cancel by that.
X5 X +3 equals 0, dividing each

term by X and then 5X is equal
to minus three, and then X is

equal to minus three over 5.

One answer X equals minus 3/5.

If we go back with X equals 0
and X equals minus 3/5, we've

lost this answer. Here we've
lost X equals 0.

Why? Because we didn't look for
a common factor at this point.

We went straight on to simply
divide throughout by the X

instead of taking it out as a
common factor. So you must be

aware of that with quadratic
expressions, were looking for

two answers to solutions to
roots. Those are if you like the

different words that we use.
Answers, solutions, roots, they

all mean. The same thing.

OK, we looked at two particular
cases. One where we did not have

an exterm and another one where
we didn't have a constant term.

So now. Let's have a look at.

One that's got both
of those in.

X squared minus 5X plus 6

equals 0. We're going to solve
this by Factorizing.

So to factorize a quadratic we
need to search for two numbers

that when we multiply them
together will give us this

number here. 6.

An when we add them together
will give us this number here

minus 5. So we look at that
and 3 * 2 would give us 6 and

minus 3 N minus two added
together would give us minus

five, but minus three times
minus two is also six, so those

look like good choices, minus
three times by minus two gives

us 6 and minus three plus minus
two gives us minus 5.

So. We take this expression
again X squared.

Instead of minus 5X, we write
this as minus three X minus two

X +6 equals 0. Now we look
at these two terms and we take

out the common factor. In this
case that's X and that leaves us

with X Times X minus three.

Now we look at these two terms
and we take out again a common

factor. Well obviously there's
two, but this is minus 2X, so it

cost. This is minus two X I'm
going to take, minus two as the

factor minus two brackets X. And
now I need to put something in

here so that I have minus two
times gives me plus six, and

that must be minus three.

Equals 0. Now I've got
one lump of algebra there. An

one lump of algebra there, and
they share this common Factor X

minus three X minus three. So we
take that out as a common

factor. I'm we're left with
the other factor as X

minus two X minus two.
Close the bracket equals 0.

And so here I've arrived at 2
numbers multiplied together and

they give me 0. So one of them
must be 0 or the other one must

be 0 or they perhaps both 0.

So if we just write that down
again, X minus three times by X

minus 2 equals 0.

Then either X minus three
equals 0 or X minus 2 equals

0. This tells us that X must
be equal to three, or this

one tells us that X must be
equal to two.

And so again, we've arrived at
two possible answers.

Let's take another one. This
time, let's take something that

does not have a unit coefficient
in front of the X squared, so

will take 2 X squared.

Plus 3X minus 2

equals 0. Now, in order
to solve this one, we're looking

for two numbers that will
multiply together to give us 2

times by minus two SO2 numbers
that multiply together to give

us minus four where looking for
those same 2 numbers to add

together to give us 3.

Well. Four times by one
would give us four and if we

made the one A minus 1 four
times by minus one, that would

make minus four and four plus
minus one would give us 3.

So these are our two numbers
that we need, so again 2 X

squared and then the X term we
can rewrite us 4X.

Minus X.
Minus 2 equals

0. And we look at these
two terms for a common factor,

and there's a 2 here and four is
2 * 2 and there's an X and an

X&X in the X squared and X with
the 4X. So our common factor

there is 2X, leaving us with X

+2. And now we want to common
factor here and that will be

minus one. Times
X +2 equals

0. Again.
We have two lumps of algebra

here and here, and in each one
there's a factor of X +2, so

that's our common factor that we
can take out X +2 times by two

X. Minus one.

So again, we've arrived at two
brackets multiplied together X

+2 times by two X minus one, so
one of these has got to be 0,

or possibly both of them.

Let's write that down again.
X plus 2 * 2

X minus one equals 0.
So either X +2.

Is 0 or.

2X minus one is 0.

This tells us that X is minus 2.
By taking two away from both

sides. This tells us that 2X is
equal to 1. By adding one to

each side, and now we divide
both sides by two. So we have X

equals 1/2. So we ended up with
our two answers for this

quadratic equation. Let's take

another one.

4.
X squared

+9 equals

12 X.
It doesn't say equals 0.

So let's write it so it says

equals 0. Four X
squared minus 12

X +9 equals 0.

Check common factor. Something
we didn't do in the last

question, but then there wasn't
one. But these numbers are a bit

bigger and there may just be a
common factor there. But as we

look there is no common factor.

Factorise it, we're looking for
two numbers that would multiply

together to give us 36 four
times by 9. Ann would add

together to give us minus 12.

Well, six and six give us 36 and
if we make them both negative.

Minus 6 times by minus six they
still give us plus 36, but now

they add together to give us
minus 12. So we've got four X

squared minus six X minus six X
+9 equals 0.

In these two terms, we want the
common factor and that will be

2X because this is four and six
and X squared and X, so will

take the two X out, giving us 2X
minus three, and then here

again. Common factor 6X9.
There's a common factor. There

are three, but this is a minus
sign here, so we want minus

three times by two X. That gives
us the minus 6X and minus three

times by something has to give
us plus 9, so that will be minus

three again equals 0.

Two lumps of algebra. A common
factor in each one of two X

minus three. Leaving us with
two X minus 3 *

2 X and two X
minus three times minus three.

One of these two or both of
them can be 0.

So we have two X
minus 3 * 2 X

minus three equals 0 and
so 2X minus three equals

0 or two X minus
three equals 0.

Adding three to both sides 2X is
equal to three, and so X is

equal to three over 2.

Of course, this is exactly the
same, so we can write down the

same stuff. Two X equals 3 and X
equals 3 over 2.

2 answers again.

The same answer twice, because
I can say it like that. It

enables me to keep up this
idea that quadratics are going

to lead me to 2 answers and
answer may be repeated, but I

still get 2 answers.

Let's take one more example.

X squared minus three
X minus 2 equals

0. We're looking for two
numbers that will multiply

together to give us minus
two and will add together

to give us minus three.

I gotta multiply together to
give us minus two. Well, that

would suggest one and minus two,
but no matter how we try and add

one and minus two together,
we're not going to get minus

three. And if we said well,
let's try minus one and two and

again and not going to add
together to give us minus three.

In short, this question does

not factorize. And because
it doesn't factorize, we're

going to have to
look for another way

of solving it.

So how are we going to solve an
equation that does not factorize

well? One option is completing
the square. I want to have a

look at this, not because it's
the regular way in which we

would try to solve an equation
that doesn't factorize, but

be'cause. In doing it, we'll see
how the next method which is

using the formula actually
works. So let's have a look at

what we've got.

X squared Minus three X minus
2 equals 0. Now in order to

complete the square, we look at
these two terms. Now this is an

X squared and this is minus 3X.
So if I want to get a complete

square which remember is like

that. And I've got to have an X
in there and I've got to have a

constant term here so that when
I do the squaring I'll end up

with minus 3X.

And we know how to do that. You
take a half of the number that

multiplies the X, so that's
minus three over 2, so that when

we multiply out, that bracket
will get X squared, will get

minus 3X, but will also have an
extra term. We will have the

square of minus three over 2

extra. In other words, we'll
have added on minus three over 2

squared, so if we've added it
on, we better take it off in

order to make sure that we've
got exactly the same value, the

same expression, and then, of
course, I've got minus two here,

so let's put that in.

So now this expression is
exactly the same as that one, so

let's keep this bit at the front
X minus three over 2 four

squared, and now what's this?
Let's simplify this here. Well,

minus three over 2 squared is 9

over 4. Take away 2

equals 0. So we've
got X minus three

over 2 all squared.

Minus, now let's write
this all over 4.

Two over 4 is 8 quarters. I'm
taking away nine quarters and

I'm taking away eight quarters.
Sign takeaway. 17 quarters

altogether. Now this now looks
more manageable. Let me just

turn over the page and write
this down again.

X minus three over
2 all squared minus

17 over 4 equals

0. Now we're going to add the
17 over 4 to both sides, so will

have X minus three over 2. All
squared is equal to 17 over 4.

And now I'm going to take the
square root of both sides,

remembering that I must get plus
or minus this side.

So the square root of this will
be X minus three over 2 equals.

To take the square root of this
I need the square root of the

top, which is going to be plus
Route 17 over 2 or minus Route

17 over 2 over the two, because
that's the square root of the

four. Now I add this bit
to both sides to three over 2,

so we have X equals 3 over
2 plus Route 17 over 2 or.

X equals. The
square root of 17 over

2. And just for the sake
of completeness, I'm going to

put all of this over this too.
So 3 Plus Route 17 over 2 or.

3 minus Route 17

over 2. So that we've
done a question that involves

completing the square.

Again, we've got 2 answers, but
there are two further things to

notice here. I've left these
square roots in. You would

normally take a Calculator and
workout an approximate value.

Of these two answers to a given
degree of accuracy, that might

be to two decimal places, it
might be to three significant

figures, but these are exact
answers. The minute you put

decimals in those answers are
not exact, they are

approximations to a given degree

of accuracy. The third thing I
want you to notice is there form

3 plus the square root 3 minus
the square root and over

something because when we look
at the next method, you're going

to see that form again.

So how else might we tackle this
equation to remember what it was

initially? It was this One X
squared minus three X minus 2

equals 0. We saw that it didn't
factor eyes, so we did it by

this method of completing the

square. So what we're going to
do now is look at the

formula. To do this, I'm going
to take it generally to begin

with. So here is our general
quadratic equation AX squared

plus BX plus C is 0.

And there is a formula which
will solve this equation for us.

And it's X equals
minus B.

Plus or minus.

And it's the plus or minus
that's going to give us the

two answers. The square root
of B squared minus four AC

all over all over whole lot
over 2A.

That's the formula for solving
quadratic equations, and

unfortunately you do have to
learn it. But if you say it to

yourself each time you write it
down, X is minus B plus or minus

the square root of be squared
minus four AC all over 2A helps

to remember. It makes it sound
stupid, but it helps to remember

it. So let's take this formula
and use it to solve the previous

equation that we just had.

So the equation that we had
was X squared minus three X

minus 2 equals 0. Let's begin
by identifying AB and see

there's One X squared. So a
must be equal to 1.

B is the coefficient of X,
and that's minus three, so B

equals minus. 3C is the constant
term, so here C is equal

to minus two. Let's write down
the Formula X equals minus B

plus or minus square root of
be squared minus four AC.

All over 2A by writing down

each time. It's a way of helping
to remember it. Now let's

substitute these values in.

So B is minus three, so that's

minus. Minus

3. Plus or
minus the square root of

B squared, minus three all
squared minus four times a,

which is one.

Times C, which is minus two,
will just extend that square

root sign all over everything
over 2A, which is 2 times by

one. Now let's tidy this up.
So we have minus minus three.

That's plus 3.

Plus or minus the square root

of. Minus 3 all squared is

9. Four times by one is 4
times by minus two is minus 8

and we've got this minus sign
here, which makes +8.

All over 2.

And what we see we've got is 3.

Plus or minus the square root of
17, all over two, which are the

answers that we had before.
Again, we can workout each of

these answers. 3 Plus Route 17
over two or three minus Route 17

over 2 using a Calculator. But
again, remember any answers you

get doing it that way. Our
approximate. These answers are

exact. Let's take one more
example of using the formula.

And the one that will take
this time, three X squared

equals 5X minus one.

It doesn't say equals 0, so we
must write it as equals 0,

'cause That's the form that the
formula demands. We have the

equation written in, so three X
squared and we gotta take 5X

away from each side.

And we've got to add 1 to each
side. That ensures that we have

that. Now let's identify AB&CA
is the coefficient of X squared,

which in this case is 3.

B is the coefficient of X,
which in this case is minus

5. See is the constant
term, which in this case

is one. Again, let's write
down the equation. X is minus

B plus or minus the square

root of. B
squared minus for

AC all over

2A. And make
the substitution.

B is minus five we want minus
B, so this is minus minus 5.

Close all minus.

Square root of B squared,
so that's minus five all

squared, minus four times a,
which is 3 times C,

which is one, and that's
all over 2A, two times

a, which is 3.

Now simplify this minus minus
five is +5 plus or

minus the square root of
minus five. All squared is

25. 4 * 3 * 1 is
12, and the minus sign is minus

12. All over 6.

So we just squeeze that one into
this space. Here we 5 plus or

minus the square root of 25 -
12, which is 13 all over 6. And

again there are two exact
answers there. 5 Plus Route 13

over 6 and 5 minus Route 13 over
6 we can workout their value

approximate value that is by
using a Calculator.

And finding out what it tells us
about Route 30, but those are

two exact answers. If left in

that form. Now basically, if a
quadratic factorizes, then you

should solve it by using
factorization, 'cause it's

clearly much quicker than this.

Don't use completing the
square unless you

specifically asked to and
use this method by using the

formula. Solving by using
the formula for those that

do not factorize.

However, right at the very
beginning we did say there was a

fourth method. And so I just
want to look quickly at what

this 4th method is.

We said this 4th method
was by graphing.

OK. Let's have a look at
what we know about quadratic

functions. If X squared is
positive in our quadratic

function, then we're going to
get a U shaped curve like that.

If, on the other hand, X squared
is negative, then we're going to

get a Hill like that.

Now. Set on a graph.

Then we might have a graph.

That looks with a positive X
squared. Say something like that

or we might have a graph that
say looked like that we might

have a graph that looked like

that. Now this is the

line. Equals 0 the value
of the function equals

0, and so here we can
see our two values of X.

Here we can see one value of X
which makes the function equal

to 0. And we've said we call
that a repeated root or a double

route, but here the graph does
not reach the X axis, it does

not reach a value equal to 0.
And that means we actually have

no roots. So there are some
quadratic equations that you

cannot actually solve well.
Mathematicians make it happen,

so to speak, they make them
solvable by inventing a new kind

of number. But we're not going
to deal with that at the moment.

As far as we are concerned, we
either get 2 routes.

A repeated root.

Or for the moment.

No roots at all.

But you will see that we can
make an equation like this have

two routes in a very special
way. Clearly what we've done for

a positive X squared term we can
do for a negative X squared

term, because again we can have
a graph say that looks like that

and does not reach the X axis.
Or we can have one that's a

looks like that and just touches
the X axis. Or we can have

another one that looks like

that. And again two values there
and there a repeated value of X

there and no values of X there.

So bearing that in mind, let's
have a look at our.

Trial equation if you like the
one that we've used to start off

each of the last two sections, X
squared minus three X minus 2

equals 0, and let's think how we
can solve this by using a graph.

Let's just look at this bit.

The X squared minus three X
minus two and think of it as a

function, not as an equation
equals 0, but just a function.

Now one of the things is that if
we put in, say, a value of X

equals what. Let's say something
like one, then we've got one.

Minus 3 - 2 is clearly negative.
If I put in a value, let's say

something like 4, then I get 16
- 12 - 2. Well that's positive,

isn't it? 'cause it's an overall
value of two. If I put in a

value of, say something like
let's say minus two, then I get

4. Minus three times Y minus two
is 6, so I get 10 takeaway 28.

It's positive, so as I've gone
from one value of X, I've got a

positive value, then a negative
and a positive value of X. So

clearly it's going across the X
axis at some point. So what I

have to do is choose a value of
X or a range of values of X that

will enable me to graph that
different behavior. First of

all, positive values, negative

values. Positive values again.

So the range of values that I'm
going to take, I'm going to take

them going from minus 2.

In unit steps.

I'm going to go
up to five.

I'm going to calculate each
term in turn.

So let's just.

Rule those off out of the way.

X squared
that will be

4101, four,
916 and 25.

The next term is minus 3X.

So I have minus three times by
the value of X, so minus three

times by minus two is 6.

Minus three times
by minus one is 3.

Minus three times by zero is
0 - 3 times by by

one is minus three, and so
on, minus 6 - 9 -

12 - 15.

Now we come to the last term
minus two and that stays the

same because it's the constant
term, no matter what the value

of X is, so we can just write
in a line of minus Two's across

our table. Now we can workout X
squared minus three X minus two,

because it's going to be the sum
of each of these columns.

So we have six and four is 10
takeaway, two is 8.

One and three is 4 takeaway, two
is 2 zero and zero and minus two

is minus 2 - 3 and minus two
is minus five and one.

Minus 4 - 6 and minus two
add 4. That's minus eight and

four gives us minus 4.

9 and minus nine is zero and
minus two gives us minus 2.

Minus 12 and minus two is minus
14 and 16 gives us +2.

Minus 15 and minus two is minus
17. Add on the 25 and that gives

us 8. Notice what we've got is
symmetric across the table. Now

that just happened to be because

I picked. Values that were going
to give me a complete symmetry,

but you always find a degree of
symmetry in the table.

Now what I would need to do now
is make a graph. I don't have

any graph paper and so
effectively what I'm going to do

is. I'm going to draw a quick
sketch because we know what the

answer to this equation now, but
I just want to show you how you

would go about doing this so

it's. Do that.

We got out X axis.

Our Y access you've got values

minus 2. Minus one.

12 3.
Four and five, and I mark these

often even numbers 2.

4. 6.
8 and minus two

and minus 4.

So. Let's just flick back for a
couple of values here.

Minus two is 8 and minus one
is 2, so I've got a point

roughly there and a point

roughly there. Then when it
was zero, I had one here

minus 2. And we go. And
then at one and two they were

both minus four, so they were
there and there.

Then at three and four, they
were minus two and two, so they

were there and there. And then
at five it was back up here at

8 again so I can get a nice
smooth curve through these

points. And then back up today.
And where are my answers while

there here, that one and that
one? Those are the values of X

for which is X squared minus
three X minus 2 equals 0.

Supposing, however, haven't been
given that equation, but I've

been given X squared minus three
X minus 2 equals 6. Now, this

shows the advantage of graphs I
took a long time to work this

out, and to draw this. And How
do I draw on it neatly? Well, it

would have taken even longer,
but because I have drawn it, I

can use all the data I've got,
not only to solve the first

equation, but to solve this
equation as well, because all

I'm looking for.

Our values of X which make the
function bit equal to six. So

where is the function bit equal
to 6? Where we go across here

and we can see these are the
points on the curve where it's

equal to six and so if I come
down, these are the values of X.

Which correspond and I can just
read off those values of X. So

by drawing the graph I've not
only been able to solve that

one, but I've been able to solve
this one as well.

What about if I have X squared
minus three X minus 5 equals 0?

Can I use this graph again?
Well, it's got the front bit the

same. It's this bit that seems
to be different, so if it's

different, well, let's make it

the same. So let's make it X
squared minus three X minus 2.

And then what do I need to keep
it as minus five? Will I need

minus three equals 0? So now
this is my function bit the

same, so I'm actually looking
for when my function bit is

equal to three. And again,
that's within the possibilities.

Here we have the value three, we
can go across.

There and then we can come down
and read off the values of X,

which gives us that. So for all
drawing a graph took along time

it actually enabled us to solve
more than just the equation for

which we drew it.

So let's just recap.

There's a basic quadratic
equation X squared plus BX plus

C equals 0. There are four ways
of solving it, first by

Factorizing and you should try
that way first.

Second, by completing the
square, but that's not advised,

so to speak. Third, if it won't
factorize, use the formula and

those two ways factorizing and
using the formula. Other prime

ways. The first methods of
recourse that you should use.

Finally, just drawing a graph.
Time consuming but may be

beneficial if you've got a lot
of the same kind of equation

that you need to solve.