In this lesson we are going to solve
quadratics using the quadratic formula.
The quadratic formula is outstanding.
It's my second favorite way. My first
favorite way is to factor because I find
factoring to be quick and easy. But as I
said before, factoring doesn't always
work and the quadratic formula works
every single time which is why it's so
great. So in the quadratic formula there's a
couple of things we need to understand
first we need to make sure that our
quadratic equation is in standard form
which means we need to make sure that
everything is on one side and zero is on
the other side. just as we would have
been trying to factor. So for instance in
this question I would subtract that 8
x over and I would subtract the 90
and that would leave me with 2x
squared minus 8x minus 90 equals 0
that's what we want just as if we were
going to factor. Step two is we're going to
use the a, b&c values into the quadratic
formula. So this is the quadratic formula
and this is great because all we're going
to do is plug in our b's, our a's, our c's and
then we're going to simplify and that's
going to give us our two answers.
Notice two because of that plus or minus.
So let's do this one together. x equals
the opposite of b. So my let's go ahead
and write my a,b&c value. my a value is
positive 2. My b value is negative
8 and my c value is negative 90. So, I'm just
looking at the three parts. They do have
to be in that order the a has to be the
one with the x square, the b has to be
the one with the x etc.
so if I'm doing the opposite of b, the
opposite of negative a is positive 8
plus or minus the square root of
negative 8 squared minus 4 times
a times c
all over to a. So a is 2 and then I'm simply
going to do whatever I can to simplify
the top and the bottom. Numerator and
denominator. 8 squared is 64.
4 times 2 times negative 90 is 
negative 720. But I have minus out here
so it's minus a negative 720 which
is plus 720 and my denominator is 2 times
2 which is 4. Then I'm going to add
those two values on the inside and I'm
going to write it here just for the sake of space.
So, I have 8 plus or minus and I have
the square root of 784 over 4 and
then I'm going to try to take the square
root of 784. The square root of 784 is 28
so I have 8 plus or minus 28 over 4 we all
know there's going to be two solutions
so I'm going to write x equals 8 plus 28
over 4  and reduce that. So 8 plus 28 is
36, divided by 4 is 9 and then I'm going to take  8 minus 28
divided by 4, which would give me negative 5.
So, my two solutions would be 9 and
negative 5. Whenever we get some nice whole
numbers like that, that means at the
beginning I could have factored and
gotten the correct answer that way as
well. Let's try this one
Same steps, everything is in standard form
already which is great. so it is already
there. Now I'm going to use my quadratic
formula which is right up here for you.
If you have a chance, Google songs for
the quadratic formula because there is one
pop goes the weasel that's very catchy.
I'm not going to sing it for you. But it is
very catchy there's also one that's very
funny that I might send a link out for
you. Because it is hilarious. But I digress
In this case I have a as 3, b is
negative 4, and c is 2. so those are
the values that I'm going to use as I
solve this quadratic equation. So x equals
the opposite of b, which is negative 4.
So that is a positive 4 plus or minus the
square root of negative 4 squared
minus 4 times a times c all over 2
a and then we're going to simplify
anything we can. So 4 plus or minus on
the inside I have 16, 4 times 3 is 12 times 2 is 24. So, 16 minus 24 over
6.16 minus 24 is negative 8. So, I've 4 plus or minus the square root of negative 8 over
6. I know that 8 is not a perfect
square, but that 8 obviously got a
negative on the inside and I don't love
that. I also know that I could break
negative 8 into the square root of
negative 1, the square root of 4 and square root of
2. which means I have an i and I'll
have a 2 and I'll have this 2 still
in the radical so I've 4 plus or minus 2i radical 2 over 6 and then of course
I should be able to simplify this.
Whenever I simplify remember the radical
the stuff in the radical stays in the
radical. So, the radicand  stays alone. I
can reduce these values. It doesn't
matter that there's an i on this one, I
can still reduce everything by 2. So
I'm going to turn 4 into 2, 2 turns into
1. So it would be 1i or i radical
2 and then 6 divided by 2 is 3. So I would
just probably leave my answer exactly
like that. If they wanted two answers I
would write it as 2 plus i radical 2
over 3 and 2 minus  i radical 2 over 3 would
be my two x values. We didn't talk about
graphing in this.
course. but if we were to graph this, that
would mean that these don't actually
touch the x-axis. Here is a another for us
to try. So again a would be 4, b would
be negative 20, and c would be 25. I'm
going to down just plug and chug. So
opposite of b would be positive 20
plus or minus the square root. b squared
would be negative 20 squared, which would be 400 minus 4 times 4 times 25 which would be 400
and the denominator would be 2a, so 2 times 4 is 8. So noitce I showed a
little bit less work on this time around
and that's okay. Now some people might
freak out a little bit on this one
because as we can see the radicand is
400 minus 400 which is 0. So what is the square root of 0, it is 0. So it is 20 plus 0 divided by 8 and 20 minus
0 divided by 8. That means we're just going
to have one solution because it's going
to be the same answer both times. So
my answer is 20 divided by 8 which of course would be 5 divided by 2.
Now in terms of what that would
look like on a graph, that would mean
that if five halves, which is halfway
between 2 and 3 that we would have a
a parabola that would come down and
touch at five-halves. Sorry that's a bad
parabola, but it would touch at five-halves and then turn around and go the
other direction. Some books might call it as a multiplicity of two because that would
be technically two solutions at that
point.
Here's another for us to try. Again we
have a is 2, b is negative 1, c is
2 and we're just plugging into our
formula. So x equals the opposite of b
plus or minus the square root of b
squared minus 4 times 2 times 2
which is 16, all over 2a which is 4.
My numerator is 1 minus 16, which should be the square root of negative 15 over 4 and what students
often do is, they just leave it like this
because 15 can't be reduced it breaks into
5 and 3 neither are perfect
squares. However, I don't ever want to
leave a negative on the inside of a
radical. So I would write that is
i radical 15 over 4 and again if I had
to write that as two solutions. it would
be 1 plus i radical 15 over 4 and 1 minus i radical 15 over 4. Here is
one for you to try. So try this first,
then press play to check your work.
With this particular question, if you'll
notice I don't have it in standard form
yet because 4 is on the wrong side. so
I do have to subtract that 4 before I
can find my a,b or c values. So 6x
squared minus 5x minus 4 equals zero
Now a is 6, b is negative 5, and c is
negative 4. So my values here the
opposite of b would be positive 5,  plus
or minus the square root of negative f
squared which is 25. Remember whenever
you square a number, that number will
be positive unless you are dealing with
an i value minus 4 times a times c. So,
4 times 6 times negative 4. So
minus negative 96 allover 2a, which
should be 12. I'm going to go ahead and
ching ching here in the middle to
combine those values together. That gives me
on the inside, I get 121 and the square
root of 121 is 11 and this one's nice
and neat. No radical is leftover and so now
I can just find my two values. So I would
never leave my answer like this. I would
if there are radicals, there are things
I could not reduce. But in this case I
would actually find the two solutions,
which would be 5 plus 11 divided by 12 or
16 divided by 12 reducing that by 4
would be four-tirds. The other answer
would be 5 minus 11 over 12 which would be
negative 6 over 12 or negative one
half. So, the two solutions would be four-thirds and the negative one-half.
