On the discussion board, you probably
determined that this quadratic equation
would be the trickiest one to solve. It
is not easily factored, nor is it easily
converted into vertex form; and graphing
wouldn't give you the precise solutions.
So today we're going to talk about a
fourth method for solving quadratic
equations: and that method is called the
quadratic formula. To use the quadratic
formula, all you need are the "a,"  "b," and "c" values of your standard form equation.
This is the quadratic formula x is equal
to negative b, plus or minus, the square
root of the quantity, b squared minus 4
times a times c, all divided by 2a. And,
again, to use the quadratic formula all
you need to do is determine your "a," "b," and
"c" values, and plug them into the formula.
It may look complicated, but you will get
used to substituting your values into
the formula. Please note you do not need
to memorize this formula: whenever you
need to use it it will always be
provided. You simply need to know how to use it; however, before we start applying
the quadratic formula we do want to
understand where it comes from. And we
can derive the formula simply by
completing the square: so we're going to
do that using these two examples. On the
left you have a normal quadratic
equation, and on the right you have
standard form. The first thing we're
going to do on the left is factor out
our leading coefficient of 2. When I
factor the 2, out of all three terms I
divide all three of them by 2. The
expression inside the parentheses is
what remains when we factor the 2 out. If
I distributed the 2 to all three terms,
we would get back to the original
equation. We're going to do the same
thing on the right, and factor our
leading coefficient of a out of all
three terms. Ax^2 divided by A is
simply x squared... and you can see
that I've taken B and C and divided them
by A as well. Going back to the left, my
expression inside the parentheses is
multiplied by two and I want to get rid
of that two. So I can use inverse
operations and divide both sides of my
equation by two; that leaves me with the
expression that was inside the
parentheses, and conveniently zero
divided by any number is simply zero.
Zero divided by two is zero: now we can
go ahead and do the same thing on the
right: I divide both sides by A and we're
left with the expression inside the
parenthesis equal to zero, because zero
divided by A is simply zero. Now we can
start completing the square. To complete
the square, I know to take positive six
and divide it by two: that gives me my
perfect square, of x plus three squared.
But, three squared is equal to nine: so I
need to subtract nine to keep my
equation equivalent to my original
equation, and then I still add the five.
While you may not have solved it this
way in your class, this is a helpful way
of showing it given what we're doing
with the standard form. I simplify my two
constants and we get X plus three
squared minus four. I can look at my
algebra tile and get the exact same
equation: we have a perfect square of X
plus three but we only have five unit
tiles so we need to take away four since
it would have produced nine. And that's
represented right here. We're now going
to do the same thing on the right: we
know we have to take plus B/A and divide
it by two to get our perfect square. Just
like we divided 6 by 2, I divide that
by 2 and so I have B divided by 2a; but
just like we subtracted nine I need to
subtract B divided by 2a squared. We
still have our plus C a now we can start
solving
our quadratic equations. I'm going to use
inverse operations to move the four to
the other side I add 4 to both sides and
we have X plus 3 squared equals 4. I'm
going to do the same thing on the right
because both of these terms are simply
numbers, they just look complicated. And
at any point during this video, you can
pause it to think about what you are
seeing on the screen. I'm going to use
inverse operations and add B divided by
2A squared and subtract C divided by A. I
also decided to square my B so you can
see B squared and square
my 2a. So we have 2 squared is 4 and A
squared is A squared and then I'm going
to subtract C/A. I want to write this
side of my equation as a fraction so I
know that to add or subtract fractions I
need to have the same denominator, so I
can multiply C over a by 4a squared
divided by 4a squared, because that
equals 1... and I haven't changed my
equation. This a will cancel out one of
the A squared's. So we are left with B
squared minus 4 times A times C because
I just rearranged the 4 the A and the
C which I can do with multiplication. Now
we're going to solve: we're going to take
the square root of both sides, and the
square root of the quantity x plus 3
squared is X plus 3, and the square root
of 4 is 2. But we know that it's really
plus or minus 2, because negative 2 times
negative 2 is also 4. I do the same thing
on the right I take the square root of
both sides on the right. You can see that
I show that I'm taking the square root
of the numerator and the denominator.
Because I can simplify my denominator,
we have X plus B over 2a equals plus or
minus the
square root of b squared minus 4ac,
because I cannot simplify that, divided
by 2a
Lastly, we write our solutions on the
left: we have X is equal to negative 3
plus or minus 2 because I used inverse
operations to move the 3 to the other
side. I could simplify this into two
solutions but that's not the point of
the video. On the right
I moved the positive B over 2a to the
other side by subtracting it and you now
see the quadratic formula. This is one
way of seeing it but you may also see it
written this way because you have two
fractions with the same denominator, so,
the numerators were combined in the
right example, and this is the form I
typically use. Now that you know where
the quadratic formula comes from we'd
like you to practice applying it to
solve. An equation this equation, is
provided on your capture sheet the first
thing you should do is figure out what
your a B and C values are so go ahead
and fill in number 1 on your capture
sheet with the a B and C values pause
this video while you do so since there
is no number in front of x squared we
know that a equals 1 we have one x
squared B is positive 5 and we know that
in standard form we add C so since we
see minus 14 we know C is negative 14 on
your capture sheet you have the
quadratic formula and you can modify it
by substituting these values into it in
step two. Go ahead and do so now pause
the video while you do. Your formula
should now look like this; wherever you
saw B, A or C, you substituted these
values into the formula. It is important
to include these parentheses around A
and C to remind you that you are
multiplying four times one times
negative 14. At this time
simplify the discriminant: the part that
is underneath the radical sign, and the
denominator in step number three. Pause
the video while you do so. Your formula
should now look like this; please note
that when you're subtracting four times
one, times negative 14 you actually have
a negative times a negative,
hence plus 56. The simplification of that
25 plus 56 is 81, so at this point we can
now create two equations to solve. For
our solutions, we have negative five plus
nine which is the square root of 81
divided by two and we also have negative
five minus nine divided by two we
simplify those expressions to get our
final answers of x equals two and x
equals seven of seven one way to check
our work would be to substitute each one
of those one at a time into the equation
and when you do so your answer should
equal zero at this time complete this
process on your capture sheet and move
on to the next activity.
