Mr. Tarrou!
If you were just watching that video about complex numbers, you saw the video run out.
Just a big BAM!
Tried to squeeze in the solution of a parabola and the last two and half minutes ran out
of time.
So since I have got to make a whole new video because I don't want to take the time to stitch
these together.
I am already doing a lot teaching after school.
And at home this processing/loading time is pretty substantial.
So let me just go a little bit slower through that parabola question and give you a better
example anyway.
So part 2 of complex numbers.
We are going to try and solve, or find out the solutions for this parabola.
Now in graphing you talk about solutions and you are talking about x intercepts.
So what I really want to find if I want to solve, or find the roots... or the solutions...
or the zeros...
they are all the same thing, this parabola I want to know where it crosses the x axis.
Now when you cross the x axis the y value is zero.
Yes.
So if this is a parabola what I am really solving in this problem is zero equals two
x squared plus twelve x plus twenty.
What is the y intercept of this parabola.
What is the solution.
We know it is a parabola because there is one squared term.
Well this might factor, but I want to go through the quadratic formula for you.
If you did watch the last video, you see a little bit of repeat here but it is only about
a minute.
When you have an equation in standard form, that is where the degrees count down by one
and it is set equal to zero like this quadratic.
This first coefficient is A, then B, then C. Then solve this using the Quadratic Formula
which is the opposite of b plus or minus the square root of b squared minus 4ac all over
2a.
Ok.
Well b is 12.
So the opposite of 12 plus or minus the square root of b^2...
Always use parenthesis when you do substitution.
It helps to save you from sign errors.
Now minus 4 times a times c which is 20.
All over 2a.
Ok.
So we have -12 plus or minus the square root... this is 144 minus 160... all over 4.
Now here is where we get into the imaginary numbers, the complex values.
144 minus 160 is negative 16.
So we get -12 plus or minus the square root of -16 over 4.
Now if it was any other math area, like earlier in the math year or in Algebra 1 or Geometry
you might just stop and say there is no solution.
But we are studying complex numbers... the imaginary solutions.
So we want to keep going.
The square root of negative one is i and the square root of 16 is 4.
So this becomes -12 plus or minus 4i all over 4.
Both of these terms have a factor of 4 that we can cancel with the denominator.
That is 4 times -3, -12 divided by 4 is -1, plus or minus...
4 divided by 4 is 1... so i... all over 4.
Those common factors cancel out.
We have our complex solution of -3 plus or minus i.
Now a couple of things before I move on and try not to run out of time again, is when
you are solving equations and you get a solution that is complex they always show up in pairs.
Why?
Because what is an imaginary number?
An imaginary number is the square root, or attempting to take the square root, of a negative
number.
When you taking the square root of a number you are always getting a positive and a negative
answer.
That is why your complex numbers always show up as conjugate pairs.
This is a binomial with the middle term changing sign.
Complex numbers a solutions always show up as conjugate pairs because you are square
rooting a number.
When you square root a number, even though it happens to be negative and that comes out
as i, you have a positive and negative answer.
So a little bit lesson than my rush job if you watched the last video.
Why does this parabola not have an x intercept?
Well depending on where you are...
If you are studying conic sections you might be looking completing the square and putting
this into what is called...
well I want to say Standard form of a parabola.
That involves completing the square and I don't want to do that because that would be
a whole new lesson.
That will be a lesson later on this year.
Let's just say, let's pull this in now...
If you want to find the x coordinate of the vertex of a parabola, the x coordinate of
the vertex is equal to negative b over 2a.
Well, this is b so -12 over 2 times 'a'. 'a' is the leading coefficient.
'a' in this case is 2.
So this becomes -12 over 4.
So the x of my vertex is going to become -3.
I am just going to write it over here because I am running out of space.
This is the x of my vertex.
How do I find the y value?
How do you always find the y value?
You plug in your x and you find your y.
So the y of my vertex is equal to 2 times -3 squared plus 12 times -3 plus 20.
Hopefully I can do this off the top of my head fairly easily.
This is 2 times 9.
This is -36.
This is 20.
I have 18 minus 36 plus 20.
What is 18 minus 36?
I believe that is -18, now plus 20 which is 2.
So the x of my vertex from this little formula you might remember for Geometry is -3 and
the y is 2.
So my vertex is (-3,2).
Ok.
Now I am going to just squeeze in a tiny little sketch here.
(-3,2) is where my vertex is.
Well do you remember or know how to tell if a parabola opens up or down?
Parabolas if the x is squared open up or down.
If you are opening up, you open up when the leading coefficient is positive.
That is a positive two.
If it were negative two, the parabola would be opening down.
If the parabola opened down I would find real solutions.
But my leading coefficient here is positive which means the parabola is opening up.
Now this is a sketch clearly because I don't know any other points besides the vertex.
But I do know because my leading coefficient is positive is opening up.
It is opening up from the point of (-3,2) which means that there is no x intercept which
is why we got a complex solution.
So BAM!
There you go.
An extended version of finding complex solutions of parabolas with the quadratic equation.