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Contents

- 1 Preparing for the Progress Assessment
- 2 All The Numbers
- 3 All The Numbers
- 4 Everything in its Place
- 5 Everything in its Place
- 6 Complex Number Manipulation
- 7 Complex Number Manipulation
- 8 Solve For x
- 9 Solve For x
- 10 Equation of a Parabola
- 11 Equation of a Parabola
- 12 Roots of the Parabola
- 13 Roots of the Parabola

Since your first midterm exam is coming up, let's talk about how to review the topics we've covered this far. >> Our journey with Grant began way back with talking about variables and expressions which we then combined into equations to help him start figuring out a plan for this glasses wiper business. First, he dealt with linear equations and inequalities, how to graph them and how to solve them. Once we encountered situations where Grant needed parabolas and their corresponding quadratic equations, we figured out how to handle those as well. Along the way, we've worked with every category of real number that we talked about at the very start of the course, as well as a new classification of number, complex numbers, which have both real and imaginary components. You've learned a ton about algebra in the course so far, and you may be wondering, how do I study for all of this material? And how do I take a test on the Udacity site? Well, we've come up with a few tips to help you out. >> Go back and look at practice questions and problem sets for Lessons 1 through 16. I also recommend watching my review sessions and redoing the problems that come after them. >> Make sure that you're comfortable using MathQuill, since you'll have to input some answers in the test using it. You can look at the MathQuill overview page on the wiki and you can go back to quizzes in the units that use MathQuill and practice typing things in there. >> Also, make sure you're comfortable working things out on paper, as well as typing them. Best of luck, we know you're going to do great. >> After this first midterm, we'll be moving on to a second case study, a new context for learning about Algebra. We're sad that our time with Grant is over, but he's going to be just fine handling his glasses wiper business now that we've taught him so much about math. And now, we get to study something new and exciting, architecture. This is going to help us learn about functions, specifically higher order polynomial functions and rational functions and give us a lot of pretty things to look at and understand mathematically.

This last lesson before the midterm, sort of stands out on its own, and w're going to do a review section to make sure you've got a grasp on this complex new material that we covered. Just as we started out the course with building up our knowledge of sets of numbers, we're going to round up this section of the class by revisiting this and acknowledging how we've built off of it. As a review of everything that we've learned about sets of numbers, here is a blank version of the full world of numbers diagram. And I would like you to type the name of each set into the proper position on the chart.

All of the numbers that we've worked with fall under the category of complex numbers. Complex numbers can have both real and imaginary components. So that means that real numbers and imaginary numbers are each subsets of the complex numbers. Within the real numbers, we have all of the same categories as we did before.

Now that we've filled in the proper sets in the proper places in the world of numbers diagram, let's put some numbers in it. As you did early in the course, please place each of these numbers in the place it belongs on the diagram. Label the diagram with each letter in the appropriate box.

I filled in the numbers in the proper spots. Irrational, integer, whole, natural, rational, complex, and imaginary.

Now that we've looked at complex numbers in our world of numbers diagram, let's manipulate them a bit. Please simplify each of these expressions so that it is written in the standard form for a complex number, a plus bi.

Remember, that when we add and subtract complex numbers, the imaginary parts are like terms and the real parts are like terms. So we end up with 1 plus 3i plus the quanity negative 6 plus 2i equals negative 5 plus 5i. 1 plus 3i minus the quanity negative 6 plus 2i gives us 7 plus i. When we multiply complex numbers, this needs to happen by distributing in the same way that we did with binomials involving, involving variables. The quantity 1 plus 3i times the quantity negative 6 plus 2i, gives us negative 6, plus 2i, minus 18i, plus 6i squared. Which when simplified, gives us negative 6, minus 16i, minus 6 or negative 12 minus 16i. And lastly, when we divided complex numbers, the final thing we need to do is to make the denominator into a real number, so that we can just let it modify the coefficients of the real and imaginary parts of the denominator. To change the denominator in this way, we multiply both the denominator and the numerator by the complex conjugate of the original denominator. So, negative 6 minus 2i in this case. Now let's look at the problem. 1 plus 3i, divided by negative 6 plus 2i, gives us 1 plus 3i, times the conjugate negative 6 minus 2i, divided by negative 6 plus 2i, times negative 6 minus 2i. Simplifying further, we got negative 6 minus 2i minus 18i minus 6i squared, divided by 36 minus 4i squared. Which gives us negative six minus 20i plus 6, divided by 36 plus 4 or negative 20i over 40, and finally negative i over 2.

Now that we've worked with expressions for a while, let's move on to some equations. Please solve for x in the following equation. 3x squared plus 18 equals 0. Although there are a couple of different methods you could use for solving this, I'd like you to try to do it without using the quadratic formula this time. Please write the solutions in as concise a form as you can.

So, when we simplify this, we subtract 18 from both sides and we get 3x squared equals negative 18. We then divide by 3 on both sides, we get x squared equals negative 6, giving us our solutions of x equals plus or minus the square root of negative 6. And when we pull out the negative 1 as an i, we get x equals plus or minus i times the square root of 6. This is two solutions, a plus i square root of 6 and a negative i square root of 6.

And now for another equation, this time of a parabola. Here is the graph of a parabola for you. Now, first of all, what is the equation of this parabola in vertex form?

Your solution is y plus 3 equals negative one third times the quantity x minus 2 squared.

What are the roots of this parabola? List them with commas separating them.

We can use a quadratic formula to solve this problem. First we put our answer in standard form, plug in 0 for y, and then substitute into the quadratic formula to obtain our two solutions of 2 plus 3i and 2 minus 3i.