Hey there. Uh, this is the video 2 of the
quadratic formula. Um, now we're actually
going to use the quadratic formula to
solve, um, some quadratic equations.
So, um, we'll
start with the, just the, the um,
the...standard
quadratic equation form. Uh, really important here.
Um, I've written it down. Y equals ax
squared plus bx plus c. You have to get
your quadratic equation in this form the
best you can, uh, before you use the
quadratic formula. And I'll show you what
I mean in the next problems. Uh, some of them
will already be set up in this form and
some won't be. The first ones actually
not going to bee. Ah, another way you can
think of it as, um,
we could put a zero in for y.
And say zero equals ax squared plus bx plus
c.
Um, and again we could think of it as zero or y.
Because remember, we're trying to find
the roots right here. Okay? So let's say
this is a root right here. Let's say this
is one. Um, the ordered pair (1, 0) right there.
And let's say our parabola crashes-crash it, um, um,
crosses through that
point right there.
We are searching for
that number, that x value. But notice what
the y is right here. The y is zero.
So we're
searching for when that y is equal to zero.
We're searching for the roots.
So. Uh, let's jump
into a problem. But remember, again, you
want to try to get it in this form right
here.
Or this form right there.
Again, really similar because we're trying to
figure out when y is zero. So, let's jump
into the first problem.
Uh, the first one
looks like this.
Uh, x squared equals 16.
Now, if you think about this. This one, um, I've
talked to you a little bit about it
already. [School bell]
I would just solve this one
algebraically.
But I want to show you how
to use the quadratic formula for it.
Because I did say in a previous video
that you can use it for any single
quadratic equation.
Um, so we're obviously
going to, um, to, to do that.
So, let's go ahead
and, um, you know, solve this.
So the first thing we want to do is we want to get it
in that ax plus b, um, ax squared plus bx plus c form. Okay?
Equals zero. So we got to
get this 16 out here. So I would subtract
16 from both sides. Alright?
And I would
rewrite it as x squared minus 16 equals zero.
Now, some you might be like, well is
that really in, um, the standard quadratic
equation form? You bet it is. Because I
could write, um,
one x squared plus zero x.
There is no b. Minus 16 equals zero. we can now go ahead
and substitute this stuff into the
quadratic, um, formula. So remember the
quadratic formula is negative b plus or
minus the square root of b squared, minus
4ac all over 2a.
So I'll show you how
we're going to get our answer here.
The one is our a. So we'll substitute that
anywhere we see a, um, an a. Again, and again,
it's just in those two spots right there.
The zero is the b. We'll substitute that
any place we see a b.
And the negative 16
is the c. And we'll substitute that
any place we see, uh, the c.
So let's go ahead
and do that.
So we have negative, um,  the b is
zero. So let's put a zero in there.
Anytime I put a number in I put
parentheses around it.
Plus or minus the
square root of, uh, b squared. Again is zero.
So we'll have zero squared.
Minus four, times
our a, in this case is one.  Let's put that
in for a. Times our c, which is negative
16. Alright.
Looks pretty good. All over
two, times our a, is one. Alright?
Let's go ahead and see what happens here.
So we have negative zero. Nothing really
happens with that. That's just zero. So I
would write zero, plus or minus.
Zero squared inside our radical symbol here.
Well zero squared we know is zero.
So it's going to be zero, plus, we have
negative four times one times negative 16. Alright?
Which is going to equal a positive 64.
I hope that makes sense. All over two
times one is just two.
Let's write it out a little bit better. I
don't really need the zero here. I don't
really need that zero there.
So it's going to be, I want to rewrite this as
plus or minus the square root of 64 over two.
Which looks nice and easy to solve
because 64 is a perfect square.
The square root of 64 is eight. So I'm going to
come over here and write.
We would have
plus or minus eight, over two, which we know is,
and, this- at this point I would break it
up into two separate problems. I would
say it's plus eight over two.
Or minus eight over two. Well eight divided by two is four.
And negative eight divided by two is
negative four.
And those are the two answers.
Now we already probably knew that.
Because if
you took the square root of, um, if you went
back to the original problem and took
the square root on both sides, we would
say the square root of 16 is plus or
minus four.
Um, and that's how we do it. So, again
this is, this is the quadratic formula
for this particular problem here.
I wouldn't use it for this problem. But I
just wanted to show it to you guys for
this one. I would just take the square
root of both numbers.
Announcement over loudpeaker: Good Morning staff and students.
Students, a couple peices of information.
Mr. Kopp: So, um,  in this video here, um.
So now we're gonna jump into a problem that, um, uh, has
the x in it. Alright? So you can see
that x right there.
So we have the x
squared and the x. This makes our lives
more difficult if we were gonna graph it.
But, uh, again using the quadratic formula it
should make our lives pretty easy.
Um, this
equation is already, um, in the standard
quadratic equation form, which is nice.
We have y equals, you know,  1x squared plus
3x minus four. So we already have it in, you
know, ax squared plus bx, min-uh, plus c form.
And the y right there is fine. We could
put a zero in right there and we're happy
with that. Okay?
So we can jump right into
the quadratic formula here. So let's go
ahead and do that.
So, I'm just gonna
substitute the numbers in. So again our
quadratic formula is negative b. Again
our three is b. So we'll put in a negative three.
Plus or minus the square root of, uh, it's b
squared, so we'll go three squared, minus four.
Times a. In this case a is one.
And times C.
Which in this case is negative four.
All over two times a. Which is two times one.
Alright. let's keep going head and solving this.
Uh, so we'll just simplify a little bit.
It's gonna be negative three, plus or minus the
square root.
And I'm gonna try to
simplify this all inside here.
Three squared
is gonna be nine. Alright? Maybe I'll write it out.
Nine. And then we have, you know,
negative four, uh, times one, times negative four.
Which is going to be a positive, um, 16.
Which is nice because when you take, and we'll
multiply the denominator. Two times one is
just two.
When we take nine plus 16
we get 25, which is a perfect square.
Which makes our lives nice. Sometimes
you're not going to end up with perfect
squares inside of there. And then that
means that, um, um, and, and, that's going to happen in the next one.
I guess uh, maybe I'll talk about that in the next one.
But when we get a perfect square, um, makes our
lives pretty easy. So we have negative three
plus or minus the square root of 25 over two.
Let's take the square root of 25 now.
The square root of 25 is just five. So we
have negative three plus or minus five
all over two. Now from here you're going to break it
into the two parts.
Okay? So we're gonna solve it two ways.
We're gonna say it's negative three plus five
over two.
Okay? Or, negative three minus five over two.
So let's go ahead and solve this.
Negative three plus five is two.
Two divided by two is one.
There's one of your roots. The other one,
I'll put the or there.
Negative three minus five
is negative eight.
Negative eight divided by two is negative four.
So that is your second root
right there. And again the roots are
where this parabola would cross the
x-axis. So if I just thought of the graph
real quick. I know ,real generic graph
here, um, it would look something like this.
It would come down, and it would hit
negative four,
and it'd come back up. So imagine
that this spot right here,
that root was negative four.
And that one right there is one.
So you can, you can put it in ordered pairs if
you wanted to as well. You could say the
roots are one and negative four.
you could also say that's (1,0), and
(-4,0). Alright?
So those are the
two roots right there. Works out really
well when you get a perfect square
inside of there.
Alright?
Let's solve one where you
don't get a perfect square inside of there.
So, um, uh, looks like this.
Y equals negative x squared plus 11x minus 16.
So, again the cool thing about this one is
already in the standard quadratic form.
Right here.
We could say y equals
negative 1x squared plus 11x. You know,
minus 16. So we got the a is negative one.
The b is 11. And the c is negative 16.
If we have a y right there that's good. Uh,
we could put a zero in right there. If there
was a number over here, like if it said five
or 10, we would want to move that over to
the other side. We have to have that
being a zero or y. In this case it's nice.
It's easy, we don't have to do anything
with it.
So let's go ahead and um, substitute
these numbers into the, um, quadratic formula and solve.
So we would have, you know,
negative 11, because that's our b. Plus or
minus the square root of, ah, this case b
square would be 11.
Squared, minus four, times a.
Which is negative one, times c which is
negative 16.
All over two times a which is
two times negative one. Alright.
Let's go
ahead and simplify this a little bit.
So we do have negative 11, plus or minus,
let's look inside the radical here.
The radical symbol would be 11 squared.
Which we know is 121. Now we have negative four
times negative one, times negative 16.
That has to be a negative because we're
multiplying three negatives together.
The negative one we're not too concerned about, um,  now
that we know it's negative. But four times
16. We know has to be 64. Alright?
All over two times negative one, which would be negative two.
So let's go a little bit further and simplify this.
121 minus, uh, 64 is 57. So we have negative 11 plus or
minus the square root of 57 over
negative two.
Okay? Notice how 57 is not perfect.
The cool thing about the
quadratic formula is we can approximate
where this crosses. We could actually
stop here and say this is the answer.
Okay? This is the most exact answer,
I should say, answers, that we could have.
Because we have the plus or minus in
there.
That's good. We're good with that.
Now, a lot of times you're gonna want to
approximate it to say where does it
actually cross, uh, the x-axis?
Um, and we can do that really well. And that's why the
quadratic formula is really, really
valuable. Because if we were gonna solve
this by graphing you would get,
you'd-you'd-you'd graph your parabola, and you say,
well it looks like a cross is about here
and here, um, but this is a more, um, precise way
to figure out what that uh, that number is.
So. Uh, We'd probably want to do is type in
the square root of 57 in calculator and
get that approximation. Um, so I have
negative 11 plus or minus, um, I believe this
number is seven point, I wrote it down,
uh, 7.5498 about.
Okay? So let's write that out like that.
All over negative two. Okay?
So then we
would just take these numbers and we
would, you know, we'd break it down into
two parts right here. Okay? And again this
is an approximation here. I could have
written more values out if I wanted.
Um, so when I take a negative, um, 11.
Uh, let's do the, the plus first.
So negative 11 plus 7.5498 divided by negative two.
So I take negative 11, plus that and I divide by, uh, negative two.
I end up with this value right here. And
I already wrote it down.
It's like 1.73 approximately.
So that's one of your roots right there.
The other one we take a negative 11 and we
have to do minus 7.5498.
And we also divide by
negative two. And so I just do the
numerators and then I divide by the
denominator. You might end up with a
different, you know, approximation if you
round a little bit differently, um, than me.
But it's pretty close to this. Um, and again,   I wrote that number down it was like
nine point, um,  it was three something I
forget what I actually wrote down.
B it's 9.3 about, it was like
maybe two seven or something.
Um, but those are approximately your two roots.
So if you want to think of it in terms
of your roots, um, or in terms of ordered
pairs, it'd be like (1.73, 0)
and (9.3, 0).
So the graph of this one, um, and would
look something like this. Uh,
in this case the a is negative so we
know it has to open uh, down.
Uh, so the
graph would look something like this.
So, would come up,  boom it hit 1.73 there and then it would
come back down and hit 9.3 over there.
So the graph of the quadratic
would look something like this with your
roots right here. And again that may be
hard to do graphing, so it's really nice
when you actually can use the quadratic
formula to actually solve that. So.
Um, I hope
this video helps out a little bit if you
have any questions feel free to ask, um, in
class.
