In this lecture, I want to talk about some more abstract problems.  The problems that we've looked at so far
with respect to quadratic equations have all been pretty concrete.  Take a specific equation or a specific word problem type scenario,
and calculate the result.
In this case, I want to be a little more abstract for those of you are maybe going
 into fields like engineering or physics where you need to be able to work
with these equations and formulas at that theoretical level as well as
a more practical level.
So
Here, I have two numbers and I want to come up with a specific quadratic equation that has those numbers as its roots
Just so you can see the reasoning here, let's
Look at a different problem.  Let's look at the equation x squared plus
five x minus 6 equals zero.
I want to think about how we solve this because, to answer this question that I've asked here, were going to follow the same process in reverse.
Our method for solving this is we say, I picked nice numbers here
numbers where this factors.  This is x + 6
times x -1
equals zero
Then we use our
procedure here were we set the factors equal to zero.  That gives us to solutions: x equals negative six and x equals one.
So now let's go to our numbers here.  I want to find a quadratic equation
whose roots are x equals three
and x equals
-2.  Look at what happened over here.
Tith these two roots came
from these two factors.
So if I kind of work these backwards.  If I set each of these two little equations equal to zero I get x - 3 equals zero and x + 2
equals zero.  This gives me the factors
to the equation that I'm looking for.  My factors are going to be x -3
an X plus two
If I multiply this out I get x squared minus x minus 6
equals zero.
You can check our work here real quick.  If you substitute three and negative two for acts, you see that that polynomial 
does equal zero.  So both of those values are roots.  Now
I want to emphasize something.  I worded the problem very carefully.  I said to find
a quadratic equation.
Let's consider this equation.  Let's say we have the equation 2 x squared minus 2 x minus 12
equals zero.
Hopefully you see what I did there.  I just took my equation, multiplied its by two and I got this different equation.
This is a different quadratic equation that also has the same two roots.  So this method is nice
in the sense that it gives me an answer.  It gives me a quadratic equation that has the roots
but you should keep in mind that it's not giving you the quadratic equation 
whose roots have those values.  In fact, to narrow it down one equation, we don't have enough information.
Basically, we know two points on the graph
with a quadratic equation, that's not enough to get one equation.  We would need a third point to do that.
So let's look at another question.  An even more abstract one here.  I have this equation with
certain parameters.  I know that for some reason the b and c values both have to be the same.
I want to know what values
of k
give me a single, repeated solution
for this equation.  Well,
anytime you're referencing the parameters, the a, b and c,
of a quadratic equation and I want to know something about its roots, the first with immediately think to go
is the quadratic formula.  The quadratic formula tells me that the roots of a quadratic equation are x
equals negative b plus or minus the square root of b squared
minus
four ac over
two a.
Now, let's think about how the applies to our equation here.  In our equation, a is one
b is k
and c is k.
o if I substitute those numbers and my equation, what does it turn into?  It becomes negative k
plus or minus the square root of k squared minus 4 times one
times k
over
two
times one.
I want this
equation to have a single
repeated solution.  That's what it means for a solution to be repeated.  It means that there's only one solution.
Now to do that
I have to somehow get rid of this plus or minus part.  The square root here because this is the part
that turns the negative k into two numbers.  So the only way I'm going to, in a sense, get rid of
this part of the expression is if it's equal to zero.
So that's what I'm looking for here.  I want a value of k that
makes the square root of
k squared minus four k equal to
zero.
Now, this is nice because this is an equation equation I can solve.  If I square both sides, 
following our process for solving a radical equation, this becomes k squared
minus four k equals 0.
Now the left-hand side, this is a quadratic equation.  If I factor this it becomes k times 
k minus 4 equals zero and this gives me two answers.  k equals zero
k minus four equals zero.  Which is the same ask k equals four.
So I got two answers and that's okay.  That often happens with quadratic equations 
and we can check these real quickly.  If I put these two numbers in
If I put zero into the equation for k, what happens?  It becomes x squared
plus zero k plus zero
equals zero which simplifies to just x squared equals zero and it does give me one solution.  X equals zero.
And if I try it again with four, let's see what happens.  This becomes x squared plus four x 
plus four equals zero and this thing factors.  It becomes x plus 2 squared
equals zero and work again ride is only give me one solution
exit polls May 2
equals zero and it works again.  This gives me one solution,
x equals -2.
Structurally, this question is very similar the previous one.  I know I'm talking about the b and the a 
the parameters from the general form of a quadratic equation and I'm talking about the roots
and the way we get to the roots is by looking at the quadratic formula. 
Now, I'm going to write as little differently than we usually do.  I'm going to actually write the roots
separately.  The quadratic formula gives us to solutions: negative b plus or minus
the square root of b squares minus four a c
over
two a and it gives me the second solution.  That's negative b minus the square root of b squaredminus four a c
over
two a.
Now this question is asking me to show something about the sum of the roots.  I'm going to be very brute force about it.
I'm going to just add these two things together.  I'll look at the sum of the roots.  That's x one
plus x two.
So that's negative b plus the square of b squared minus four a c
over two a
plus
negative b minus
the square root of b squared minus four a c
over
And this is nice.  We got a little lucky here.  The denominators are the same already so I can just add these two together.
It becomes negative b plus the square roout of b squared minus four a c
plus negative b minus the square root of b squared minus four a c all over
two a.   Now here comes the cool part.  Look at what happens to the square roots.  I have plus b squared minus
four a c minus the square root of b squared minus four a c.  These two
cancel.
All I'm left with is minus b plus minus B.  That's minus two b still over two a.  The two's cancel
and I'm left with negative b over
a which is exactly what I was asked to show
So, first I want to think about what we did here.  These are kind of 
classic techniques for answering questions
about the roots of a quadratic equation.
I hate to say all the time because there are always going to be unique situations but a lot of the time to answer to this kind of question
you're going to go to the quadratic formula.  You're going to work with it, manipulated somehow
to get the solutions that you're looking for. One other point that I did want to make here, let's say we try to apply this here
to the special case you usually talk about in algebra classes.  Let's say we have this case x squared
plus b x plus c
equals zero.  When you're first learning to factor quadratic polynomials, this is were you usually start.  
You start by looking at the special case where a is equal to one.
For this polynomial, let's look at what our formula says.  In this special case,
the sum of the roots is negative b over one which is just
negative b.
I want to think about how you learned tofactor these.  There's a process that you learn where you're looking for numbers
whose product is
the c term 
and whose sum, remember were talking about the sum of the roots here,
is the b term.  And we see here why the sum of the roots is equal to, is equal to the opposite of that the term but when we make the factors,
remember if we have, thinking back to the first problem we did, if we have x equals three
as a solution then x -3
is the corresponding factor.  So this idea that the sum of the roots is equal to this formula based on a and b
is the theoretical basis behind that method that everybody learns an algebra for factoring this special kind
of quadratic polynomial.
