BAM!!!
Mr. Tarrou.
In this video we are going to talk about solving
quadratic equation.
That is equations with a power or a degree
of two by Completing the Square.
Now when you solve equations whether Quadratic
or not, linear equations, cubic equations,
you can solve them with graphing...
We are of course going to focus on second
degree equations.
You can graph them to find out where they
cross the x axis, you can solve them with
factoring, you can solve them with Completing
the Square, and you can solve them... at least
second degree equations... you can solve them
with the Quadratic Formula.
Great thing about the Quadratic Formula is
it is going to give us a solution to this
Quadratic equation regardless of what kind
of value it is.
It could be a whole number answer which are
usually easy to find with factoring, it could
be a rational answer... again those are usually
solvable with factoring and completing the
square, but it also finds solutions that are
irrational.
Like maybe the square root of seven.
And of course it will let us know when there
is no solution at all.
So why bother with learning to solve equations
with factoring and completing the square.
It is because those mathematical techniques
are used in other areas of Algebra.
And just in mathematics in general.
When you go to learning something graphing
conic sections like your circles, your parabolas,
ellipses, and hyperbolas completing the square
is going to be an integral process in those
problems.
So we have to learn it.
And we are going to learn Completing the Square
in this video for just solving equations.
We are going to first...
We are going to rewrite the equations so the
x terms, or whatever variable is in your equation...
a, b, whatever, but we are going to rewrite
the equation so all the terms with the variables
in them are going to be on the left hand side
and the constant is... take that s out of
there... the constant is on the right hand
side.
So variables on the left and constant on the
right.
Now you cannot complete the square unless
the leading coefficient is equal to one.
So we are going to...
With the leading coefficient equal to one
we are going to complete the square by adding
b/2 squared to both sides of the equation.
Now I teach PreCalculus and like... in this
year I have already done videos on graphing
conic sections.
So I wrote a note here, but I am probably
going to do my third example...
I am doing three by the way... a little bit
differently than what I have here in the note.
This is more for when you are setting up to
graph conic sections.
I will point this out, but you might want
to ignore it.
x^2+bx.. and this yellow is us adding the
(b/2)^2 term to both sides of the equation.
Again the x terms on the left and the constant
on the right.
We are going to let that number, that coefficient
of the x to the first term... the linear term...
take that value of b and divide it by 2, square
it, and then add it to both sides of the equation.
That is going to set up something called a
Perfect Square Trinomial.
I am getting ahead of myself here.
That...
I read was written in step three.
And factoring a Perfect Square Trinomial is
just sort of a pattern.
I will get to that in a step 3.
But this adding b/2 squared to both sides,
that is the completing the square process.
It sets up a perfect square trinomial on the
left which makes the rest of the problem much
easier.
Now in green over here I have got this k out
front because I have factored out... you know...
we need a leading coefficient that is equal
to one to complete the square.
And in PreCalculus when you are graphing conic
sections, your only option is to factor that
out of those x terms, find your value of b...
the number in front of the linear term.. the
x term, divide it by 2, square it, and then
add it to both sides.
And you won't have the luxury...
You will have to remember what you factored
out of those parenthesis and apply that multiplication
to both sides because whatever you are going
to add inside here is going to be added this
number of times.
Now I might show the last example both ways,
but if you are just doing Algebra 2 and you
are just learning how to solve equations,
you might be better off just...
If there is a 2x^2, just divide everything
in the equation by 2 instead of having it
in that factored form.
Again that is more like for Precalculus or
the end of Algebra 2 when you are graphing
conic sections.
That type of notation.
Factor the perfect square trinomial on the
left.
We are going to do that by square rooting
the first and last terms and keeping the middle
sign to make your squared binomial.
Actually I did not write it.
The reason why you can factor the left hand
side through this pattern of just square rooting
the first term, square rooting the last term,
and keeping the middle sign.
You will see exactly what I mean by that in
the examples here in a second.
It is because this x^2+bx+(b/2)^2, that trinomial
on the left hand side again like I just said
is a perfect square trinomial and thus they
factor just with a pattern.
You can actually factor them like you learned
out to factor, but this is going to be a little
short cut just to have a process that always
works.
Then square root both sides and solve it for
x.
Enough of my talking.
Let's get to some examples.
BAM!!!
First example We have x^2-4x+12 equals 0.
Now you won't know this in your textbook when
you are doing these problems, but this is
roughly what the graph looks like.
You will see that there is no x intercept.
So when we go to try and solve this with Completing
the Square, Factoring, Quadratic Formula,
whatever you use, you're not going to get
a real answer.
That is because there is no x intercept, so
thus there is no Real solution.
So with the steps that I just gave you we
are going to isolate the x terms and move
the constant to the right.
This is a plus 12.
Whenever you move across the equal sign you
want to use the inverse math operation to
make that happen.
We are going to subtract both sides by 12
and get x^2-4x, leave some room for completing
the square, and then minus 12.
Now that I have my two x terms alone on the
left, my leading coefficient is equal to 1,
my constant on the right, I am ready to complete
the square.
Now to do that we take half of b...
And it is good practice to take the sign along
with you just to recognize that this the coefficient
of x.
And it is negative four, but you are going
to square it and that is going to cancel out
the negative anyway.
-4 divided by 2 is -2 squared.
That is equal to 4.
Now I can do this in my head very quickly
and hopefully you can too.
But maybe these numbers are a little bit bigger
or just... you are tapping along on the calculator
because you are used to using them and you
are not really thinking.
Let's not forget, if you have a calculator
that allows you to see the text as you type
it in.. your syntax it is called... if you
were to leave off the parenthesis and type
-2^2 into your calculator it will give you
-4.
That's because you are only squaring the positive
two and the negative sign is going to float
out there and give you an answer of negative
four or negative whatever it happens to be.
So please to make sure that you watch those
parenthesis and use them correctly and get
those positive answers that you should be
getting from this completing the square process.
So negative two squared is four.
We are going to add 4 to both sides.
Get this out of my way.
And this is a perfect square trinomial.
Now I have video for going over factoring
techniques so I am just going to note that
completing the square makes a perfect square
trinomial and remind you that to factor these
perfect square trinomials you square root
the first term, you square root the last term,
and you keep the middle sign.
So this is going to be the square root of
x squared is x, so x minus... the square root
of 4 is 2... squared.
And if you properly factor this you would
get (x-2) times (x-2) which is (x-2)^2.
This equals negative twelve plus four is negative
eight.
In the directions that I gave you I said the
next step after you finish this factoring
process is to square root both sides of the
equation.
Well when you go to square root both sides
of this equation, I am going to attempt to
take the square root of a negative eight.
And there is no real square root of negative
eight.
There is no number times itself that is going
to give you a negative answer.
Let's say it was -9 temporarily.
The square root of -9.
Well you might think it is 3, but 3 times
3.. a number multiplied by itself is going
to be positive nine not negative nine.
Oh well then it is -3.
Well -3 times -3 is positive 9 again.
We cannot square root negative numbers.
And we do teach negative numbers but the results
are what is called imaginary numbers.
And we are not dealing with this today with
these problems.
That parabola had not x intercept and thus
the completing the square process will not
be able to give us a real solution.
And indeed no method of solving this quadratic
is going to give us a real solution because
there is no x intercept.
Next example.
For our next example we have solve x^2+8x+2
equals 10.
Now if I got this equal to zero I am not actually
sure if it would factor or not.
But again we are learning the Completing the
Square process not factoring.
So we are going to keep following the steps
that I gave you, get the x terms and get the
constant on the right.
We are going to do that by subtracting by
2.
So we have x squared plus 8x equals 8.
We are going to remember to leave space to
complete the square, which I did not.
Alright.
There we go.
And complete the square.
That means take half of b, square it, and
add it to both sides.
So half of b, square it...
That is going to be 8/2 is 4.
Keep using those parenthesis in case you make
a sign error.. to not make a sign error.
We don't want to be dependent on the calculator,
but a lot of kids use them a lot.
Positive four squared is 16.
We are going to add 16 to both sides of the
equation.
That is going to give us x^2+8x+16 is equal
to 24.
This is a perfect square trinomial.
I just keep saying that to let you know what
it is.
You can look that up in your textbook if you
would like to read up more on that.
But for factoring because it is a special
type of trinomial, square root the first term,
square root the last term, and keep that middle
sign.
So the square root of x squared is x, middle
sign is positive, square root of 16 is 4.
And this binomial you are creating is squared.
x+4 squared is this trinomial again factored.
If you don't believe me, write x+4 times x+4,
distribute them together and you will get
that as your answer.
This equals 24.
We are going to square root both sides to
get the power of two away from this x+4.
We don't want to take x+4 and FOIL it back
together, expand it back out, because we would
be going reverse order of how we started this
problem.
We would be getting back to the beginning.
We are going to square root both sides of
this equation.
The square root and the power of two are inverse
functions so they cancel each other out.
When the textbook gives you a square root
symbol, they are expecting... the author is
expecting you to only use the positive answer.
That is the way that is in the textbook that
I use.
If you introduce your own square root symbol,
you do need to account for the fact that there
is a positive and negative answer.
The square root of 9 is 3 and negative 3 because
-3 times -3 is 9.
So anyway, plus or minus the square root of
24.
Now we are not going to be happy with decimals
right.
We are not going to be happy with square symbols
that can be reduced.
There is a perfect square that is in 24.
It is 4 times 6 and 4 is a perfect square.
So I am going to do a little bit of reduction
here, a little bit of simplification.
x+4 is equal to plus or minus the square root
of 4 times 6.
4 is a perfect square, so that comes out as
2.
We have x+4 is equal to plus or minus 2 square
root of 6.
Now you can finish solving for that x.
Your teacher might be writing two separate
equations, one for the positive answer and
one for the negative.
I am just going to keep this written as one.
Subtract both sides by 4.
Oops.
And get as a final answer x equals plus or
minus 2 square root of 6 minus 4.
Now that is great if your book, or your teacher,
or your test asks for your answers in exact
form.
If I type square of six into my calculator
I get a crazy run on decimal that never ends
and never repeats.
It is an irrational number.
So any form, any decimal form of this answer
even if you go out to 10 decimal places has
been rounded off and not in exact form.
However if I was trying to graph this quadratic
and see where that parabola crosses the x
axis... well it does cross the x axis here...
but if I was trying to graph it, I cannot
look at this and tell where the decimal versions
are.
So approximate answers, type this into your
calculator... once for the positive square
root and one for the negative to get those
decimal answers.
If you are trying to graph this, you want
those answers so you know roughly where they
go along the x axis.
But this is what your answer looks like in
exact form.
One more example.
Last example.
There is that leading coefficient that is
not equal to one.
The more I think about this, I am going to
do this example a little bit differently than
the notation that I used in my notes as I
started this video.
That is again....
That is me just thinking about graphing conic
sections and it is not really the best way
to do that notation for solving equations.
So x terms alone, constant on the right.
We are going to take the six and add it over
to the right.
I need my leading coefficient to be one.
It is 2, so I am going to divide everything
by 2.
Now you might think...
OH... wow you know...
I kind of set this up nice.
All of these are equal because I knew that
I was going to divide by two.
But I am not going to avoid fractions entirely
because my b value is equal to 7.
There are fractions in math right.
So again we can't avoid them all the time.
Yes I forgot to leave room for completing
the square again.
We are going to take half of b, square it,
and add it to both sides.
So negative seven divided by two squared.
That is -7/2 times -7/2, and -7 times -7 is
49.
Of course 2 times 2 is not equal to 2.
It is equal to 4.
So I am going to add 49/4 to both sides of
the equation.
Ok, so...
We take one last step here and get x^2-7x+49/4
equals...
Ok I need a little bit of scratch work.
Right.
Because what is 3/1 plus 49/4?
So we are going to have 3/1 plus 49/4.
I am doing this little section right here.
Of course you guys know right that you need
common denominators to add fractions.
That is going to be 12 over 4 plus 49 over
4.
And 49 and 10 make 59, and another 2 makes
61.
So we have 12 plus 49 is 61/4.
Just so I am not making a video with a mistake,
let me double check it.
Yes...
Ok!
Get that erased.
Now we are going to factor this perfect square
trinomial.
OHHH.
I hate fractions.
Ok.
Deal with it.
We are going to square root the first term,
keep the middle sign, square root the last
term which is always positive so I don't need
to keep including that plus sign.
When you square root a fraction, just square
root the numerator and square root the denominator
separately like they are two small little
questions.
We get the square root of x^2 is x, keep the
middle sign of negative, square root of 49
is 7 and the square root of 4 is equal 2.
You might notice a pattern.
You kind of get those... that little bit of
scratch work... you keep getting as you square
root the constant of your perfect square trinomial
and that is a legitimate pattern.
It does keep showing up.
Oops.
What did I forget?
I did not add my...
Excuse me.
Yes I did.
Scratch that.
So 61/4.
Now...
Doing your notes, sometimes I see the notes
on the side of my vision as I am talking and
I think I see a number that is not really
there.
Ok.. so..
I have got my binomial squared is equal to
61/4.
I am going to square root both sides of the
equation, not forgetting to include plus or
minus because I am introducing my own square
root.
Remember to square root a fraction you can
square root the top and bottom separately
as I have written here.
And the square root of 4 is equal to 2.
I am going to erase that, the square root
of 4 is equal to 2.
Now I have already got common denominators
on both sides.
I am going to have to of course add 7/2 to
both sides of the equation to get it away
from the x.
And our final answer is x equals plus or minus
the square root of 61 over 2 plus 7/2.
That can be rewritten as plus or minus the
square root of 61 plus 7 all over 2.
By the way the decimal answers to our previous
question were .9 and -8.1.
And this is again our exact values, the exact
form of this answer of where this parabola
crosses the x axis.
I am Mr. Tarrou.
This is completing the square.
BAM!
Go do your homework:D
