This is Susan Poss. We have in the past looked at
three methods for solving quadratic equations,
that is second degree equations. And now today in
Section 10.2, we're going to look at a fourth
method.
For solving quadratic equations, and it's called
using the quadratic formula. But before we do
that, let's quickly review the techniques. The
first was called factoring.
The arrangement is really important in these
because each arrangement is a little bit
different. So we put everything on one side and
put zero on the other side...
Where all the terms were located except for the
zero and we factored it and solved. The second
technique we looked at was the square root method.
Where we put the second degree term on the left
and the constant on the right and then we took the
square root of both sides.
The third technique we looked at was called
completing the square. And again, look at the
arrangement on this one, we put the second degree
and the first degree on the left we put the
constant on the right.
And then we had to make sure that we had a one
coefficient in front of the x squared. So, we
divided through if there was a number there other
than one on that second degree x we divided
through by that.
And then we filled in these empty spaces here with
a particular number after we did the completing
the square process, and you'll remember that.
So now today for the fourth method, which is
called using the quadratic formula, here is the
quadratic formula highlighted for you here in
yellow.
And what we have to first of all do is make sure
we understand about the arrangement. The
arrangement of the quadratic formula is very
similar to factoring, you put everything on the
left.
And you put zero on the right, so the arrangement
is critical, you have to have that done first,
everything on the left. Zero on the right. Before
you start identifying these variables in the
formula.
So variable a and let's look at this example that
we have. In this problem, a, the value of a is
three, because that is on the second degree
variable.
The value of b is negative five, because that is
on the first degree variable, and the value of c
is the constant and that is negative two.
Now again, you have to have it arranged in
standard form, which is everything on the left,
zero on the right before you identify a, b, and c.
So I've already written down the formula.
And I like to just kind of go through the formula
and create these pockets. These sort of like
little receptacles where I'm going to plug in each
of those numbers.
Be sure you draw that great big long fraction bar
up underneath all of that. Alright, so I've left
out all the letters A, B, and C. And now I'm going
to replace those with numbers.
So the B value is negative five, and that goes
here and here, the a value is three, and that goes
here and here. And the c value is negative two,
and that goes in that spot right there.
So all I did was look at the formula and plug in
the letters for a, b, and c, just like the formula
guides me so now I do the arithmetic. This become
postive 5. I square that negative 5 and get 25.
And this is where you have to be careful about
signs folks, I multiply negative four times
positive three times negative two. Looks like I
drew this radical. A little bit too long. Let me
go back and fix that.
Whoops, sorry, I see a multiplication mistake I
made here. It should be a positive 24. Here, I
don't know how I managed to get that wrong. But I
did. Alright, so let me do a quick double check
there.
Make sure I have that right. So here is the
positive 5. Here's the positive 25 here's 12 times
two is 24. And that's a positive 24. And on the
bottom, I have six.
Alright, so continuing on, I think my arithmetic
now is correct. This 25 plus 24 gives me 49. And
of course that 49 is up underneath the radical. So
I'm going to be taking the square root of 49.
Let me swing out here where I have a little more
space to finish. So this is five plus or minus
seven, that's the square root of 49 sitting over
six.
And again, since this is a second degree equation,
we are expecting two answers. We may not get two
answers, we may only get one we may get nine in
the set of real numbers.
So here's how you want to finish this up. In one
case, you want to do five plus seven divided by
six is 12 divided by six, which is two. And in the
other case, you want to do five minus seven.
Which is negative two over six, which is negative
one third. So again, when you have that plus or
minus, that's where you account for the two
expected solutions to the second degree.
Alright, I have two more I want to work for you.
So let me just scroll along past some things here.
This particular one was already in the arrangement
that I wanted.
It's got everything on the left and zero on the
right. And you see where I've already gone through
and I've identified a, which is the coefficient on
the second degree term.
I've identified B, which is the coefficient on the
first degree term and other identified C, which is
the constant. So I've already written the formula
down here.
And again, let me just go through and create some
spots. Didn't quite that long enough. Here's the
big long fraction bar sitting over two times a.
Alright, so filling in here.
B is negative 12, a is negative one, and c is
negative 36. So now I'm just gonna do the
arithmetic. So I see a positive 12 here,
underneath the radical negative 12 squared is 144.
And then negative four times negative one times
negative 36. I don't know what that is, that's
positive four times negative 36, which is 24,
which is 144, negative, negative 144. All sitting
over negative two.
Alright, so continuing on look under the radical,
I can do a little more arithmetic under there.
This is the square root of zero. And finally,
that's a plus or minus zero when I take the square
root of zero.
So again, swinging out here, x is 12 plus zero
divided by negative two and x is 12 minus zero
divided by negative two. And I was expecting two
answers, and I only got one answer.
That happens sometimes when you're working with
second degree equations you expect two but you
don't always get two solutions. So that was why I
wanted to show you that one.
And then scrolling down, oops, I missed it right
here. This is the other one I wanted to show you.
Again, notice it's not in the correct form.
So you put it into standard form everything on one
side, zero on the other. I'm going to go out here
and identify a and b and c. And I didn't write
down the formula this time I'm not sure why it let
me do that.
Alright, so here I go, just sort of filling in
making those openings so that I can substitute in
the correct numbers for the letters. Alright, so B
is what I want first, so I'm going to fill in
values for B here.
B is one, a is two, c is two. So I see negative
one plus or minus the square root of one minus 16
over four. And again, I'm going to swing out here,
make a little more room.
Minus one or negative one plus or minus the square
root of negative 15 over four, and of course you
remember from chapter nine. This takes me into the
realm of complex numbers. This is not a real
solution.
This is a complex number solution. So this is
going to become i times the square root of 15
sitting over four and generally what we do with
this is we separate it out into a real part and
imaginary parts.
So there's one of the complex solutions and this
would be a minus here. So, there are my two non
real complex solutions. So, again, I wanted to
show you each one of those, let me quickly review.
So, what can happen when you solve these, you can
either get two distinct solutions as I did on
first one, you may get only one solution, as I did
on the second one.
Or you may get to non real solutions, or in this
case, two, imaginary two complex numbers,
solutions.
