We're going to look
at solving the quadratic
equation using the
quadratic formula.
Now the steps we're
going to go through.
We always need to write the
equation in standard form.
That means you collect
all the terms on one side
of the equation and
have a 0 on the other.
Once it's in that form, you
can identify the a, b, and c,
which are the co-efficients
and the constant.
Those get plugged, those are
the numbers that get plugged
into the formula, and then we
have to simply our solution,
and this can often be
the most difficult part.
So we'll take a look
at that in our example.
Here's the example
I want to cover.
This is the equation
we're going to solve,
and I really had two objectives
in choosing this example.
The first one was to
show you what do to do
when you see fractions
in a quadratic equation,
and the second one is really
just to give you an example
of an equation with solutions
that were not real
numbers, complex numbers.
So fractions to start with.
You can actually use fractions
in the quadratic formula.
You'd want to put this
in standard form first,
and then you'd have, like,
a three-fourths for the a,
a negative one-half on the
left side here for the b,
and a one-fourth for the c.
I do not recommend solving it
that way.
If you do it correctly, you
will get the right answers.
However, we had a method for
solving equations with fractions
which involved multiplying
both sides of the equation
by the least common
denominator of all
of the fractions
in the equation.
So it's pretty clear that the
LCD here is going to be 4,
and we're going to multiply both
sides of this equation by 4.
Four on the right,
4 on the left,
and this technique is called
clearing denominators.
So let me go ahead and write
this out with the 4 distributed,
and even put it as a 4 over 1.
So it'll be clear
how this cancels.
So here it is.
Multiplied by every term in
the equation, and you can see
on the left that
the 4's just cancel.
So you're left with just 3x
squared times 1 all over 1.
On the right side here, the
first term, you can cancel a 2.
That leaves you with
a 2 on the top.
So 2 times x all over 1, and in
the last term, the 4's cancel.
That just leaves
you with 1 over 1.
So minus 1.
Now the fractions are gone.
This is kind of a more
typical quadratic equation
where you just have
integer co-efficients.
So now let's kind
of just start over.
Let's pretend like this
was our starting equation.
This is equivalent to the
one with the fractions.
Has the same solutions, but
this will be a lot easier
to work with.
So if we had just
started with this,
the first thing we would want to
do is put this in standard form.
So we need to get a 0 on
one side or the other.
I like my x square
term to be positive.
So I'm going to leave
this one on the left side
of the equation, and I'm
going to subtract 2x,
do the opposite here to
move that to the left side,
and add 1, and that'll give
us 3x squared minus 2x plus 1
on the left side
of this equation,
and a big 0 on the right
side, or a little 0.
It doesn't matter.
And this equation
is in standard form.
Now we can identify a, b, and
c, and start plugging them
into the quadratic formula.
A is the co-efficient
of x squared.
That's going to be 3.
B is the co-efficient of
x. So we have negative 2,
and c is the constant
term here, which is 1,
and these are the values
that we are going to plug
into the quadratic formula.
So go ahead and rewrite this.
Here's what we have
from the previous page.
So we will continue now by
plugging these values of a, b,
and c into the quadratic
formula.
Our variable in this equation
is x. So we will start
with x equals [inaudible]
big fraction bar here.
Everything here goes on the top.
So we start with the opposite
of b. So opposite of negative 2.
I'm going to go ahead and
write the double negative,
minus negative 2, plus or
minus square root b squared.
So negative 2 squared,
parentheses are very
important here,
minus 4 times a times c.
That's 4 times 3 times 1.
All over 2a.
So 2 times 3.
So here's what the
quadratic formula looks
like for this particular
equation.
We're going to start
simplifying now.
I'll start by doing
the obvious things
like the double negative
here becomes positive 2,
inside the square root negative
2 squared is 4 minus 4 times 3
times 1 is minus 12,
and 2 times 3 is 6.
So now being very careful
with our signs here,
4 minus 12 is negative 8.
So I have a square
root of negative 8.
So here we are at the point
where we would now want
to start simplifying.
So find the radical and then
simplifying for common factors,
but it's important to notice
that we have a negative
inside the square root.
And that changes
things a little bit.
I now have to make a distinction
between my introductory
algebra class
and my intermediate
algebra class.
Most introductory
algebra classes,
you haven't really dealt with
the square roots of negatives.
We haven't defined the i equals
square root of negative 1 .
So we would actually stop here
and just say there
are no real solutions.
The solutions are
not real numbers.
You don't have to try
to simplify or anything
at this point once you have this
negative inside the square root.
For intermediate algebra
or higher, you're going
to take this step further,
and we will write this,
these two solutions
as complex numbers.
Let me remind you where
we were in our steps.
We're simplifying the radicals,
simplifying our solutions
rather,
and we want to simplify
the radical first
and then cancel common factors.
Radical first, and then
cancel common factors.
So that's what we're
going to do here.
We're going to simplify the
square root of negative 8.
Let me kind of just do this off
to the side so we don't have
to rewrite the whole
thing every time.
The negative I'm just
going to factor that right
out front here, and now
the 8, I want to factor
out the largest perfect square.
That's a factor of 8,
and that's going to be 4.
That leaves us with 2.
So I've rewritten negative 8
as negative 1 times 4 times 2.
Splitting the square root up to
each of these factors, right,
there's a multiplication
property for radicals.
I can take the square root
of each factor separately.
Now we have square
root of negative 1,
that's i. That's the definition
of i. [Inaudible]
the i's a symbol
that represents the
square root of negative 1.
Square root of 4 is 2 and
square root of 2 is what's left.
So i times 2 times
square root of 2.
Most likely, you'll see that
written either 2i root 2
or 2 root 2 times i. All three
of these are the same thing.
I'm just switching the order
on the multiplication here.
I'm going to go ahead
and just pick this one.
Be careful if you
put the i last.
Just make sure you do not put
the i inside the square root,
alright.
The square root is gone on that
part because the square root
of negative 1 is i. So it's
outside of the square root.
When I rewrite this
expression for my solution,
I'm going to replace
the square root
of negative 8 with 2i root 2.
So 2i root 2 all over 6.
Now the square root
is simplified,
and the last step was to
cancel the common factors.
You have to be able to cancel
common factor out of each
of the two terms in the top
and the term on the bottom.
So the only common
factor I see there is 2.
I prefer to factor the
common factor out of the top.
So 2 divided by 2 is 1, and
2i root 2 all divided by 2,
we're just going to divide
it by the co-efficient.
So 2 divided by 2 is, again, 1.
One times i times
square root of 2,
and you can even factor
the 2 out of the 6.
Two times 3.
So you can see that this factor
of 2 cancels top and bottom,
and that leaves us with 1 plus
or minus i root 2 all over 3.
Usually we just leave
it in this form,
but when there's an i there,
we will often take
it a step further.
Let me show you why
we're going to do this.
I'm going to split
this fraction up.
So one-third plus or
minus i root 2 all over 3.
I'm actually going
to put root 2 over 3,
and then put the i
on the other side.
So this expression is
equivalent to this expression.
They're exactly equal.
This represents the two
solutions, one-third plus root 2
over 3i and one-third
minus root 2 over 3i,
and the reason we split the,
split it up when we have the i
in there is because
complex numbers have two,
they're two dimensional numbers,
they have two components.
They have a real component, and
then an imaginary component.
So this shows you
the two components.
The plus or minus, again,
gives us two separate answers,
and if you go back and
review your complex numbers,
you'll see that these are
conjugates, complex conjugates.
One-third plus root 2 over 3i,
one-third minus root 2 over 3i.
So that wraps up this example.
What I want to add or finish
this video with is a discussion
that we started in
the first video.
We started talking
about the discriminant.
So here is the discriminant.
It's the expression that's
underneath the square root
in the quadratic formula.
B squared minus 4ac.
We take the square root
of this whole thing
when we find the solutions
to those equations.
And so that gives us information
about the types of solutions
that we're going to
find in the equations.
In this equation, our
discriminant was negative 8,
and because it was negative, and
we had to take the square root
of that, that gave us
these imaginary components,
which made our solutions
complex numbers.
So I just want to compare
that to two examples we had
in the first video where we
had positive discriminants,
9 in this example and 28 in
this example, and what we looked
at in those two examples
are what we focused
on was whether these were
perfect squares or not
because they were both positive.
So our solutions
were real numbers
because we were taking
square roots of positives.
When it was a perfect square,
we ended up with these
rational solutions
because we could take the
square root of 9 and get a 3.
So there was no square roots
in our answers, and we talked
about how that meant that
the expression over here
on the left was factorable.
When the discriminant is not
a perfect square, you're going
to be left with a square root in
your answer after you simplify,
and these are irrational
solutions, and we talked
about how that meant that this
expression was not factorable.
So just to summarize the results
for the discriminant here,
if the discriminant
was positive,
you get two real solutions
because you're taking
the square root
of a positive number,
either plus or minus.
If it was a perfect square,
they are rational numbers.
If it was not a perfect
square, they were irrational.
In this example, we saw how if
the discriminant was negative,
we got two complex
solutions, and they happen
to be conjugates of each other.
Complex conjugates.
And, finally, we haven't seen
an or done an example of this,
but if the discriminant is
actually 0, well, if you think
about what happens in
the quadratic formula,
you take the square root of 0.
That's equal to 0, and whether
you add 0 or subtract 0,
you get the same answer,
the same final answer.
So the two solutions
end up being
up just one repeated
solution, and it happens
to be a rational number.
And that's all we can say
right now about discriminants
and the quadratic formula.
