In this video we're
going to look at examples
of solving quadratic equations
using the quadratic formula.
Here's the quadratic formula.
You will have to
have this memorized.
It gives you solutions to an
equation that's in this form.
This is called standard
form for quadratic equations
which means you have all the
terms collected on one side
of the equation, and
zero on the other side.
Intermediate algebra students
and higher should understand
where this formula comes from;
how we solve this equation
with the A, B, and C by
completing the square,
and we end up with
the quadratic formula.
I'm not going to cover
that in this video.
We're going to look at what
a quadratic equation is,
and how to use the quadratic
formula to find the solutions.
A quadratic equation always has
a term with a variable squared.
It usually has a term with
a variable to the 1 power,
and it usually has
a constant term.
These last two could
be missing or zero,
but the squared term
is always there.
That means that the
coefficient A is never zero,
but B or C could be zero,
and those are the
only possible terms.
Here's a couple examples
of quadratic equations.
We are going to solve
examples 1 and 2.
3 and 4 I just included
so you could see a
couple other examples.
The first two are
pretty straightforward.
You have your squared terms,
you have your X to the 1;
your Y to the 1 terms, and
you have your constant terms.
The variable can
be anything as long
as you have the same variable
throughout the whole equation,
so X, Y, T. Standard form
over here on the right.
I collected all the terms
on one side of the equation,
and the coefficients
give me my A, B,
and C. A represents the
coefficient of the squared term,
so it's 1 in this case.
Be represents the
coefficient of X-- negative 4--
and C is the constant,
which is negative 3.
3 is interesting because
there's a squared term,
but there is no T
to the 1 power,
so its coefficient is zero.
B equals zero.
And number 4-- sometimes you'll
see equations that may not look
like quadratic equations
until you simplify them.
You would distribute
this X to both
of these terms giving you 2X
squared, X, subtract the 145
to get the standard form,
and there's your A, B,
and your C. So, let's go ahead
and look at those two examples.
The steps we will be using--
the first step is always
to write the equation
in standard form if
it's not already.
Then you can identify the
values of A, B, and C,
the coefficients;
plug into the formula,
and then we will
simplify our solutions.
That usually involves
several steps.
That can often be the
hardest part of the problem,
and we will look
specifically at how
to do those steps
in these examples.
All right.
Here was example 1.
This was not in standard
form, because I've got one
of the terms on one side of the
equation and the other terms
on the other, so the easiest
way to fix this would be
to subtract 3 from both sides.
And that gives me X squared
minus 4X minus 3 equals zero.
That is our standard form.
Now, I can identify
my A, B, and C values.
A-- coefficient of
X squared is 1.
B is the coefficient
of X, negative 4,
and C is the constant
which is negative 3.
That gives me A, B,
and C. Remind you
of what the formula looks
like as I plug it in here.
My variable is X, so we're
going to have X equals
[ Silence ]
This is one big fraction here.
So, in the numerator I start
with opposite of B. So,
opposite of negative 4 plus
or minus the square root
of B squared, negative 4 squared
minus 4 times A times C. A is 1.
C is negative 3.
And the denominator
is 2 times A.
So here's what the
quadratic formula looks
like in this example.
Now we will start simplifying.
I'll start by simplifying
the double negative.
So we get a positive for--
be careful with the
square root here.
The first one will always be
positive because it's a square,
so 16 is negative 4 squared.
Here we have a double negative.
So, 4 time negative 3 minus
negative 12, or plus 12,
and the denominator is 2.
[ Pen Scribbling ]
So we're at 4 plus or minus the
square root of 28 all over 2.
And here's where I
want to remind you
of the steps we were looking at.
We've plugged into this formula,
we've kind of simplified the
initial pieces, and now we're
at the point where we need
to simplify the radical
before we try
to cancel any common factors.
It's very important.
All right, it's a common
mistake at this point to try
to cancel the 2 into the 4.
That would not be okay, because
there's actually two separate
terms here.
Whatever we want to cancel
we have to be able to factor
that out of the two terms.
And before we can do that, we
need to simplify the radical.
So, if you only cancel
with one of them,
you're not doing it correctly.
Let's simplify this
radical first.
Square root of 28--
you might need
to review simplifying
radicals at this point--
largest perfect square that
divides 28 is 4-- 4 times 7.
Split up the radical, and that
gives you 2 times the square
root of 7.
So I'm going to 2 root 7 to
replace the square root of 28.
[ Pen Scribbling ]
And now I can look to
cancel common factors.
Let me rewrite this.
[ Silence ]
First we'll identify which
factor we want to cancel
out of the three terms here.
In this case we're going to
cancel a 2 out of each of these.
So, identify the common factor
you want to cancel first,
factor it out of the top.
[ Pen Scribbling ]
And it's not idea to factor
it out of the bottom as well.
[ Pause ]
Factoring the 2 out of the
numerator-- 4 divided by 2 is 2,
plus or minus, and 2 root
7 divided by 2 is 1 root 7.
So there's the 2 factored out of
the top and out of the bottom,
and I'm just left
with 2 times 1.
There's not much
to factor there.
Now I have a factor of 2
in the top and the bottom,
and I can cancel that.
So we are left with 2 plus
or minus the square
root of 7 all over 1.
We can drop the 1
in the denominator.
Those are the two solutions of
this equation, and this is fine.
You can use this compact
notation for your solutions,
but be aware that this
represents two separate answers.
2 plus root 7, and
2 minus root 7.
The 2 minus root 7
would be the smaller;
2 plus root 7 would
be the larger.
So, if you were using your
solution set notation,
two separate solutions.
These, by the way, are
examples of irrational solutions
because we have the
square root of 7.
Any time you have a
square root in the answer
after you've simplified it,
you have irrational solutions.
All right, let's look
at the next example.
Again, we have an example
that's not in standard form.
Step 1 was to write this
equation in standard form.
I like to have my squared
term be positive, so I'm going
to add the 2Y squared
to both sides--
move everything to
the right side.
That gives me zero equals
2Y squared plus Y minus 1.
And my A, B, and C-- we
now get positive 2 for A,
positive 1 for B, and negative
1 for C. So you might ask
at this point what if
I had moved everything
to the left side
of the equation?
Subtracted the Y; added the
1-- that's perfectly fine,
you could solve it that way.
You would get-- if you did it
carefully you would get the
exact same answers we're
going to get right now.
The difference would be each of
these would have opposite sides.
You would have a negative 2,
a negative 1 and a positive 1.
I just think it's a little bit
easier if the A is positive.
So, plugging into this equation.
Here it is again.
Our variable this time is Y, so
we're going to have Y equals--
opposite of B is
now negative 1 plus
or minus 1 squared minus 4
times 2 times negative 1--
all over 2 times 2-- 2A.
So, here's what the
quadratic formula looks
like in this example.
Let's go ahead and simplify.
We have our negative 1 plus
or minus square root of 1.
Again, we have a
double negative,
so this becomes plus 8 all
over 2 times 2 which is 4.
The inside of the radical is 9.
So now we're at the
point when we wanted
to simplify the radical before
we try to simplify the fraction.
Because we have a
perfect square,
simplifying the radical
is just going to be--
evaluating the square root
of 9, which is 3 all over 4.
We can simplify this
answer further than we could
in the previous, because we
have like terms in the top.
In the last example, we
had a square root of 7
and we couldn't combine
that with the other number.
You do not want to leave any
plus or minuses in your answer
when you have a nice
whole number here--
when the radical disappears.
You're going to simplify
this down to two answers--
two separate numbers:
Negative 1 plus 3 all over 4--
[ Pen Scribbling ]
-- and negative 1
minus 3 all over 4--
[ Pen Scribbling ]
-- you do not want to leave a
plus or minus in your answer
when you have these like terms.
You can combine these.
So that becomes positive 2
over 4, which is one-half,
and the second one
becomes negative 4
over 4, which is negative 1.
So our solutions here are
negative 1 and one-half.
That's the solution
set brackets.
These are examples of
rational solutions.
These are rational numbers--
a fraction and a whole
number-- or an integer.
So that wraps up the two
examples that I wanted
to cover in this video.
I'd actually like to finish
this video by discussing kind
of the connection between this
method and the factoring method
that we've seen previously.
If you look at this example,
we started with 2Y squared
plus Y minus 1 equals zero--
or that's what we had
for our standard form.
Now, if you imagine if we
had been asked to solve this
by factoring-- and you really
have to pay attention, you know,
at this point now
when you're asked
to solve a quadratic equation,
which method you're asked
to use when you solve it.
If we tried to factor this,
here's the two factors we'd get.
I'll leave that to
you to double check.
And each of those factors, then,
using the zero product
rule, gives us a solution.
We said each one equal to zero.
So, one-half, the negative
1, the same solutions we got
from the quadratic formula.
But there's a connection
between these solutions
and these factors, which
you can clearly see how--
when we solved for these.
When you have rational
solutions, like this,
you're going to have nice
factors with coefficients
that are rational
numbers or integers.
That's what we usually
consider factorable.
So there's a connection
between the solutions
of the equation and the factors.
Let's compare both
of the examples here.
Here's our two examples.
Here is the one with the
nice rational solutions.
That one was factorable.
Here's the one that had
the irrational solutions.
And that's one that we
say is not factorable.
We call that an irreducible
quadratic.
I like to just add
the condition though--
it's not factorable with
rational coefficients.
That's because to factor this
trinomial, you would have
to have one factor for
each of these solutions.
Just like the last
one had one factor
for each of these solutions.
But in this case, the factors
would have these irrational
numbers in them.
That's something we'll
do in college algebra,
but at this level, we
consider that not factorable.
We could not have
solved this by factoring.
I have two reasons for
discussing factoring here.
One is just to try to
connect some different ideas
and techniques from
different parts of the course
and show you some of
the bigger picture.
I think that's important.
And secondly, forget about
solving equations for a minute,
and think back to when we
were factoring trinomials.
Sometimes we'd get stuck
and think maybe a trinomial was
not factorable, but it was hard
to be sure, and this will
give us a test that we can use
to determine if a trinomial
is factorable or not.
Because if you remember, when
we were using the quadratic
formula, the thing-- the
part that determined whether
or not we got rational solutions
or irrational solutions
was this square root.
If we had a square root left
in our answer, they
were irrational.
If we didn't, they
were rational.
How do you know whether you're
going to have a square root left
in your answer or not?
This expression, B
squared minus 4AC.
That's-- that whole thing is
underneath the square root.
That has a special name.
It's called the discriminant,
and the discriminant
in the first case was 9.
I'll leave that for you to
double check or to go back
and take a look at that, but
we had the square root of 9,
and that turned into a 3 which
gave us the nice solutions.
And in the second example,
our discriminant was 28.
Our B squared minus 4AC was 28,
which reduced into square root
of 7-- or simplified
into square root of 7.
So, perfect square, factorable;
not a perfect square,
not factorable.
So that can be summarized
with this.
Here's the definition
of the discriminant,
and the corresponding
trinomial can only be factored
if the discriminant
is a perfect square.
There are other things we can
say about the discriminant based
on whether it's positive
or negative and how
that affects the
solutions, but we'll look
at that in the next video.
