In this session we're going to see how
to use the quadratic formula
to solve quadratic equations.
And, the quadratic formula actually comes
from completing this square in particular
or I should say in general
because this is generalized,
solving ax squared plus bx plus c equals to 0.
And, I tried to make it fast.
But, it seems too fast.
So, we'll just do this step by step.
But, you can relax and just kind of watch.
This is in just about every textbook out there.
Remember that, when we're completing this
square, we want that first coefficient to be 1
and we also want that constant
to be on the right hand side
because we're only completing the
squared for the variable terms.
So, if I divide everything by a
and then subtract the constant,
I get x squared plus b/ax is equal to -c/a.
And notice, I'm leaving that gap after b/ax
because I'm going to add the right thing
to both sides to complete the square.
To figure out what to add, we multiply
linear coefficient, b/a by 1/2.
That gives us b/2a.
And, squaring that gives us the value that
we're actually going to end up adding.
In this case, b squared/4a
squared, we're going to add
that to both sides.
On the left hand side, we have
x plus, as you may recall,
what goes inside the parentheses matches
what we got when we cut b/a in half.
Right? And the sign matches that term.
So, we have x plus b/2a the
quantity squared is equal to,
now the -c/a plus b squared/4a
squared needs common denominators.
Right? In this case, we're going to get
that by multiplying numerator
and denominator by 4a squared.
I'd like to write the positive term first.
So, I have b squared.
I'm going to write minus 4ac all over
because we've got common denominators,
right, 4a squared.
At this point, we have squared
equals thing in an equation.
Right? so, we can square root both
sides, not neglecting the plus or minus.
The square root cancels out the squared on the
left hand side, we end up with x plus b/2a.
On the right hand side we get plus or minus.
We're going to square root the numerator and
get nothing pleasant, b squared minus 4ac.
But, square rooting the denominator,
4a squared gives us 2a.
Finally, we're going to subtract
b/2a from both sides.
Since we have common denominators, I can
write that as -b plus or minus the square root
of b squared minus 4ac all over
those common denominators of 2a.
And, this is our familiar quadratic
formula which is just a generalization
of solving by completing the square.
Now, there's a part of the quadratic formula,
the part that's inside the radical,
it's called the discriminant.
The discriminant is b squared minus 4ac.
And, there are 3 distinct possibilities.
The discriminant can either
be positive, 0, or negative.
If the discriminant is negative, that means
we've got a negative inside the radical.
And that's going to give us complex
solutions because of the plus
or minus we're going to have 2 of them.
If the discriminant is 0, then, we're only
going to have 1 solution because if you add
and subtract 0 you don't get anything different.
And that solution is going to be real.
If the discriminant is positive, this
breaks down into more possibilities.
First off, if the discriminant is
positive, we're going to have real solutions
because it's a positive radicand.
But, because of the plus or minus,
we're going to have 2 of them.
The further breaking down that I mentioned
are; if it's a perfect square, then,
we can fully simplify the radical
and our solutions are actually
going to be rational solutions.
We'll have 2 of them because
of the plus or minus.
If it's not a perfect square, then we won't
be able to do away with the radical entirely
and we're going to have irrational solutions.
But again, 2 of them.
If our discriminant's a perfect square and
we end up with these rational solutions,
that means we could've factored
the whole thing to begin with.
And, the reason I get into this is, if you've
got to solve a quadratic equation and it turns
out the discriminant's a perfect square,
then, you can solve by factoring.
It's a quick check to figure out whether you
should even bother to trying to factor or not.
In any case, now let's get on to our examples.
For our first example we've got 2x
squared minus 5x minus 3 is equal to 0.
And, I'm going to rewrite the
quadratic formula for reference here.
So, I've got x equals -b plus or minus the
square root of b squared minus 4ac, all over 2a.
And, it might be good to
do some color coordination
when we try to flush out the formula.
So, I'm writing a in blue,
b in red, and c in green.
So, when I flush out the quadratic
formula I might think about that a is 2.
In this case, be is -5.
Yeah the sign of the number goes
along with that letter, okay.
And, we could finally say that c is -3.
Substituting for a, b, and c, we're going to
end up that x equals the negative of -5 plus
or minus the square root of -5 squared,
minus 4 times 2 times -3, all over 2 times 2.
Let's explore this discriminate that we have.
It's in evaluate to 25 that
negative negative is going
to be a plus because that's multiplication.
24 which is equal to 49.
Since that's a perfect square we know that
this is something that we could've factor.
But, we're instructed not to.
We're instructed to use the
quadratic formula for practice.
This is just for general
knowledge let's call it.
Anyway, when we simplify the quadratic
formula, we have x equals 5 plus
or minus the square root of, we've
already figured out the discriminant's 49,
so we might as well put square
root of 49 all over 4.
Continuing to simplify, we get
x equals 5 plus or minus 7, over 4.
And, this should be split up into 2
expressions because we can continue with these.
And, I'll show you the distinction in
a few minutes when we can't continue.
x is equal to now 5 plus 7, over 4 or 5 minus 7, over 4.
Taking each of these in turn, we
have 5 plus 7, over 4 is 12/4 reduces to 3.
5 minus 7, over 4 is  -2/4 reduces to -1/2.
So, x equals 3 or -1/2 are the
solutions to this quadratic equation.
Next, we have 3w times the
quantity of w plus 2 is equal to 5.
I'd like for you to get all the terms on the
same side, say it equal to 0, identify a, b,
and c, and try to solve this
using the quadratic formula.
So, here, I've written quadratic formula with
the a's and b's and c's, and for me to be able
to use it I need to know what a, b, and c are.
So, first thing I'm going to do is distribute
the 3w into w plus 2,  then I'm going to subtract 5
so that I have everything equal to 0.
And the result of that's going to be 3w
squared plus 6w minus 5 is equal to 0.
And, using the color coordination that we had
before, when I flush out the quadratic formula,
I'm going to write down -6
plus or minus the square root
of 6 squared, minus 4 times 3
times -5, all over 2 times 3.
Now, we're going to take a moment
and consider this discriminant again.
In this case, it's going to simplify and
become 36 plus 60 which then becomes 96.
96 is not a perfect square, but it is positive.
So, my solutions are going to be real.
I'm going to have 2 of them.
But, they will not be rational.
In other words, there's no chance
I could've factored the trinomial.
And again, just call it knowledge
of tool in your toolbox.
In any case, when we continue with simplifying
the expression, we have w is equal to
-6 plus or minus the square root of 96 all over 6.
And, you may be tempted to
try to reduce the 6's.
Don't do it because you can't divide out
a term you can only divide out factors.
So, our correct point of action is going
to be to simplify inside the radical.
And, as it happens, 96 is equal to 16
times 6, 16 being a perfect square.
So, that's going to come
out of the radical as a 4.
So, I have -6 plus or minus 4 times
the square root of 6 all over 6.
At this point, I want to reduce a fraction.
And, the way we do that is
we factor and then divide.
So, I am going to factor
out what I see is common.
In this case, that's 2.
So, I get 2 times the quantity of -3 plus or
minus 2 root 6, over 2 times 3 if you want to factor
that denominator even and dividing out
the 2's gives me w is equal to -3 plus
or minus 2 times the square root of 6 all over
3 which I would accept as a final solution.
If you wanted to further break this apart,
then, you might separate the numerator.
And, you end up with -3 and then over 3, right,
plus or minus 2 root 6 but that's also over 3.
So, that would become w equals
-1 plus or minus 2 root 6, over 3.
That's as simplified as we can get it.
And this is what I was eluding
to over the previous example.
There's no point in separating the plus or minus
because I can't further simplify
those answers if I do.
So, 1 fraction or 2, I'll accept
either of these as your final solution.
Next we have 12x minus 5 equals 9x squared.
Solve using the quadratic formula.
So, do what you've seen so
far, give this a shot.
So, what I've done is written down
the quadratic formula for reference.
And, I subtracted 12x and added 5 to both sides.
It doesn't matter if the
equals 0 is on the right
or on the left and I prefer it on the right.
So, the quadratic formula, x equals
-b plus or minus the square root
of b squared minus 4ac all over 2a.
I'm really just repeating that for
emphasis to help you try to memorize it.
We notice that a is 9, b is -12, and c is 5.
And, when I put those into
the quadratic formula,
I'm going to end up with x equals the
negative of -12 plus or minus the square root
of -12 squared minus 4 times
9 times 5 all over 2 times 9.
So, x is equal to 12 plus or
minus the discriminant is going
to be 144 minus 180, all over 18.
So, that discriminant becoming
negative means we have x equals 12 plus
or minus the square root of -36.
And, as soon as we know that
the discriminant is negative,
we know that our solutions
are going to be complex.
And, complex solutions always come in pairs.
x is equal to 12 plus or minus,
now, that square root of -36,
that's the 6i, all divided by 18.
And, whenever we have complex solution, this is
just a math thing, you've got to deal with it,
we're going to want that answer in a plus
bi form, a plus or minus bi form is fine.
But, it's technically referred
to as a plus bi form.
In other words, if you see
i's, split the fractions.
x is equal to 12/18 plus or minus 6/18
and then stick that i out as a factor.
So, 6/18 times i. I see these
fractions can be reduced.
We're going to divide 6 everywhere.
So, x is equal to 2/3 plus or minus (1/3)i and
that's how we want to see a complex solution.
In our next example, we've got (1/4)m
squared minus m plus 1/2 is equal to 0.
And, I really threw this one in just
to emphasize you don't ever need
to unless you really want to put
fractions inside of the quadratic formula.
Instead, what I'm going to do is multiply
by the least common denominator throughout
so I can clear all the fractions.
Then, I'm going to proceed as before.
So, why don't you pause the video
and do all that on your own?
So, I've written down the formula for reference.
And, after distributing the 4 to
both sides, we're going to end
up with m squared minus 4m plus 2 is equal to 0.
I see a is 1.
That's where we're always there when
we don't have a coefficient it's 1.
And, b is -4.
And, c is 2.
So, when I flushed out the formula, we
get, in this case, m equals negative
-4 plus or minus the square root of -4 squared
minus 4 times 1 times 2, all over 2 times 1.
Now, if you haven't already done so,
take a moment and exam the discriminant.
That discriminant becomes 16 minus 8 which is 8.
It's positive.
We're going to have real solutions.
There are 2 of them.
And, notice, since 8's not a perfect
square, they're not going to be rational.
So, m is equal to 4 plus or minus, we
already figured out the discriminant is 8.
So, the square root of 8 all over 2.
And now, we can simplify m
is equal to 4 plus or minus.
Remember 8 factors into 4 times 2,
4 being a perfect square means it's
going to come out of the radical.
We're leaving 2 in the radical.
So, 4 plus or minus 2 root 2, all over 2.
Now, I want to reduce this fraction which
means we're going to factor and then divide.
So, the numerator turns into
2 times the quantity of 2 plus
or minus the square root of 2 all over 2.
The 2's divide out.
So, we have m is equal to 2
plus or minus the square root of 2
is our solutions to this equation.
Finally we have what we call a literal equation.
We're supposed to solve a equals 2pi(r) squared
plus 2pi(r)h for r. Whenever we have a situation
like this and they give us a variable
to solve for, I'm highlighting that.
And, we pretend everything else is a number.
So, what I want to do is express this in the
form ax squared plus dx plus c is equal to 0
so that I can flush out this quadratic
formula with r playing the role of x. So,
I'm going to subtract a from both sides.
And, that'll give me 2pi(r)  squared
minus 2pi(h)r minus a is equal to 0.
And, comparing with the equation
right above it, we're going to see
that our coefficients are variables.
So, we have that a is 2pi, b is, don't
forget that negative, -2pi(h), and c is -A.
So, flushing out the formula, we get not x
equals but r equals the negative of -2pi(h) plus
or minus the square root of -2pi(h), 
the whole thing squared,
minus 4 times a which is
2pi times c which is -A,
based on that radical, all over 2a which is 2pi.
I took the time to actually write all that out
so that it wouldn't be really
cumbersome to look at.
And, what we'll do, if we
can, r equals 2pi(h) plus
or minus the square root
of 4pi squared h squared.
now, the -4 times -A is going
to become positive.
Throw in that 2pi, we have 8pi(A), all over 4pi.
Now, inside the radical, that discriminant,
folks we're not going to talk about positive
or negative or anything like that.
What I am going to point out is that
we have a factor of 4 in both terms.
So, we're going to factor that out.
And, it's going to end up looking
like r equals 2pi(h) plus or minus.
In the radical, we have 4 times the quantity of
pi squared h squared plus 2pi A all over 4 pi.
That 4's going to come out of
the radical as a factor of 2.
It's going to give me r equals 2pi
h plus or minus 2 times the radical
that has pi squared h squared
plus 2pi(A) ,all over 4 pi.
And, now we can factor out the
2's and then divide out the 2's.
I'll cut to the chase here.
r equals pi(h) plus or minus the square root of
pi squared h squared plus 2 pi A, all over 2 pi.
And there's nothing I can do to
further simplify this answer.
I don't like the answer.
But, I don't always have to like it.
This is that equation solved for
r. That equation, by the way,
if I can scroll all the way back
up here, happens to be the formula
for the surface area of a cylinder.
I know you care about that.
