Hey, guys.
Welcome to Algebra I.
Today's lesson
is going focus on
solving quadratic equations
algebraically.
Do you remember when you learned
how to factor trinomials?
That will come in handy
when doing these problems.
You ready? Let's go.

(Describer) Standing at a screen, she gets a stylus.

Okay.
Before we start
factoring quadratics,
I want to jog your memory--
er, solving quadratics--
I want to jog your memory
about how to factor trinomials
completely.

(Describer) The screen is shown.

So, remember, a trinomial
in standard form
is written in this form:
ax squared plus bx plus c.
In this case, we have
x squared plus 5x plus 6.
And if you remember, when we
wanted to factor a trinomial,
we started out by finding
factors of that c term,
in this case, 6.
So, we'd come off to the side.
Let's get the pen. We can
come off to the side and work.

(Describer) She writes.

We would start listing
the factors of 6.
For example,
1 times 6 will give me 6,
2 times 3
will give me 6.
That's it as far as whole
numbers go for factors of 6.
Now we'd look back
to the trinomial,
find the factors of 6
that would combine
to give us our middle term.
We're focusing
on a coefficient.
In this case,
we want a +5.
Play around with the numbers,
and I'm sure
it jumps out to you
that you need this pair here.
If this were a +2
and this a +3,
multiplying those together
would give me +6.
Adding those together
would give me +5.
I come back to my trinomial
and write my factors
as (x+2) and (x+3).

(Describer) ...in parentheses.
...in parentheses.

And I would have factored
the trinomial completely.
Now, solving quadratic equations
algebraically,
one of the strategies
involves factoring.
I wanted to make sure
that you remembered
how to do that process.

(Describer) She changes the screen.

What we're getting into now--
a quadratic equation.
It's basically
a trinomial in standard form,
but it's set equal to 0.
The solutions to a quadratic
are the values of x
that, when you substitute them
into a quadratic,
it gives it a value of 0.
We also call the solutions
to a quadratic "zeros,"
because if you substitute
their values in there,
your quadratic
will have a value of 0.
They're also called "roots."
They have one more name
I'll tell you later.
For now, think about them
as zeros or as roots.
This is a quadratic equation.

(female)
Standard form

(Describer) Standard form of a Quadratic Equation: a x-squared plus b x plus c equals zero.
She changes the screen to show Example 1: x-squared plus x minus 20 equals zero.

of a quadratic equation:
ax squared + bx + c = 0.
Okay. Let's keep going.
Let's solve one. Okay.

(female)
x squared + x - 20 = 0.

(instructor)
The first step
when solving a quadratic
equation algebraically,
try factoring, right?
I'm ignoring
that negative temporarily,
if you think about
how we would factor.
Let's focus on the 20.

(Describer) She writes it.

If you need more
factoring practice
so that these problems
don't become too intense,
check out that old lesson
to review factoring trinomials.
For now, let's keep going.

(Describer) {for now let's keep going.
She draws two lines.

So, factors of 20:
1 and 20, 2 and 10, 4 and 5.

(Describer) She writes under the lines.

I know that
I'm trying to get -20,
so one of these factors
needs to be positive,
one's gonna be negative
for the right pair.
I'm looking for the pair
that will combine
to give me a +1.
If you do your mental math
and search through these,
you'll find that it is
this last pair.
You'll need that 4 to be
negative and that 5 positive.
If you multiply those,
-4 times 5, that's -20.
And if you combine those,
-4 plus 5, that is +1.
So, now you'll go back
to your quadratic,
and let's rewrite the left side
as a product of its factors.
We're going to write it as

(x-4)(x+5) = 0.

(Describer) ...in parentheses.
...in parentheses.

How we actually figure out
what the values of x
are going to be
for the solutions
to this quadratic--
let's think about something
for a second.
Let's just think
about properties.
Remember that if we have--
Let's just pretend--
we'll come off to the side.

(Describer) She writes.

We have
a times b equals 0.
Just playing around
for a second.
If I multiply two numbers
together and the answer is 0,
that means that one of these
numbers has to be 0, right?
We use that same thinking
in order to figure out
the solutions to our quadratic.
Because these are
two factors, right?
This is one factor,
this is another factor.
If I'm multiplying
these two factors together
and getting 0,
that means that either
this first factor is 0
or the second factor is 0.
In other words,
that means either x-4
is going to have to equal 0,
or that means that x+5
is going to have to equal 0.
One of those factors
is going to have to equal 0,
if when I multiply them
together, the answer is 0.
What we have to figure out
is these values of x.
We solve--I call them
little mini equations
or one-step equations--
We'll solve each one for x
and see.
If we solve that left one--
we'll add 4 to both sides,
so x equals 4.
And if we solve the right one,
we're gonna have to subtract 5
from both sides,
so x equals -5.
These are our two solutions
to this quadratic.
If x is 4, this quadratic
will have a value of 0.
If x has a value of -5,
it will have a value of 0.
Both of these make up
what we call the "solution set"
to our quadratic.
And let me show you
how we represent that.

(Describer) She moves the screen.

Let's get this out of our way
for a second. Okay.
If you're asked to represent
your solution
using set-builder notation,
this is what that means.
If you stopped here,
you're not necessarily wrong,
depending on how you
must represent your solution,
but you need to be familiar
with representing your solution
in set-builder notation.
Basically, it means
"add some braces"
for this kind of problem.
If my solutions, or my roots,
are 4 and -5,
then I'll represent that
as [4,-5].

(Describer) ...then a comma.
...all within braces.

I can look at that and tell,
"Okay, the solutions
to this quadratic
are x is 4 and x is -5;
one or the other."
All right, let's keep solving
a few more quadratics.

(female)
Example 2 is
x squared - 3x - 10 = 0.

(Describer) Example 2 is x-squared minus 3 x minus 10 equals zero.

Okay, so,
find the solution set.
If you remember
the first thing we did,
try to factor it, right?
Let's come off to the side--
factors of 10.

(Describer) She writes.

1 and 10, 2 and 5.
I need a -10,
so the pair I need,
one will be negative
and one positive.
I need them to combine
and give me a -3.
That means I need
the second one.
+2 and -5
is what I'm gonna need.
I'm gonna rewrite
the left side
as a product of its factors.
So, (x+2) and (x-5)...
equals 0.
Then I'm just gonna
separate these
and solve
the two mini equations.
Remember, I called them that.
So, x plus 2 equals 0
and x minus 5 equals 0.
So, if I solve the left one,
subtract 2 from both sides...
so x equals -2.
Then if I solve the right one,
add 5 to both sides,
x equals +5.
If I represent that
in set-builder notation...

(Describer) She writes a left brace.
...then a comma.
...then a right brace.

[-2,5].

And you're all done.
You factored it.
You solved the mini equations.
You got your solution set
for your quadratic. Okay?
Let's keep going.
You should
try a couple on your own.
Press pause and work through
these problems.
To compare your answers
with mine, press play.

(Describer) Title: Solve the quadratic equations.
Number One: x -squared plus 2x minus 8 equals zero.
Number Two: x-squared minus 7x plus 12 equals zero.

(female)
Title: Solve
the quadratic equations.
Number one:
x squared + 2x -8 = 0.
Number two:
x squared - 7x + 12 = 0.

(instructor)
Let's see how you did.
For the first one--
let's switch
to the pointer tool.
I'll move this.

(Describer) She uncovers the answer.

Your solution set was [-4,2].
And for the second one,
the solution set was [4,3].

(Describer) She uncovers it.

If you want to see how
I got these, let me show you.

(Describer) The screen just shows the first equation.

Okay. So, I started out
by factoring.
Switch back to my pen.
Let's get the factors of 8:
1 and 8, 2 and 4.
I needed a -8,
so the pair I need,
one will be negative,
one positive.
I need them to combine
to give me a +2.
That means I need this pair,
the -2 and a +4.
If I combine those,
I will get a +2.
If I multiple those,
I will get a -8.
I'll rewrite the left side
as a product of its factors,

(x-2)(x+4) = 0,
and I'm gonna split those up
into the mini equations
and solve those.
So, x minus 2 equals 0,
x plus 4 equals 0.
Okay. Now, I'll solve.
So, +2 plus 2, x equals 2,
and -4 minus 4,
x equals -4.
So, that's how I got
my solution set
of [-4,2].
You may have even started
to pick up on a pattern.
Whatever your factors are,
like how that was x-2,
the solution is a +2.
And that factor was x+4.
The solution is a -4.
That's another relationship
between the factors
and solutions.
They're the same numbers
but with opposite signs.
Another pattern.
Let me show you
how I got that second one.

(Describer) She changes the screen.

Okay. So, begin
by factoring the 12.
So, 1 and 12,
2 and 6, 3 and 4.
I need a +12,
and I need the numbers
to combine to give me a -7.
Whichever pair
is the right pair,
both of the numbers
are negative.
If you run through these,
you'll see it's the last one.
You need a -3 and a -4.
I'm gonna rewrite
the left side
as a product of its factors.
So, (x-3)(x-4) = 0.
I'm gonna split this up
and solve the mini equations.
So, x minus 3 equals 0,
x minus 4 equals 0.
Okay?
Add 3 to both sides here.
So, x equals 3.
And add 4 to both sides here.
Okay. So, x equals 4.
And there is
your solution set, [3,4].
Okay?
All right,
before we leave quadratics,
there's one more thing
you need to know
about solving
quadratic equations.
If you find somebody
that took Algebra I,
I bet they might remember
a little thing called
"the quadratic formula."
Before I show it, I warn you
that it looks intense,
but when you
start running through problems
using the quadratic formula,
it just breaks down.
It's easy to work with.
So, a disclaimer first.
You have to have
a quadratic in standard form,
and "a" cannot equal 0.
Basically, you have to have
a leading coefficient
is what this is saying.
If you have that situation,
besides factoring,
another method you could use
to solve your quadratic equation
is to use the quadratic formula,
and here it is.
Okay. Your solution's "x."
They equal -b
plus or minus the square root
of b squared minus 4ac
all over 2a.
Okay? Take a deep breath.
Let's say it one more time.
So, quadratic formula:
x equals -b
plus or minus the square root
of b squared minus 4ac
all over 2a.
So, in the situation where we're
using the quadratic formula,
we identify a, b, and c
in our quadratic equation,
we substitute it
in this formula,
and we can get our solutions
that way as well.
Let me show you some examples.
Seeing a quadratic formula,
you're probably thinking,
"Why would I ever want to use
that formula?
It looks intense.
Why don't I just factor?"
Sometimes your numbers,
they don't lend themselves
to be factored easily.
Maybe you'll have a problem
that involves decimals,
or it involves fractions,
or it involves numbers
that are quite large
or quite small.
Or you may be asked to show
that you know how to use
the quadratic formula.
You can always choose
whichever way you want to use.
I have to make sure
that you know how to use
the quadratic equation.
When I took Algebra I,
my teacher made me memorize
the quadratic formula,
and I still know it now.
But anytime you need to use it,
write it down
to help you memorize it
and keep everything organized.
So, before I get going,
I'm going to write
the quadratic formula,
which I have memorized
from way back in the day,
but you'll have to look
back and forth.
Before you know it,
you'll have it memorized too.
So, x is gonna equal -b
plus or minus
the square root
of b squared minus 4ac
all divided by 2a.
So, we're gonna use
the quadratic formula
to solve this equation.

(Describer) The next equation is x-squared plus 7x plus 10 equals zero.

(female)
The next equation is
x squared + 7x + 10 = 0.
First, let's identify
what a, b, and c are
in our quadratic equation.
Remember, when we don't see
a coefficient,
there's an imaginary one.
Right?
I know that a is 1,
b is 7, and c is 10.

(Describer) She writes them all down.

Okay, half the battle
you have taken care of
right there
by just identifying
a, b, and c
in your quadratic equation.
Now it's all
just about substituting
and keeping everything
organized,
and then you'll be fine, okay?
Let's get more space up here
and start doing that.
So, x = -b, and b is 7.
Anytime you
substitute a value in here,
use parenthesis.
I'm going to put -(7).
Write the x so you can follow
what I'm doing.
Plus or minus the square root,
b squared, and b is 7,
so 7 squared
minus 4 times a times c.
Okay, so a is 1,
c is 10.
Make that a little longer.
Come down a little bit.

(Describer) She scrolls down.

And this is all divided by
2 times a,
and a was 1.

(Describer) She writes 2 times 1.

So, let's get things
a little more spaced out here
so it doesn't become
overwhelming,
or any more overwhelming
than it probably already is.
Let's just bring that
down a bit.
Keeping it spaced out will help
you keep everything straight.
Okay. Now it's about
just cleaning up this monster
that you see in front of you.
Let's get the pen back.
So, -b,
that's basically like
a -1 times 7.
Don't think too much into it.
It just became -7.
Plus or minus.
All right.

(Describer) She draws the radical sign.

Once we get that cleaned up
underneath that radical,
it's gonna make everything
seem so easy.
I already see I'm getting
a little scrunched in here,
so I'm just gonna move this down
a little more.
Okay.
So, 7 squared,
I know that's 49. Right?
Let's get the pen back here.
All right. So, 49 minus--

(Describer) She writes under the new radical.

Okay, so 4 times 1, that's 4.
4 times 10, that's 40. Okay?
So, 4 times 1 is 4,
4 times 10 is 40.
So, this is 49 minus 40.
All right. More space.
And this is all divided by
2 times 1.
2 times 1 is 2. Okay?
It's already looking like
less of a monster.
We've got it cleaned up.
So, let's keep going.
Keep simplifying.
A little more space here.
All righty.
We've still got that -7
plus or minus the square root--
So, 49 minus 40,
that's just 9. Right?
And this is all divided by 2.
Right? Okay.
Gonna keep scrolling down here
so you can keep following
my steps.
Now we're gonna clean up that
square root of 9, simplify that.
So, I have -7 plus or minus--
I know the square root of 9
is 3,
so plus or minus 3,
all divided by 2.
Now you're at the point
where you split this up
into two cases.
You've heard me over and
over again say, "Plus or minus."
That literally means one time
we're gonna add on the top,
and one time
we're gonna subtract on the top.
That's how we get
our solution set.
So, I'm gonna write an arrow
to say this implies
-7 plus 3 divided by 2.
We're gonna get that answer.
Let's scroll down.
Then we're also gonna get
the subtraction case.
We're gonna get -7 minus 3
divided by 2.
That's what
that plus or minus means.
The first time,
we're gonna add -7 and 3.
Second time,
we're gonna subtract -7 and 3.
So, -7 plus 3,
that's -4 divided by 2.
-4 divided by 2.
That's -2. Right?
Now the subtraction case.
So, -7 minus 3,
that's -10 divided by 2.
-10 divided by 2.
That's -5.
Okay? There we go. There are
our two values for "x."
That's our solution set.
So, we could write
our final answer,
if you wrote it
in set-builder notation,

[-2,-5].

And there you go. Okay?
Those are the steps
we would take
to use the quadratic formula
to solve a quadratic equation.
Let's do one more together
to get you
more comfortable with it.

(Describer) The next equation is 3x-squared minus 11x minus 4 equals zero.

(female)
The next equation is
3x squared - 11x - 4 = 0.
Here we go.
We're gonna use
the quadratic formula
to solve
this quadratic equation.
Okay? So first,
I'm gonna write the formula.
So, x equals -b
plus or minus
the square root
of b squared minus 4ac
all over 2a.
All right. We've got
the formula written down.
Now let's come off to the side
and identify a, b, and c.
In this case,
a is 3, b--
because it's subtract 11,
we're gonna take that sign
with that coefficient.
So, b is -11,
c is -4.
Okay? All right.
Half the battle's already done.
We wrote down a formula,
we've identified a, b, and c.
Let's start substituting
and cleaning things up.
All right. Scroll a little bit,
get a little space.
I do want to keep things
spaced out.
Okay, so -b.
So, -11
plus or minus the square root
b squared,
that's -11 squared
minus 4...
a is 3,
c is -4.
Okay. Now let's
scroll down a bit.
And this is all divided by 2a,
and I believe a was 3.
Yes. So, 2 times 3.
Okay. We've got everything
substituted in there.
This is where it starts to look
kind of monstrous,
but we're gonna clean it up
and everything will be fine.
Let's get more work space here.
Okay.
The opposite of -11.
That's how you can
kind of interpret that.
Opposite of -11 is +11,
plus or minus the square root--
And I'm noticing things
are getting crowded,
so I'm just going to scoot
this down a little bit.
Okay. And get the pen back.
All right.
So, -11 squared. That's 121.
I'm gonna take
that negative sign with the 4
to help me do the math.
So, -4 times 3, that's -12.
-12 times -4, that's +48.
Okay? So, plus 48.
And that is all divided by
2 times 3,
which is 6.
Okay. Let's just
keep cleaning this up.
Keep scrolling.
Keep breathing.
Okay.
All right.
So, we have 11
plus or minus the square root--
121 plus 48 is 169,
and that is all divided by 6.
All right.
Looking better, looking better.
Keep simplifying.
A little more space
up here. Okay.
So, we have 11 plus or minus--
Okay, the square root
of 169, that's 13.
All divided by 6.
Once we've gotten here,
we're at the place
where we do addition
and then we do subtraction
to get the two solutions.
Okay.
So first, 11 plus 13
divided by 6.
Then we'll do 11 minus 13
divided by 6. Okay?
All right,
so 11 plus 13 is 24,
divided by 6.
24 divided by 6, that's 4.
Okay.
And then the subtraction case,
11 minus 13, that's -2.
-2 divided by 6.
We can reduce this fraction,
and it will reduce to -1/3.
Okay? And there we go.
We've got our solution set.
So, let's write it
in set-builder notation.
Our solution set is [4,-1/3].
All right?
Okay, I do believe
it's time for you to try one.
So, use the quadratic formula
to solve
this quadratic equation.
When you're ready to compare
answers with me, press play.

(Describer) The equation is 2x-squared plus 5x plus 3 equals zero.

(female)
The equation is
2x squared + 5x + 3 = 0.
Let's see what you got
for your solution set.
If you got [-1,-3/2],
then you're good to go.
Let me show you how I got that.
Remember, we were using
the quadratic formula,
so we always begin
by writing it down.
Let's get the pen back first.
I think I'll go
over here this time.
So, x equals -b
plus or minus
the square root
of b squared minus 4ac
all over 2a.
And in this problem,
our a is 2,
b is 5, c is 3.
Now let's just start
substituting.
So, -b would be -5,
plus or minus the square root,
b squared would be 5 squared,
minus 4,
a is 2, c is 3,
and this is all divided by
2 times a, so 2 times 2.
Okay. Just substituting should
have looked something like that.

(Describer) Minus five plus or minus the square root of 5-squared minus 4 times 2 times 3, all over 2 times 2.

(female)
-5 plus or minus
the square root of 5 squared
minus 4 times 2 times 3
all over 2 times 2.
All right, let's keep going,
Keep simplifying.
Come down a good bit.
The opposite of 5,
so that's -5,
plus or minus the square root.
5 squared, that's 25,
minus 4 times 2, that's 8,
8 times 3, that's 24.
So, 25 minus 24.
All divided by 2 times 2,
so that's 4.
We should start feeling good
when we get about right here.
It's starting to look simpler.
Let's keep going.
Now our goal is to simplify
underneath the radical there.
We'll have -5 plus or minus,
25 minus 24 is 1,
so that's just the square root
of 1 divided by 4.
Let's kind of move that
down a bit.
Starting to get crowded.
There we go.
Make this page longer. Okay.
All right. Now, the square root
of 1, that's just 1.
Get the pen back here.
So, we have -5
plus or minus 1
divided by 4.
Okay. And this time I'll do the
two cases just right underneath.
Because we're at the point
where we need to see
what -5 plus 1 divided by 4 is,
and we need to see what
-5 minus 1 divided by 4 is.
Okay.
So, -5 plus 1, that's -4,
divided by 4.
-4 divided by 4, that's -1.
-5 minus 1, that's -6.
Divide it by 4, and this
is a fraction you can reduce.
It will reduce to -3/2. Okay?
That's how we got
our solution set
of [-1,-3/2].
All right?
Okay, good job, guys.
You are now very familiar
with how to solve quadratic
equations algebraically
by factoring or by using
the quadratic formula.
Hope to see you back soon
for more Algebra I. Bye.

(Describer) Accessibility provided by the U.S. Department of Education.

Accessibility provided by the
U.S. Department of Education.