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>> Okay. So in this video,
we're going to be using
a quadratic formula
to find solutions to
quadratic equations.
I mean you could
do this many ways.
You could find solutions
to quadratic equations
by completing the square.
You could do it by factoring.
And you could also do it
by the quadratic formula
which is what we're --
it's how we're going to go
about finding them in this video
is using a quadratic formula.
So before I start
doing examples.
Let me go ahead and derive
the quadratic equation for you
so you could get a sense
of where it came from.
Now here I have a quadratic
equation in standard form.
And zero's already on one side.
So I'm going to go ahead
and derive the quadratic
formula using this equation.
So the first thing I'm
going to do is go ahead
and divide this coefficient
off the leading term.
[ Silence ]
Don't forget the zero.
So then you get x squared plus e
over a x plus c over
a equals zero.
So now I'm going to go ahead
and complete the square.
So I'm going to subtract
this constant over.
So then I get x squared plus.
Now I'm going to
leave a space here.
I'm going to subtract
the constant over.
And then I'm going to
leave another space
because when you complete
the square you have to add
that term onto both sides.
So to complete the square,
this guy right here.
Whatever's in front of this
variable to the first degree.
So it's b over a. we're going
to multiply that by one half.
And then we're going
to square that.
That's the term that's going
to get added onto both sides.
So this will become b over
2 a and when you square
that you get b squared
over 4 a squared.
So this is a term that I'm
going to add on to both sides.
B squared over 4 a squared.
[ Silence ]
B squared over 4 a squared.
B squared over 4 a squared.
Okay. So now whenever complete
the square, it factors nicely.
So I'm going to go ahead
and factor this side.
It's going to become x
plus b over 2 a squared.
That's how that factors as.
And now on this side, let's
go ahead and get an LCD.
So the LCD looks
like 4 a squared.
So I'm going to multiply
the top and bottom
of this fraction by 4 a.
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Okay. So now we have an LCD
of 4 a squared and on top,
this [inaudible] multiply.
So you'll get a negative
4 a c plus b squared.
Okay. I think I'm going to
have to move onto this side.
Now because this
[inaudible] is kind of long.
So now we're trying to
solve this variable x
because everything else
is a constant here.
Right? A, b, and
c are constants.
So I'm going to go ahead
and take the square root
of both sides right now.
So I'll take the
square root over here.
And I'll take the square
root of this whole thing.
Okay. On this side, we just get
what's inside the parenthesis
which is x plus b over
2 a. And on this side,
since we're taking the
square root we got to account
for the plus and minus.
Right? So we get plus or
minus the square root of --
now I'm going to write
this b squared first just
because that looks
cleaner instead
of starting off with negative.
So I'm going to write
b squared minus 4 a c.
[ Silence ]
Okay. And this is going to
be all over the square root
of 4 is 2 and the square
root of a squared is a.
So that's all over 2 a. Okay.
Still trying to solve
for x. So all you have
to do is just subtract over
the b over 2 a. So I get
that x is equal to
plus or minus --
[ Silence ]
And now I'm going to subtract
over the b over 2 a. Okay?
Notice that we already
have an LCD of 2 a.
So let's just write it
as one single fraction.
So then we finally get
that x is equal to --
I'm going to go ahead and
write the negative b first.
So negative b plus or
minus the square root
of b squared minus 4 a c and
that's all over the new LCD
which is 2 a. And that right
there is your quadratic formula.
[ Silence ]
So the quadratic formula
x which is our solution.
That's equal to negative b
plus or minus the square root
of b squared minus
4 a c and everything
over 2 a. Not just
the square root part.
So now that I've
derived it for you,
we're going to go
ahead and use it.
All right.
So actually before we
start doing some problems
with the quadratic, I wanted
to mention something to you
about the quadratic equation
which is the discriminant.
It's the part that's
under the radical.
So the discriminant is
just this guy right here.
So the discriminant is b squared
minus 4 a c. So I'm just going
to refer to the discriminant
as d. So here's the three cases
that you would expect when
using a quadratic formula.
A discriminant's going to shed
some light into our solution.
So you have d here right?
If d is greater than zero,
we're going to expect
two real solutions.
If d is equal to zero,
we're going to only
expect one real solution.
And if d is less than
zero, then we're going
to expect two complex solutions
because the discriminant,
we said is what's under
the radical, right?
So if we get something
that's less than zero
under the radical,
that would be negative.
And when you have a negative
under the radical you
get complex solutions
with I's [phonetic] and stuff.
Right? When the discriminant
equals zero,
the square root of zero is zero.
So we get rid of
the square root part
and we only get one solution.
And when it's greater than
zero, the square root remains
and therefore, the plus or minus
or however it happens,
we get two solutions.
So yeah, that's the
discriminant.
And, you know, beforehand
you don't want
to do the whole process, do
the whole quadratic formula,
you could just jump straight
to the discriminant just
to know ahead of time what
kind of solution you expect.
All right.
So we're going to go ahead
and do the first example here.
Here's the example right here
and the -- I can't think lately.
Directions are to solve this
using the quadratic formula.
So you need to know what a,
b, and c are before you jump
into the quadratic formula.
So here I've written
out the standard form
of a quadratic equation.
And you know you
do want to get it
in standard form
before you start.
It'll just make it
easier for you.
You don't have to but it'll
just make finding a, b,
and c. So here I
could see that b is 2.
C is 8. And a is 1, right?
A can't just be nothing.
So we've figured out a, b,
and c. So we got
that a is equal to 1.
B is equal to 2.
And c is equal to negative 8.
And I'm just going to go ahead
and just plug and chug all
that into this formula
and see what happens.
So we get x is equal to
negative -- and my b was 2 --
plus or minus the
square root of b squared.
Our b was 2.
Minus 4 a was 1.
Our c was negative 8, I believe.
Okay. And this is all over
2 times a which was 1.
So then we get that x is
equal to negative 2 plus
or minus the square
root of 4 minus.
Two negatives make a positive.
Eight times 4 is 32.
And this is all over 2.
So then we get x is
equal to negative 2 plus
or minus the square
root of 36 all over 2.
We get x is equal to the
square root of 36 is 6.
So we negative 2 plus
or minus 6 all over 2.
Now this is like -- these
guys have the same LCD.
You could split up
the LCD if you like.
That's how you like doing it
[inaudible] or you could just
that you're really just
going to divide this 2
into the negative 2 and to the 6
which is actually what I'm going
to do or maybe I should
write it down first.
So you could write this as a
negative 2 plus or minus --
you know just split up the LCD
if you're more comfortable
doing it that way.
Okay. So then you
get x is equal to --
this will become negative 1.
Plus or minus and
that will become 3.
So now the place where the plus
or minus comes in is you have
to -- you could just write out
two different solutions with it.
So one of our solutions
is going to be x is equal
to negative 1 plus 3
from the positive group.
And the other one's going to be
x is equal to negative 1 minus 3
from the other group, from the
negative group or whatever.
Okay. So this, 3
minus 1 would be 2.
And this would be negative 4.
So the two solutions I got
for this were x equals 2
and x equals negative 4.
[ Silence ]
So notice that we got two real
solutions and they were rational
which means that we could have
factored this instead of going
by doing the quadratic
formula but you know,
this was a good example.
