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>> Shannon Gracey: Hi there,
my friends.
We are in the last section
of Chapter Six.
We are almost done with --
in Beginning Algebra
or Introductory Algebra,
whatever we're calling it.
This is Section 6.6
from Robert Blitzer's
Introductory
and Intermediate Algebra
for College Students,
and I am Professor Shannon
Gracey from MiraCosta College.
Today we'll be talking
about solving quadratic
equations by factoring --
so, so important,
when you're done
with your homework,
you should be able
to use the zero
product principle.
Notice zero product.
No negative one product.
No 256 product.
There's only a zero product
because if you recall,
zero times any number is zero,
so there's a very,
very powerful tool
that we can use called Hey,
the Zero Product Principle
to solve quadratic equations.
We will use our factoring
skills to solve these
equations,
these quadratic equations,
and then we'll solve some
applied problems using
quadratics, so go ahead.
Pause the movie
and do the movie, and we'll --
I'll be right back.
[ Silence ]
All right.
Let's see how you did
on the warm ups,
so the very first one factor,
so if we do our friendly X,
we have an overall product
of positive 7
and negative 8 is what we get
when we combine like terms,
so positive means
that we have the sign,
which means this is the result
of addition.
So the two numbers we come
up with are negative 7
and negative 1.
So here we go.
This yields a result
of X squared minus,
7Xs minus 1X, plus 7,
so we have our middle term has
been broken up to
in two negatives, 7X minus 1X,
and we have our red group.
Uh oh. That wasn't red at all.
We have our red group,
and we have our blue group
so when we factor the greatest
common factor
from the red group,
which is X, we are left
with X minus 7
and then be careful.
We'll factor out a negative 1
and be left with X minus 7,
so this was the results
of the red group here
and the results
of the blue group here.
Now our --
we did come
up with a common binomial
factor of X minus 7,
so to complete the
factorization we will have a
result of X minus 7 times X
minus 1.
How'd you do?
I bet you kicked butt
on that one.
Oh no, I said
that out loud didn't I?
Guys, I apologize.
It's late,
and Professor Gracey is having
a glass of wine along
with the lesson
tonight [chuckles].
Okay, so here's kind
of a blast from the past.
This is I believe a Chapter
Two concept.
We're solving a linear
equation and one variable,
so remember our goal here.
We've got an X to the 1,
and we have no other variables
so we want
to isolate the variable.
We will add 7 to both sides,
and we will get X is equal
to 7 and roster notation,
meaning that 7 is the only
answer that will work
for our variable X. I bet you
did a fabulous job
on the warm ups.
How do you feel about those?
I mean after all those hard,
hard problems we've been
working on,
I bet that was a lot easier.
Okay, here we go,
definition
of a quadratic equation.
A quadratic equation --
oops -- in X is an equation
that can be written
in the standard form.
[ Silence ]
AX squared, plus BX plus C --
and remember we were dealing a
lot with that in factoring,
but now there's an equal sign,
and it's equal to zero,
where A, B,
and C are real numbers
with A not equal to zero.
Otherwise we would not have
something that was quadratic
in form.
A quadratic equation in X.
[ Silence ]
Is also called a second degree
polynomial equation in X.
[ Silence ]
All right, so here we go.
Solving quadratic equations
by factoring,
so consider the quadratic
equation X squared,
minus 8X plus 7, equals zero.
How is this different
than the first warm up?
So if you recall,
in the first warm
up we had X squared minus 8X
plus 7, and we were --
our job was to factor it,
so what do you think?
How is it different?
Well good, yes.
There's an equal sign.
[ Silence ]
That means, you know,
we'll be able to solve
for the variable X. Aren't you
excited [chuckles]?
Okay, so here's how you do it.
We can factor the left side
of the quadratic equation X
squared minus 8Xs,
plus 7 to get X minus 7,
times X minus 1.
If a quadratic equation has a
zero on one side and a --
and by the way,
can I just emphasize that zero
on one side and a --
or a factored expression.
[ Silence ]
On the other side it can be
solved using the Zero
Product Principle.
All right,
the Zero Product Principle.
If the product of two
or more algebraic expressions
is zero, notice zero, zero,
only zero,
no other number [chuckles] are
you getting a clue
as to mistakes people make?
It has to be zero on one side
and a quadratic on the other.
Then at least one
of them is equal to zero,
so in Mathese what we have is
if A times B equals zero,
then A must be zero and,
or B must be zero.
So, okay, don't flip your
page yet.
Hang on. Let's check it out.
Let's go back to our warm up
and let's --
instead of keeping it just
to factor, let's equate it
to zero, so it's a new
problem, a new problem
that we've equated to zero.
Now, what you found
from your warm up is
that the left side factors,
so it's X minus 7 times X
minus 1, and that's equal
to zero.
Now what that Zero Product
Principle says is that hey,
this factor --
I mean notice that each
of these are factors, right,
that I've kind of grouped,
tried to carry.
They're each factors,
so what that means is
that either X minus 7 is zero,
or X minus 1 is zero, or both.
Hey, that works, too,
so if you solve, you know,
adding 7 to both sides,
which I think we did already.
You get X is 7 and then
on the second one you would
have -- you'd need to add one
to both sides, so make sure,
you know, your skills are good
with your solving linear
equations and one variable
and you get X is 1,
and then hey
in our roster notation,
meaning that only a result
of X equal 1 are X equals 7
will satisfy the
quadratic equation.
So you've solved your first
quadratic equation.
How do you feel?
I bet you feel
fabulous [chuckles].
Here we go.
Solve the following equations,
so I will leave you.
Pause the movie and do
at least A and B,
and if you feel very,
very confident go ahead
and do Part C, as well.
On your mark.
Get set. Go.
[ Silence ]
Okay. Let's see how you did.
Part A we need to add 11
to both sides.
We have a quadratic equation
and one variable,
so we will get 2X is equal
to 11, and then if I multiply
by 1/2, then I will get 1X,
which is X, is equal to 11/2,
and then over here I will need
-- I will write my --
and my roster notation,
I'll write the results
of 11/2s and put my happy box
around the solution.
Part B I need to subtract 1
from both sides.
Uh oh. There we go,
and then I will get X is equal
to negative 1,
and then in the roster
notation we will get
negative 1.
Now, okay,
so that was earlier stuff,
Chapter Two stuff,
which hopefully you feel
like you have mastered.
Part C you're just adding
another teeny little layer.
It's just a little,
little bit that you're adding.
Now you use that --
what was it called?
It starts with a Z. It was
a property.
You got it, the Zero Property.
We're using the Zero Property
to solve something that's
going to end
up being quadratic in form,
so here I've got, you know,
each of these are factors,
and they're equal to zero
so therefore
to solve this equation,
I would set 2X minus 11 is
zero, or X plus 1 is zero.
Didn't we just solve those
up there?
So notice that each
of those had a single solution
for X. Well now,
they were multiplied together
and they were set equal
to zero, so the solution
for these is going to be both,
so I put them
in ascending order,
so negative 1 and 11/2,
so you end
up with the two solutions
found the same way
as we found up there.
I mean, do you see this was 2X
minus 11 is zero
and that's what we found here,
so we used our Zero Property
to solve a quadratic.
I bet you feel fabulous right
now [chuckles].
All right.
Here we go.
All fun aside.
Here's some steps
for solving quadratic equation
-- a quadratic equation
by factoring.
If necessary,
rewrite the equation
in standard form,
and this is important.
A lot of times people get
so excited they decide to make
up their own 2 property
or negative 8 property.
They forget to get zero
on one side of the equation.
All right,
so this standard form is AX
squared plus B,
times X plus C, equals zero.
There's got to be zero
on one side, moving all terms
to one side thereby obtaining
a zero on the other side.
Factor, starts
with an F. It's not a
bad word.
I promise.
Apply the Zero Product
Principle setting each factor
equal to zero.
Solve the equation set forth
in the last step,
and it's always, always,
always advised
to check the solutions
in the original equation.
I'm going to leave the
checking to you guys.
All right,
but that's a great last step.
Okay. Here we go.
Solve. All right.
Can I just warn you guys
about something?
Your exam is going
to be comprised
of Chapters Five and Six.
The last page
of your exam has you solve
quadratic equations.
You will not find a value
for X. In the other ones
you'll be factoring,
so don't make up your own math
and start solving equations
earlier in the exam.
It's only the last page,
I promise you.
Okay. A lot
of times people get so excited
with their factoring skills
and then now their new solving
equations skills using the
Zero Product Principle
that they decide
to apply the Zero Product
Principle to everything.
Don't do that please.
Please don't do that.
Okay, here we go,
but now we can
because we have an equal sign,
and we have zero on one side
and guess what?
We have one factor,
two factors, so we say "Oh,
X must be zero,
or X plus 9 must be zero.
Now we are
down to a linear equation
and one variable,
so this one's done.
The X is isolated.
This guy we got to subtract 9
from both sides,
and we get a result
of X is equal to negative 9,
so the result
for the overall quadratic
equation will be negative 9
and zero.
You put your happy box
around your answer
and you're done, okay.
Okay, next step factor.
Factor. Factor,
and most importantly zero,
so we're good to go.
That means that 8 is zero,
or X minus 5 is zero,
or 3X plus 11 is zero.
Okay, now I told you I'd had a
little wine, but even I know
that 8 is not zero,
so guess what?
That must not be the zero
factor, and we throw it out.
All right.
Now coming to this guy,
if I add 5 to both sides,
I will get a result of X is 5
and then let's complete
this guy.
Subtract 11 from both sides,
and we'll get 3X is
negative 11.
Now I need
to multiply both sides by 1/3
and I will get 1X,
which is just X,
is equal to a negative 11/3,
so at the end
of the day my answer is
negative 11/3 and 5.
Both of those will satisfy
the equation.
Okay. Here we go.
Now we're getting
into the meat and potatoes
of what's going on.
Are these factors?
No. Good. These are terms.
We have not gotten these
into factored form,
so what you need to check for,
there's zero on one side,
so I've got a zero
so that's awesome.
Okay, but here you know what?
It starts with F. It's not a
bad word.
You got it.
We need to factor this bad
boy, so you're going
to use your mad
factoring skills.
There's an overall product
of negative 42, a difference
of positive 1.
The two numbers
that fit the bill are negative
6 and positive 7.
I'm going to factor
by grouping.
We will get X squared minus 6X
plus 7Xs minus 42 equals zero.
My middle term was rewritten
like so.
I've got my red group
and my blue group.
Sound familiar?
Now factoring
from my red group I get X
times X minus 6,
and then plus 7 times X minus
6, equals zero.
See that equals zero is the
different part
of what's going on,
so completing the factoring I
have a common binomial factor
of X minus 6.
I will factor that out.
I have X minus --
plus 7 equals zero.
Now what is that property?
You got it,
the Zero Product Principle.
[ Silence ]
Okay, so what does that mean?
I can set each factor equal
to zero, and I will --
on the first one I'll get X is
equal to 6 or X is equal
to negative 7,
so in my roster notation I've
got negative 7,
6 and those two solutions
for X satisfy the equation.
Okay, now D I got
to tell you right is a bit
of a sucker problem.
People want to divide out Xs.
Don't do it.
You need to get zero
on one side.
At this point
in your math life,
just get zero on one side.
That will help you,
so we will have X squared
minus 8Xs is zero.
Factoring,
we'll get X times X minus 8
is zero.
Using the Zero Product
Principle we'll have either X
is zero or X minus 8 is zero.
Again, that is the Zero
Product Principle.
[ Silence ]
And solving --
the first one is
already solved.
To isolate X
in the second guy we just need
to add 8 to both sides
and we get an end result
of either X is zero or X is 8.
Awesome. Okay,
now if you haven't done
so already, I would advise
that you pause the movie
and try these on your own.
Just remember you need
to have a zero on one side
and you need
to have everything
in factored form, so again,
if you haven't done
so already, pause the movie.
See how you do.
All right.
On your mark, get set, go!
I know you can do it.
[ Silence ]
Okay. Let's see how you did.
Part E, I don't have zero
on one side.
That's all I know,
so I'm going to subtract 12X
and add 9.
You can do it one at a time
if it -- if you
like that better.
I will end up with --
all right,
so this will give us
in descending powers of X --
we'll have 4X squared minus
12Xs, plus 9,
is equal to zero.
Now when we make our X
over here,
the overall product --
be careful,
is 36 because we're doing the
4 times the 9.
That gives us the 36,
and then we have a middle
product of negative 12 and,
you know, this is positive
so we're going to have a sum
to get that middle term,
a sum of two negatives,
so the two numbers
that fit the bill I believe
will be negative 6
and negative 6,
so we will have 4X squared,
minus 6Xs, minus 6Xs plus 9,
equals zero and then doing,
you know, our two groups we've
got our red group
and our blue group,
so if I factor
out I can factor out a 2X
from the red group
and I'm left with 2X minus 3,
and then if I factor
out a negative 3 here I'll be
left with 2X minus 3
and that's equal to zero
so then factoring
out we now have a common
binomial factor of 2X minus 3,
so when we factor
that out we're left
with 2X minus 3,
so this one was a perfect
square trinomial.
This is the one we factored
out, so notice that, you know,
you can go
through the whole shebang,
but notice you're going
to get the same result,
2X minus 3 is zero
or 2X minus 3 is zero,
so we'll just solve one
of them.
We'll basically have an answer
that's got a multiplicity
of 2, which you'll learn
about when you go
into Calculus,
so we're adding 3
to both sides
and we'll get 2X is 3
and then dividing both sides
by 2, we'll get X is 3/2,
so at the end
of the day we get one answer
of 3/2 that makes it work.
Again, I'll leave the checking
of the result
to you License to kill.
You are not spies.
You're math students,
so you do not have a license
to kill, especially you do not
have a license
to kill your math stuff.
So this is a very,
very deceptive problem.
We actually need
to multiply all this stuff
out, move that bad boy seven
over and get zero
on that side,
so let's go ahead and get zero
on that side first
so then we'll get X plus 3,
times 3Xs plus 5,
minus 7 is zero.
Now in order to factor
as I'm sure you well know,
we need to multiply all this
stuff out, so we're going
to have X times 3X plus 3,
plus 3 times 3X plus 3.
I'm using the distributive
property minus 7 equals zero.
We will get 3X squared plus
3Xs, plus 9Xs plus 9,
minus 7 is zero.
We'll get 3X squared plus
12Xs, plus 2 is zero.
All right,
got zero on one side.
Now, what do we need to do?
Starts with an F. It's not a
bad word.
Factor, you got it.
Overall product of 6 --
and you know,
I bet something went
wrong here.
Let's see what I did
because it was --
oh, you know what,
it wasn't 3X plus 3.
Thank you.
I'm going to keep it in there
and put it in red.
This was supposed
to be 3X plus 5.
This was a 5.
This was a 5.
This would give me 3X squared
plus, 5X, plus 9X,
plus 15 minus 7,
so I will get 3X squared plus
14Xs, and then I'm going
to have 15 minus 7,
which is 8.
Sorry about that,
so here we go.
Overall product of 24,
positive 24 and a sum
of positive 14, so it seems
to me like positive 12
and a positive 2 fit the bill,
so we will have 3X squared
plus 12Xs, plus 2Xs,
plus 8 is zero,
so that middle term --
the new middle term, 14X,
was split into these two.
We've got our red group,
and we've got our blue group
and here we go.
We can factor out a 3X
from the red group,
and we're left with X plus 4.
If we factor out a 2,
positive 2,
from the blue group we're left
with X plus 4
and it's all equal to zero.
We have a common binomial
factor of X plus 4.
Here we go, so we will end
up with 3X times X plus 4.
Nope. I apologize, not 3X.
We will have X plus 4 is
factored out.
[ Silence ]
And then we'll be left
with 3X plus 2
and that's all equal to zero
and then we'll get X plus 4 is
zero, or 3X plus 2 is zero
and then we need to subtract 4
from both sides for this guy,
and we get X is negative 4
and over here if we subtract 2
from both sides we will get
3Xs is negative 2.
Now we got to multiply it
by 1/3, and we will get 1X,
which is X, which is equal
to negative 2/3,
so our end result is negative
4 and then negative 2/3 Okay.
Here we go.
X cubed minus 4X is zero.
Factoring we get X times X
squared minus 4 is zero.
We're not fully factored.
Be careful.
You got a difference
of squares,
so you'll get X plus 2 times X
minus 2.
It's all equal to zero,
so X is zero,
or X plus 2 is zero,
or X minus 2 is zero,
and then subtracting 2
from this guy,
adding 2 from this guy we will
get X is negative 2
and X is positive 2,
so our three solutions are
negative 2,
zero and positive 2
and we're done.
Well, sort of done [chuckles].
We've got one more practice
problem, and then we've got an
application
and then we'll be done
with Chapter Six.
Woo hoo. Okay.
We've got a binomial all
right, so we've got something
that's considered
to be quadratic in form,
so this is you know
on the harder end.
Personally I'd rather just
factor it as is.
Some people use substitutions.
You've got 1 times something
squared, plus 2 times the same
something minus 8 is zero.
Okay, so let me write
in the one, so if you were
to make your X your overall
product would be 1 times
negative 8,
which is negative 8
and then you'd have a
difference of 2 so isn't
that going to be negative 2
and positive 4
to get the right
numbers there?
Let me do the same colors I've
been doing.
Okay, so here we go.
What does that mean?
So we will have X minus 3
squared instead
of two these X minus 3's we'll
have minus 2 of the X minus 3s
and them plus 4
of those X minus 3's
and then we're going
to have our negative 8
and that's all equal to zero.
So our middle term was split
into these two terms, okay,
so just hang in there.
So now we will get X minus 3,
the quantities square --
oh, I apologize.
We need to start factoring I
guess, so here's our red
group, and here is our
blue group.
So what did we factor
out from our red group?
Good, a factor --
one factor of X minus 3.
What are we left with?
Well, we've got another X
minus 3 and then we've got a
minus 2, okay.
All right and then
from the second one I believe
we can factor out a 4,
and then we're left
with an X minus 3 minus a 2.
Okay, so this was our red
group and this was our blue
group, but it was a little
more complicated than normal.
If you don't mind,
I'm going to do one more step
to simplify the inside,
so we'll have X minus 3 times
-- X minus 3 minus 2 --
do you see is X minus 5.
Then we'll have a plus 4
and again,
we'll have another X minus 5
and that's equal to zero,
so now we have it,
this beautiful common binomial
factor of X minus 5,
so we'll get X minus 5
and that's going
to be multiplied
by X minus 3 plus 4.
That's all equal
to zero still,
so then simplifying
that X minus 3 plus 4,
so this again,
was our common binomial factor
and now I'm simplifying X
minus 3 plus 4 to be X plus 1
and that's equal to zero.
I now use my new zero
property, and I get X minus 5
is zero, or X plus 1 is zero.
When I add 5 to both sides
on this guy,
I will get X is 5,
and on the other one
when I subtract 1
from both sides I will get X
is negative 1 and at the end
of the day I will have
negative 1 and 5
as my solutions
to this quadratic equation.
Okay, so here we go
with our application.
An explosion causes debris
to rise vertically
with an initial velocity
of 72 feet per second.
The formula H equals negative
16 times T squared plus 72T
describes the height
of the debris
above the ground H in feet,
T seconds after the explosion.
How long will it take
for the debris
to hit the ground?
So the ground is
at height equals zero,
so here we go.
We would set negative 16T
squared plus 72T equal
to zero, believe we can factor
out a negative 16T
and we get 1
and this would be --
let's see, 16 times --
oh maybe I can't do 16.
Sorry, 16 times 5 is 80,
so I don't think I'll be able
to do 16.
Let's try negative 8.
I can always refactor
if it doesn't work,
so negative 8 would give me 2T
squared and then I'd have
to do negative 9T is zero.
That looks good,
so we would get negative 8 is
zero, not unless we've been
drinking, right?
So we cross that out, or --
and this is no longer --
this is -- ahh guys, okay.
[ Silence ]
Okay, so at height zero is
when you hit the ground,
so the way we would set this
up is that we would say zero
equals negative 16T squared
plus 72Ts.
So zero equals --
I'll factor out a negative 8T.
I'll be left with 2T minus 9,
so either negative 8T is zero
or 2T minus 9 is zero,
so T would be zero,
or 2T would be 9
and multiplying both sides
by 1/2 we would get T as 9/2.
All right,
so here it doesn't --
you know it doesn't really
make sense at time zero,
right, for the debris
to hit the ground,
so I think that we would go
with the result of --
actually we can say it
in words.
We would say
that it will take...
[ Pause ]
...the debris.
Nine halves and that's
in seconds to hit the ground.
[ Silence ]
All right.
When will the debris be 32
feet above the ground?
Well, what we would do is we
would set the height to 32,
so instead
of the height being zero,
we would say 32 equals
negative 16T squared plus 72T,
but we need zero on one side,
so we'd have
to subtract the 32
from both sides,
and we would get zero is
negative 16T squared plus 72T
is minus 32
and then solving we'll get
zero is.
We'll have to factor.
Let's see.
What goes into all
of these guys?
I believe 8 still goes
into all, and I'll factor
out a negative 8,
so this would give me 2T
squared minus 9T plus 4,
so zero is negative 8.
I'll use trial and error,
since I've got a prime,
and my other factors --
I need a negative 9T
in the middle,
so I think if I do 4
and 1 it'll work,
and this would be a negative
and a negative, so negative T
and negative 8T is negative
9T, perfect.
So I'll get negative 8 is
zero, which we know is not
true, or 2T minus 1 is zero,
or T minus 4 is zero,
using our zero property,
so that would be 2T is 1.
T is 1/2 or T is 4,
so to answer our question,
the debris will be 32 feet
above the ground.
[ Silence ]
After half a second
and 4 seconds, because it goes
in a parabolic arc, so it --
so there's two points
where it'll often --
where you'll get the same --
where you'll get something --
a number that will satisfy the
solution --
satisfy the equation.
All right, so --
oh boy we are done
with Chapter Six.
Have a fabulous,
fabulous evening.
It is 9:11,
and I hope all is well.
Bye.
[ Silence ]