>> So far we've solved
the quadratic equation,
which of course has this
form, either by factoring
if it was factorable, by square
rooting both sides if it was
in a correct form, or
by completing the square
and then getting it ready
to square root both sides.
Now we're going to do it
by the quadratic formula.
So, what is the quadratic
formula?
It's basically a formula
that helps us find solutions
to this quadratic equation
rather quickly by plugging
in numbers into a formula
and getting solutions.
How to get the quadratic
formula?
Well, we get it by completing
the square method used
on this general equation.
So, let's get started.
We're going to start with this
general quadratic equation.
We're going to move the
constant to the other side.
Remember; leave some space
because you're doing
completing the square.
The leading coefficient
is a, we want it to be 1,
so we divide everybody by a.
Now the leading coefficient is 1
and we have negative c
over a. Now we are ready
to do the half business
from completing the square.
So, we're going to take half
of the coefficient of x,
multiply across b over 2a.
Remember, we save this guy and
then we immediately take it
and square it, so
we get b squared
over 4a squared and
we use this guy.
Well, how do we use it?
We add it to both sides
and when we do that,
we can factor the left side into
2 binomials that look identical,
x and x. And remember the guy we
saved is always the second term
in my binomial and since
it's completing the square,
both binomials look the same.
On the right, I need to get a
common denominator, so I'm going
to multiply top and
bottom by I guess a 4a
to make the denominators
look the same,
so I'm able to add those 2 guys.
That's going to be
negative 4ac over 4a squared
and b squared over 4a squared.
So, I have 4a squared
downstairs, same denominator,
and up above just put
the numerators together
because they're not like
terms and you cannot add them.
Now, I am almost there.
I have a binomial
squared equals--
I'm going to put the b squared
first because I just like it
to start with a positive term.
Notice now I'm ready to
square root both sides
because now it looks
like the goal that I was
after to get a binomial squared.
So, I have x plus b over 2a that
pops up and then I have plus
or minus b squared minus 4ac
up on top and the bottom pops
out a 2 and an a. There's a link
here, so I can't have a b pop
out because the b squared
is linked with the 4ac.
For the same reason, I
cannot have a 2 pop out.
Now I'm going to isolate x.
So, move this to the other side
and I have x equals
negative b over 2a plus
or minus this business.
And since they have
a common denominator,
I can put them together.
Negative b plus or
minus square root
of b squared minus
4ac all over 2a.
And this most of you
probably remember
as the quadratic formula,
but basically where it comes
from is just completing
the square.
So, if you have the
quadratic formula memorized,
you basically plug in
the a, b, and c values
from your quadratic equation
and get your 2 solutions
after you simplify.
So, this is the main thing
we'll be using in this section,
the quadratic formula.
So, we're going to solve
using the quadratic formula.
First thing I have to do is
put this in standard form.
You can't get your a, b,
and c unless you have
the 0 on one side.
So, a is 2.
B is 11 and c is negative 6.
Make sure you pay
attention to signs.
Formula says x is equal
to negative b plus
or minus square root of b
squared minus 4ac all over 2a,
so I'm going to go
plug those in.
plug everything in into b into
b and to a into c and into a
and don't do anything in
your head to make mistakes.
Now go simplify these.
So, negative times negative is a
positive, so I get a positive 48
and add it 169, which
is a perfect square,
so I have negative 11 plus
or minus 13 pops out over 4
and that will give me 2 answers.
Negative 11 plus 13 over 4 or
negative 11 minus 13 over 4.
Once I simplify those I
get 1/2 and negative 6,
so those are my 2 solutions
to this quadratic equation.
In other words, if I were
looking at this parabola,
these 2 would be
the 2 x-intercepts.
Let's solve this using
the quadratic formula.
I have to get it in standard
form first, so I'm going
to distribute and then I'm
going to combine like terms
and I'm going to move the 5
while I'm at it to get my 0,
so I'll have my a to be 1.
My b to be 2 and my
c to be negative 5,
then I go plug them in.
So, I have negative 2 plus or
minus 2 squared minus 4 a is 1
and c is negative 5
all over 2 times a. So,
I simplify everything
and I'm up to here,
but 24 could be further
simplified, so I'm going
to break up 24 as 4 times 6
and I have all of that over 2.
This one pops out a 2, the
6 remains and now I see
that they're all divisible by 2.
So, I really have
negative 1 plus
or minus just the 1
radical 6 as my 2 solutions.
Next thing we're going to look
at is analyzing the discriminant
in the quadratic formula, which
is really just this piece.
So, we're just going to analyze
some signs of the discriminant.
B squared minus 4ac.
If this is positive, I'm
going to get 2 real solutions.
Like the examples
we've just done.
If the discriminant is less than
0, in other words, negative,
I'm going to get 2
complex solutions
out of my quadratic equation.
And the last case of b squared
minus 4ac is equal to 0,
hen I'm going to get
one real solution
because this piece completely
disappears and I have negative b
over 2a, which would
be the vertex
of the parabola being my only
solution and if I'm thinking
of it graphically, also
that one x-intercept only.
So, let's look at an example.
It says use the discriminant
to analyze whether this parabola
has 1, 2, or no x-intercepts.
In other words, when
I make y equal to 0
to find the x-intercepts,
instead of going
and finding the x intercepts,
let's just use the discriminant,
b squared minus 4ac to
predict what might happen,
what kind of x-intercepts
I may have.
So, there's a, b, and c. So,
b is 1-- 10 squared minus 4.
A is negative 3 and
c is negative 5.
Let's see what that gives me.
One hundred.
That's going to give me a
positive 12 times a negative 5,
so 100 minus 60, which
looks like it gives me 40,
and as long as that's a positive
number, it looks like I'm going
to have 2 real solutions.
So, I'm going to
have 2 x-intercepts
for this parabola
if I go look at it.
Let's look at another example.
Given the graph of this
parabola, whatever a, b,
and c may be, we don't know.
Here's the parabola.
Is a positive or negative?
Well, since this thing opens
up, a must be positive.
That's all we have to answer.
A is positive because
my parabola opens up.
Part b says solve this equation.
Well, if I'm setting
the equation equal to 0,
I'm making y equal
to 0, so I'm looking
for x-intercepts in my graph.
Well, when I go to the x-axis,
there's no way this parabola is
ever going to touch the x-axis
because it keeps going up and up
and up in the other direction,
so there are no x-intercepts and
if there are no x-intercepts,
this equation has no real
solutions because I wasn't able
to find where it
crosses the x-axis.
Part c says is the discriminant
positive or negative.
Well, if I'm looking
at the discriminant,
which is b squared minus 4ac,
and if I'm getting no solutions
because that's what
the graph tells me
and that's what part b tells
me, if I have no x-intercepts,
that must mean that b
squared minus 4ac is negative
since I have no real solutions
coming out of the equation
in part b. So, even though I
don't know what a, b, and c are,
I can tell all of this
information visually.
First of all, my
parabola opens up.
Second, there are
no x-intercepts,
and if there are no x-intercepts
that means my equation
doesn't have real solutions.
If it doesn't have
real solutions,
it must have complex solutions.
If it has complex
solutions, then that must mean
that the discriminant
is negative
because I don't see
any x-intercepts.
Solve by any method.
So, any method-- if
I can factor easily,
I always choose that first.
And it looks like if I put this
in standard form
it is factorable
and if something is factorable,
might as well go ahead and pop
out your solutions right away.
So, this I chose
to do by factoring,
just because it was an
easy problem to factor.
Part b, again, I'm not going to
use the quadratic formula, a, b,
and c. Why go through that?
If I just isolate the x-squared
piece, notice what happens.
I can easily use the square
root property and get x
to be plus or minus 3.
So, I did this by
square root property just
because it was easier
to do it that way.
Now, part c, if it's any method
and it looks a little ugly,
I'm just going to
choose to go ahead
and use the quadratic formula
because quadratic
formula basically comes
from completing the
square anyway,
so that was my third
option usually,
but now the quadratic formula
makes things easier for me.
I don't like those fractions.
Because it's an equation, I'm
going to multiply everybody
by the LCD, which is 4, so I get
a 2x squared plus a 3x equals
negative 4.
I've got to put it
in standard form,
2x squared plus 3x plus
4 equals the 0 I need.
So, now I have my a to be 2,
my b to be 3, and my c to be 4
and I'm going to
go plug them in.
So, I got them plugged
in, now I'm going
to go ahead and simplify.
OK and it looks like I get
a 9 minus 32, which is going
to be a negative 23
and notice that's going
to create complex solutions,
so I have negative 3 plus
or minus-- I'm going to pull
out an i. The 23's going
to remain all over 4.
There's nothing more
to simplify,
so I ended up with
2 complex solutions,
which means if I am looking
at the graph of this parabola,
if I set this equal to
y equal to my parabola,
then my graph will have no
x-intercepts because I ended
up with complex solutions.
So, solve by any method.
I'm going to go ahead and use
the quadratic formula again
because I would rather just plug
in some values really quickly,
but I need to get everything
in standard form first
and get a 0 on one side.
So, 5x squared plus 13x
plus 6-- actually minus 13x.
So then my a is 5, my b is
negative 13 and my c is 6.
I'm going to go plug those in.
So, we plug them in.
Notice it's a negative part
of the quadratic formula
and then a negative
also part of my b,
so double negatives are
going to make a positive.
Looking good so far, 169
minus 120, that gives me a 49,
which is a perfect square, so I
have 13 plus or minus 7 popping
out over 10 and that's going
to give me 2 solutions.
So, once I simplify each piece,
I get 2 solutions, 10 and 3/5,
so those must be my x-intercepts
if I were to graph this
and make it look
like a parabola.
And since I have 2 real
solutions and they both look
like rational numbers
and not any square roots
and irrationals involved,
it means this problem was
factorable to begin with,
but I just went ahead and
used the quadratic formula.
