the quadratic formula.  If you recall a
quadratic equation
in standard form can be expressed in a
very general way as
aX squared plus bx plus c is equal to 0
Now if we were to take this quadratic
equation in standard form
and perform a completing the square
just using the variable terms
the result would be the quadratic formula.
And the quadratic formula is written below 
X equals negative b
plus or minus the square root of b squared minus 4ac
all of that over 2a
Well now, let's look at solving the same
quadratic equation in a variety of
different ways,  First we're going to use factoring
then we're going to use completing the
square, and  finally
we'll use the quadratic you know and you
know
as you'll see will get the same result each
time.  So first let's factor
This trinomial, if it's factorable, should factor out 
to two binomials.
And I believe X squared minus 2X minus 3
would factor and as
X minus 3 times X plus 1
because you know we're looking for two
numbers are multiplying give you -3
but those same two numbers must and
give you negative 2.
And using the 0 factor property, we know that
X minus 3 is equal to 0
or
X plus 1 is equal to 0
So that would give us that X equals 3
or X equals negative 1
Therefore our solution set
would be negative 1, 3.
so now let's solve the same quadratic
equation
using completing the square.  Well with
completing the square remember our goal is 
to create this perfect square trinomial.
We know that
3 needs to be on the other side
because it's not the last term that we
would need in order
to have a perfect square trinomial.  So
we'll add 3 to both sides 
and that will give us X squared minus 2X
equals three
Now, the next step to completing the square is to figure out 
figure
what term would complete the
square
And so in this case, remember 
we would want to take half of the middle coefficient.  The middle coefficient is negative 2.
So we'd like to take half of that middle
coefficient and square it.
Half of negative 2 is  negative 1.
And negative 1 squared is positive 1
So we'll add 1
to both sides
Now we'd like to take this perfect
square trinomial, and rewrite it
as a binomial square.  So what binomial could we square to get X squared minus 2X plus 1?
 
It would have to be X minus 1.  3 added to 1 is 4.
So we went through all this trouble just to get it in a form
that would enable us to
use the square root property.
Using the square
root property
we have X minus 1 equals
plus or minus
square root of 4
We know the square root of 4 is 2, 
So we get X minus 1 equals plus or minus 2.
Adding 1 to both sides we get
X equals 1
or
plus or minus 2
And now we can write this in the longer
form.  So we know that
either X equals 1 plus 2 or
X equals 1 minus 2.
well 1 plus 2 is 3.
and 1 minus 2 is negative 1
So as you see, we get the same result.
Now I know the
thing probably comes to mind immediately
is that the factoring was so much easier
why in the world would wanna use
completing the square?
Well we have to remember, every
quadratic expression is not factorable,
and so when that's the case, then
up until now we were forced to use
completing the square.
Completing the square works on any trinomial, 
whereas factoring only works on the ones that are factorable
Well finally, let's use the quadratic
formula
So when we look at the quadratic formula
the first thing we need to make sure of, as
with the other methods, is that
this is standard form.  And it is.
And we need to identify A B and C.
so A in this case would be 1.
B is negative 2, and C is negative 3.
Once you identify A, B and C, it's a matter
of just substituting
into the formal.
So we start out with a negative B. so
that would be the negative of negative 2,
plus or minus
Square root of B squared
That would give us negative 2 squared
minus 4 times A, A is one, times C
which is negative 3.
All this
is over to 2 times A.
Well, when we simplify that, we get negative of negative 2 will give us 2, 
plus or minus the
square root of
let's see, 2 squared is 4
and then we have negative 4 times 1  times negative 3, 
well, that would give us positive 12.
all over 2
so then we would get X equals 2, plus
or minus the square root of 16
but the square root of 16 is 4
over 2
Well again, we can write this in the
longer form without using plus/minus
so we know that either X is equal to
2 plus 4 
divided by 2
or
X equals 2 minus 4
divided by 2
Well, 2 plus 4 is 6, and 6 divided by 2 would 
give us 3
And 2 minus 4 is negative 2, and negative 2 divided by 
positive 2 is negative 1.  So you can see we
get the same solutions, negative 1 and 3.
Let's solve another equation.  This time
we'll just use the quadratic formula.
So we X squared minus 4X equals 1.
The very first thing we need to do is
to make sure that
our equations is in standard form.  You know standard forms just means
all the terms on one side and 0 on the
other.  So we would add 1 to both sides
and that would give us X squared minus  4X
plus 1
equals zero.
And now that we have all our terms on the
same side, we can identify
AB and C.  So in this case A would be 1,
B is negative 4
and C is 1.
Next we substitute
those
values into the formal so we would have
X equals negative B
so negative of negative 4
plus or minus
the square root of B squared
so that would be negative 4 squared
minus 4 times A, A is 1.
times C
and C is also 1.
And this is all divided by 2 times A
So that gives us X equals 4
plus or minus the square root of, let's see we would have negative 4 squared which is
16
and 4 times 1 times 1 is 4, so 16
subtract 4
would give us 12.
all of that divided by 2.
Now, the square root of 12 can be simplified.  We know that square root of 12 
can be rewritten as the square root of 4
times the square root of 3.
And the square root of 4 is 2.  So we would have
2 square root of 3.
So we could rewrite square root of 12 as 2 square root of 3.
And this is all dived by 2.  Finally we
know 2 is the denominator for both
so we have 4 divided by 2
plus 2 square root 3
divided by 2
and this would give us X equals 2
 
plus or minus the square root of 3
and that should be our solution set, 2 plus or minus
the square root of 3
Well that's our proposed solution
so next we would go ahead and check this.
And you can see that if we would have tried to factor on this one that it wouldn't have been
possible because we would have been
looking for two members that multiplied
and gave us 1, but added and
gave us negative 4, and so there are no two numbers that do that
So again,  with the quadratic formula, as well as completing the square,
you can use those on any quadratic
equations
Well let's use the positive version
So we'll have
2 plus the square root of 3
squared
minus 4
times 2 plus the square root of 3.
Well, let's go ahead and square 2 plus the square root of 3.
we know that would just mean to multiply it
times itself
so using FOIL we would have 2 times 2 is 4, that's first,
Outer would be 2 times square root of 3 which would be 2
square root of 3.  Inner would be square root of 3 times 2
which would give us 2 square root of 3 again.  And root 3 times root 3 would be
square root of 9, but we know the square
root of 9 is 3.
Finally, that gives us 7
plus 4 square roots of 3.
so we know that 2 plus the square root of 3  
squared
is 7 plus 4 square root 3.
then we would distribute negative 4
on 2 and root 3
so that would give us negative 8
minus 4 root 3
Well,
we can see here
that
negative 4 root 3 and positive 4 root 3 
would cancel out
and that would leave us with 7
and negative 8.  And when we combine those two
we get negative 1.  So you can see we have
negative one on the left
the right was negative 1.  So everything checks out.
Let's look at one more example.  Again we're going to solve this using the
quadratic equation.
Again, we's like to get our terms in
standard form so
we'll and 3x and 2 to both sides, and that will give u
6X squared plus 3X
plus 2
equals 0
and now that we have our equation in
standard form we can identify
AB and C.   So A is 6, B is 3
and C would be 2
So next we substitute into the formula
we would have X equals negative B
so negative 3 plus or minus the square root of
B squared, 3 squared
minus 4 times A
times C
All of this
over 2 times A
2  times 6
This will would result in negative 3
plus or minus
the square root of 9 minus
Let's see, 48
all over 12.
And so from there we would have negative 3
plus or minus
square root of negative 39
all over 12
and we know that the square root
of negative A is
I times the square root of A.  
So we would have X equals negative 3
plus or minus I times the square root of 39
divided by 12.
Now, when we look at square root of 39, square root 39
is not a perfect square
and it doesn't have any perfect square
factors.  The only factors
for 39 would be 1 and 39 or 3 and 13.  So therefore we would
leave it under the radical
as it is.  So our solution set would be
negative 3 plus or minus I
square root of 39
all over 12
Some now the fun part, checking.
So we have 6, and we use one of those
solutions
let's use negative 3 plus
I square root 39
all over 12, squared
So, of course we would have to square the
numerator first
When we square the numerator, it will be negative 3 plus
I square root 39
times itself.
Use a format that negative 3
times negative 3 is positive 9
 
Negative 3 times I root 39 would be negative 3 I root 39
I root 39 times negative 3 would give the same thing negative 3
times I root 39
And finally I square root 39 times I square root 39 would give us
I squared.  Now root 39
times root 39 would be
39.  I think we've done enough of these by now that 
you should be able to see
that's the case
So if we combine like terms: here we would have 9 minus 6 I
 
square root of 39
and then I squared is negative 1, so we would have negative 1
have -1
times 39, so that will give us negative
39
Finally,
combining 9 and negative 39, would have negative 30
minus 6I
 
square root of 39
so we would have 6
and we know the numerator would be negative 30
minus 6 I
times the square root of 39
Now all this over
12 squared.  And instead of writing this as 12 squared let's just write this as 12 times 12.
the reason why I chose to write it that way
is because we would have some
cancellation there
that might helps us not have to use such large numbers
So you know this is 6 over 1, or 6 divided by 1
and let's see, a common factor of 6 between numerator and denominator, 
6 would divide itself one-time in here
6 will divide 12 twice
So that would leave us with
negative 30
minus 6 I
 
Square root of 39
all of that
do that divided by 24.
Now before we go any further let's go ahead and work on the other side because remember
the whole goal is to see if these two
sides come out to be the same
I wouldn't want to simplify too much and then the other side may
be the exact expression that we already have.
So on the other side we have negative 3
and
we're using negative 3 plus
I square root 39
all that over 12
minus 2
So, here
we would just take negative 3 and
we could distribute it
on the numerator
so if we do that, we would have negative 3
multiplied by negative 3, that will give us
positive 9
and then negative 3 times
i root 39 would be negative 3
I root 39.
And all of this is over 12 minus 2
so we need a common denominator.  The common
denominator in this case would be 12.
So we would have 9 minus
3 I root 39
divided by 12
minus, and we know 2 is the same as 2 over 1, 
so if we needed a denominator of 12 we would
multiply numerator and denominator by 12
with you
which would give us 24 over 12.
So now putting numerator and denominator
together we'd have the
9 and then the minus 24 so 9 from 24,
9 minus 24 would be negative 15
minus
3 I root 39
all that over 12
Now when we compare these two expressions, you
should be able to see that
if we were to multiply
numerator and denominator by 2 on the
right hand side then
you would get what's on the left hand side.
Or, if we were to divide numerator and
denominator by 2
using the expression on
the left the we would get the same thing as on the right.
So it doesn't matter which you do it.
Remember the goal is just to show that
we get the same thing.
So let's the numerator and denominator
by 2
on the expression on the left.  So if we
divide the numerator by 2, 
then that would give us negative 15
minus 3 I root 39
and dividing denominator by 2, 
24 divided by 2 is 12, so you can see we get
the same expression on the left is on
right
so everything checks out
