Hello, and welcome to
Bay College's video
lectures for
intermediate algebra.
This is section 7.2,
where we introduce
the quadratic formula.
Now, in the previous
section, we learned
how to solve quadratics.
And we discussed how to
solve them using factoring,
the method of the
square root property,
and then we looked at
completing the square.
Now, if we have a quadratic
equation in standard form, ax
squared plus bx
plus c equal to 0,
we can identify the coefficients
of the x squared term, the x
to the first power term, and
the constant, so a, b, and c.
Now, we can use a tool
called the quadratic formula.
Now, honestly, I listed these
in this order for the reason
that factoring should
be your go-to method
to solve a quadratic.
If it factors, this
is the method to use.
If you can isolate
a squared factor,
use the square root property,
the square root method.
And if that fails, use
completing the square.
Only when a, b, and
c, your coefficients
in your quadratic
equation, are tedious
and maybe we don't want
to use the other methods,
this would be a last resort.
That's how I feel about
the quadratic formula,
because completing the
square will work every time,
just as the quadratic
formula will work every time.
The only problem with
this is, so many times,
we make sign errors.
When it comes to
mathematics, we're
all going to make sign errors.
And if we look at this formula,
we see we have negative b.
So we have to change the
sign of this coefficient
when we put it
into this formula.
That's a place where we
may make sign errors.
Then we have plus and
minus this quantity here.
There's an opportunity
to make sign errors.
Here, we have negative
4 times a times c.
Well, if a is negative
or c is negative,
we have the opportunity to make
a sign error when multiplying
a negative times a
negative times a negative.
So lots of opportunities
to have those sign errors.
And that's why I
really recommend
this to be a last resort to
solve a quadratic equation.
So if this quadratic formula
works every single time,
well where did it come from?
Who derived it?
Why does it exist?
And how do I know
it works every time?
Well, I had said completing the
square will work every time.
So what I'm going
to do for you is I'm
actually going to show you
how this formula was derived.
And you'll see that
completing the square
is the tool that I
prefer, because it
is what's used to
derive this formula.
So what we're going
to do is we're
going to take a quadratic
formula in standard form--
or not formula, but equation.
ax squared plus bx
plus c equals 0.
Now, if I were to
complete the square,
the first thing I want to do
is isolate the variable terms.
ax squared plus bx, I have to
subtract that constant of c.
It's just a number.
I move it to the other
side of the equation.
Now, to complete the square,
this coefficient has to be 1.
So I'm going to divide
all the terms by a so
that this coefficient is 1.
So now I have x squared plus
b/ax equals negative c/a.
Now we're ready to
complete the square.
Well, to complete
the square, we take
1/2 of the b term and square it.
Well, my b term
is now a fraction.
And sometimes that happens.
So if I multiply this term
together, I get 1b over 2a.
And I square that
factor, which is
going to give me b
squared over 4a squared.
I have to square all the terms
within these parentheses.
Now, even though this
doesn't look nice,
it is the value
that I need to add
to both sides of the equation.
So I'm going to add b squared
over 4a squared to this side.
And I'm going to add b squared
over 4a squared to this side.
What I do to one side of the
equation, I do to the other.
This is now a perfect
square trinomial,
because I completed the square.
So I know that it's going to
factor to a perfect square.
Now, if you look at
these coefficients,
they're nasty looking, right?
Well, what does this factor to?
Well, if you recall,
when we introduced
completing the
square, what's in here
before you squared it
is what it factors to.
So it's going to be x and
this value, b over 2a.
So I was able to factor
it without having
to do some major brain work.
We just have to refer back to
some work we've already done.
Now, hopefully,
we recall, if I'm
going to combine
terms like this,
it's a fraction
minus a fraction.
In order to subtract
fractions, they
have to have a
common denominator.
Well, this has a factor
of 4 and two a's.
This only has one a.
So to make a common
denominator, I'm
going to multiply the
negative c/a by 4a/4a.
I essentially just
multiply it by 1.
And that's going to give
me a common denominator
of 4a squared.
So I have b squared minus
4a times c [INAUDIBLE]
squared minus 4a times c, 4ac.
So now we have the quantity,
this perfect square,
x plus b/2a squared, equals b
squared minus 4ac all over 4a
squared.
Now I can use the
square root property.
If I use the square
root property,
I take the square
root of both sides
to give me the
factor being squared
equals the square root of this.
Well, when we introduce
a square root,
we have to remember
plus or minus.
And we can use the
quotient rule for radicals.
I could take the
square root of the top
and the square
root of the bottom.
Well, the square root of
the top doesn't simplify.
It's just the square
root of b squared
minus 4ac divided by the
square root of the bottom.
Well, this is a perfect square.
The square root of 4 is 2.
And the square root
of a squared is a.
Now I can solve for x.
To solve for x, I
need to subtract b/2a.
So I subtract b/2a
from both sides.
So we get negative b/2a plus
or minus the square root of b
squared minus 4ac all over 2a.
And since these have
a common denominator,
I'm just going to write them
over that same denominator.
And you can see that
x equals negative b
plus or minus the
square root of b
squared minus 4ac all over 2a.
This is our quadratic formula.
This is where it came from.
Somebody took the
time to complete
the square on the
general equation
to give us this formula.
Now, like I said,
there's opportunity
to make sign errors.
So if you're using
this, make sure you're
watching your signs.
Don't make those sign errors.
But this is how the
equation was derived.
So let's look at an example
where we can use it.
In order to use the
quadratic formula,
we need to have our equation
in standard form, which
means we set it equal to 0.
Well, if we look at this,
it's not set equal to 0.
So I'm going to
set it equal to 0
by subtracting 1
from both sides.
Now it's in standard form.
And maybe you want to move
these values to the other side
so you have some
positive coefficients.
Either way you do
it, it's just fine.
Now we're ready to use
the quadratic formula.
Well, if we have it memorized,
x equals negative b.
Well, a negative negative
6 is a positive 6.
You can see where
that opportunity
to make a sign
error can come from.
Plus or minus the
square root of b
squared-- well, b is negative 6.
Negative 6 squared is 36, minus
4 times a, which is negative 3,
times c, which is negative
1, all divided by 2a.
Well, 2 times a, 2
times negative 3.
So essentially, I just plugged
in my b term, my b term,
my a term, my c term, and
my a term into that formula.
Where I see those coefficient
values, I just plug them in.
I did a little bit
of simplifying.
Now we have to
simplify even more.
And if we look at all
those negative signs, lots
of opportunities to
make sign errors.
So I'm going to be very
careful with my signs.
So under this radical, I have
negative 4 times negative 3,
which is a positive 12.
Positive 12 times a
negative 1 is a negative 12.
2 times negative
3 is a negative 6.
Now I can do this math here.
And we get 36 minus 12
is going to give me 24.
So it's getting a little nicer.
And now we can
simplify this radical.
And I'll just do
that right here.
We're going to get 6
plus or minus-- well,
24 has a perfect square in it.
It would be 4, of course.
So 4, the square
root of that is 2.
And the remaining
factor would be 6.
Now that we have 6 plus or
minus 2 square root of 6
over negative 6, we still
need to simplify that.
6 over negative 6 is going to
be negative 1 plus or minus
2 over the negative 6
would be a negative 1/3.
But since it's plus
or minus, we'll
have both instances there.
So we'd get 1 square
root of 6 over 3.
So we have negative
1 plus or minus
the square root of 6 over 3
after we've simplified it.
And we can see, this
was a lot of work
and a lot of simplifying
and lots of opportunity
to make sign errors.
So we have our solution.
We should always
check it just to make
sure we didn't make any
errors along the way of using
this formula.
Now, I just want to
point out, I honestly
believe that, had I
completed the square on this,
it would have gone much faster.
And I would have come to
the same solution and less
opportunity to make
the sign errors.
But a lot of students, they
like the quadratic formula,
and they utilize it.
All right, let's look
at another example.
Here, we have x squared
minus 4x equals negative 4.
Well, to use the
quadratic formula,
I have to have this
in standard form.
And I do so by adding
4, in this case.
And now I can go ahead and
use that quadratic formula.
I can identify a, b, and c.
So x equals negative
b-- so I have
to change the sign
of that negative 4--
plus or minus the square
root of b squared-- well,
negative 4 squared is 16-- minus
4 times a, which is 1, times c,
which is 4, all over 2 times a.
So we're ready to
do some simplifying.
And this one isn't so bad,
because we have negative 4
times 1 times 4.
That's negative 16.
16 minus 16 is 0.
The square root of 0 is 0.
So x equals 4 over
2 times 1 is 2,
because plus or minus 0 isn't
going to change anything.
And then that's just a
fraction that I can simplify.
4 divided by 2 is 2.
x equals 2.
I found one solution here.
And we can check that easy
enough by plugging it in.
2 squared is 4.
2 times negative
4 is negative 8.
4 minus 8 is, in
fact, negative 4.
So it works in the
original solution.
So x equals 2 is my solution.
All right, let's look
at this example here.
Here, we have 2x squared
plus x plus 1 equals 0.
And if we were to complete
the square on this,
we'd have to divide
through by 2.
That would give us 1/2
here as a fraction.
Then we'd have to
take 1/2 of 1/2,
if we recall how to
complete the square.
And that could be tedious.
So maybe the quadratic formula
is the way to go for this one.
So what I'm going
to do is I'm just
going to use the
quadratic formula.
I'm going to identify a, b,
and c, and plug them in. well,
negative b would
be a negative 1,
because that coefficient is 1.
The square root
of b squared is 1,
minus 4 times a, which is 2,
times c, which is a positive 1,
all over 2 times a,
which is 2-- 2 times 2.
So now I'm going to simplify it.
I get negative 1 plus or minus
the square root of-- well,
negative 4 times 2 is negative
8, times 1 is still negative 8.
1 minus 8 is a negative 7.
Hopefully, we recall, a
negative under the square root
is in the complex number system.
That is a value of i/4.
So if we simplify that,
negative 1 plus or minus
i square root of
7-- we just take out
that square root of negative
1 and call it i-- over 4,
because this is an imaginary
number, a complex unit,
it is proper notation to
write it in a plus bi form.
Hopefully, we recall that
from the previous chapter.
So negative 1/4 plus or minus
the square root of 7/4 times i.
So we identify the
coefficient a to be
negative 1/4-- or the real
part-- and the coefficient of b
to be plus or minus
the square root of 7/4.
Hopefully, we remember our
imaginary complex number
system.
So we get negative 1/4 plus
or minus the square root
of 7/4 times i.
We can check this,
but for time's sake,
this is the solution.
It will work.
It would be a little
tedious to check that.
So what we notice here is,
of these three solutions,
we had one, the very first one
gave us two real solutions.
The second example we did
only gave me one solution.
And the last one I had gave
me two imaginary solutions,
the negative 1/4 plus the
square root of 7 over 4i
and negative 1/4 minus
the square root of 7/4i.
Now, there is a tool we can
use to determine what type
of solution we're going to
get and how many of them.
Are we going to get two real
solutions, one real solution,
or two imaginary solutions?
And what we can use
to determine that is
called the discriminant.
In the quadratic
formula, this is
the value under
that square root.
And it's that value that
determines the quantity
and type of solution
that we're going to get.
So using the discriminant, if
the value under the square root
is greater than 0, essentially,
if it's a positive value,
we're going to get
two real solutions,
like we saw in
the first example.
If the discriminant, what's
under that square root,
is less than 0, it's
a negative number,
we're going to get
two complex solutions,
two solutions containing an i.
And the last case here,
b squared minus 4ac,
that discriminant,
if it's equal to 0,
well, if we add or subtract
the square root of 0,
it doesn't change anything.
We're going to get
one real solution.
And this is what's called
a repeating solution,
because it came from
a squared factor.
So let's go back to
our first example
and see what we have here.
If we look at this
discriminant here,
the value that was under
the square root, when
we simplified, it
was a positive value.
This value is greater than 0.
We found two real
solutions, negative 1
plus the square root of
6 over 3 and negative 1
minus the square
root of 6 over 3.
Even though it's not a
nice integer solution,
it's still two real
values, two real solutions.
The next one, if we look at that
discriminant, if you recall,
this was 16 minus 16.
What was under the
square root was 0.
And I found one
solution, because this
was a perfect square.
If we were to recognize
that right away,
we could have factored it.
This is a perfect
square trinomial.
Perfect square trinomials
have one solution.
And that would be this would
factor two x minus 2 squared
equals 0.
Well, using the 0 factor
theorem, 2 minus 2 is 0.
So 2 would be the solution.
And that's what we found.
Obviously, factoring
would have gotten
us there a lot faster, right?
So that's a method
that you want to use
before you go into
the quadratic.
Don't jump right straight to it.
See if the other
methods will work.
So because the discriminant
was 0, we got one solution.
But it's a repeating
solution, because it's still
from two factors.
All right, and then,
here, well, when
we worked through
the quadratic and we
looked at the discriminant,
what was under that square root,
it was negative.
And hopefully, at
this moment, we
said, hey, this is in the
complex number system.
This contains an i.
And that's what we found
when we simplified it.
So we can use the discriminant.
If we're told to find the
real solutions to a quadratic,
we can check that discriminant,
that b squared minus 4ac.
And if we say,
hey, it's negative,
I don't even have to
solve this equation,
because it has no
real solutions.
It's solutions are complex.
So that's a tool we can use
to save ourselves some time
and maybe be a
little convenient.
And also, it gives
us the ability
to say, all right, if I can
assess this before I even
simplify it and
work through it, I
know I'm looking for
two real solutions, two
complex solutions,
or one real solution.
If I know what I'm looking
for before I find it,
I can rest assured that
I'm on the right track
when I get to that final answer.
So what I recommend
is do the homework.
This is a skill that you
only build with practice.
Watch those sign errors when
using the quadratic formula.
And good luck.
This is section 7.2.
Thank you for watching.
