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.