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Contents

- 1 Grants New Decorations
- 2 Grants New Decorations
- 3 How many real roots
- 4 How many real roots
- 5 Fitting the Fountain Equation
- 6 Fitting the Fountain Equation
- 7 Defining the Discriminant
- 8 Defining the Discriminant
- 9 Special Aspect of the Discriminant
- 10 Special Aspect of the Discriminant
- 11 Roots and Signs
- 12 Roots and Signs
- 13 Non negative Discriminants
- 14 Situating Roots of Negatives
- 15 Situating Roots of Negatives
- 16 i
- 17 i
- 18 Exact and Complete Answer
- 19 Exact and Complete Answer
- 20 Exact and Complete New Answer
- 21 Exact and Complete New Answer
- 22 Several Solutions
- 23 Several Solutions
- 24 Playing with i
- 25 Playing with i
- 26 Another Equation
- 27 Another Equation
- 28 One More Equation
- 29 One More Equation
- 30 No Touching
- 31 No Touching
- 32 Plug in for x
- 33 Plug in for x
- 34 Where does it go
- 35 Where does it go
- 36 Relating Sets
- 37 Relating Sets
- 38 What is a complex number
- 39 What is a complex number
- 40 Real and Imaginary Numbers
- 41 Real and Imaginary Numbers
- 42 Complex Plane
- 43 Find the Coordinates
- 44 Find the Coordinates
- 45 Pick the Proper Point
- 46 Pick the Proper Point
- 47 Adding Complex Numbers
- 48 Adding Complex Numbers
- 49 Subtracting Complex Numbers
- 50 Subtracting Complex Numbers
- 51 Multiplying Complex Numbers
- 52 Multiplying Complex Numbers
- 53 Special Multiplication
- 54 Special Multiplication
- 55 Complex Conjugates
- 56 Complex Conjugates
- 57 Product of Complex Conjugates
- 58 Product of Complex Conjugates
- 59 Make the Denominator Real
- 60 Make the Denominator Real
- 61 Rewrite as a+bi
- 62 Rewrite as a+bi
- 63 Dividing Complex Numbers
- 64 Dividing Complex Numbers
- 65 Rationalizing the Denominator
- 66 Rationalizing the Denominator
- 67 Rationalize and Simplify
- 68 Rationalize and Simplify
- 69 Find the Roots
- 70 Find the Roots
- 71 More Root Finding
- 72 More Root Finding
- 73 Complex and Concise
- 74 Complex and Concise
- 75 Rooting out the Solution
- 76 Rooting out the Solution
- 77 Number of Solutions to Quadratic Equations
- 78 Number of Solutions to Quadratic Equations
- 79 Solutions with Imaginary Components
- 80 Solutions with Imaginary Components
- 81 Find that Vertex
- 82 Find that Vertex
- 83 What are the roots
- 84 What are the roots
- 85 Fitting the Fountain Better
- 86 Fitting the Fountain Better
- 87 Get a root of 0
- 88 Get a root of 0
- 89 Equation of the New Parabola
- 90 Equation of the New Parabola
- 91 Where will the water land
- 92 Where will the water land

We've helped Grant a ton throughout his journey building up his glasses wiper and nozzle enterprise. He's doing algebra all the time now to solve problems he encounters in his business and he knows how to do all of it because of us. He's been so successful that he was able to afford new office and he's trying to decorate his new place with art that reflects his products. Although he has different sculptures and paintings and other masterpieces strewn throughout his new space, one of the things he is most excited about is this fountain that looks like a giant wiper nozzle. He thinks he's worked out an equation for the trajectory of the water coming out of the fountain and then going up, falling down on the ground. However, he wants us to check that equation for him. Let's help him see if it's right. Just like we had with Grant's slingshot a few lessons ago, let's set up a coordinate plane so that the fountain is positioned at the origin, just like this. Now, I realized that the fountain is not as small as its single point on the coordinate plane but for the purposes of this question, let's just pretend like the entire thing is located exactly at the origin. From this position of the origin, the water is going to shoot up out of the fountain and then fall and end up slightly over to the right. Here's the equation that Grant came up with. Y equals negative 7x squared plus 14x minus given moment, and y stands for a corresponding vertical position for that same drop of water. So, every drop of water of this fountain should follow about this equation. Remember, this is just an approximation. We're sort of idealizing the situation, so just go with it. Now, if Grant is right, then one of the roots of this equation should be where the fountain is located, at 0. And the other should tell us where the water lands, over here. In other words, if we saw the quadratic equation that goes along with this equation for a parabola, then we should get those two solutions, those two roots. Let's use the quadratic formula to see if Grant is right. For now, I'd just like you to simplify the parts of the formula that belong in each of these boxes. Just put a number in each spot.

When we plug in the proper coefficients to the proper spots in the quadratic formula, we can simplify to end up with, x equals negative 14, plus or minus the square root of negative 84 over negative 14.

Now that we have an expression for x, we are moving closer to simplifying and finding a final solution. How many real roots do you think the equation will have? Try simplifying this expression we have x equal to, and see how many answers you know how to get from that.

There will be zero real roots. Looking at what we have for x, we need to take the square root of a negative number. And right now, we don't know how to do that. So we can't solve this.

Thinking about what we just discovered about the roots of this equation, do you think that this equation that Grant came up with accurately describes the path of the fountain water? Remember to think about the picture and what kind of parabola we need to produce this.

No. This equation is not going to cut it. Let's take a peek at the graph to compare. This is really interesting. Our equation clearly graphs a parabola, and we have numbers in every spot in our quadratic formula, so it looks like we should be able to get some solutions. The question is what kind of solutions are they. We can see that the maximum of the graph here is below the X axis. So no point on this parabola is every going to cross the line Y equals 0. What does it mean to have a root then? We clearly have some pretty important stuff to talk about.

At the end of the last lesson and now with Grant's latest issue as well, we've seen that sometimes parabolas don't have two real roots. Remember that a root is a value for x that satisfies a quadratic equation. Now, there's a single part of the quadratic formula that we can look at to determine how many real roots of parabola will have. That expression is called the discriminate. Just to remind you, here's the general form for a quadratic equation, and the quadratic formula for calculating the roots of that equation. Knowing that the discriminate on it's own is what determines the number of real roots, which of the following expressions down here do you think is a discriminate? Is it negative b, negative b plus or minus the square root of b squared minus 4ac, b squared negative 4ac, b squared minus 4ac, or 2a?

The discriminant of a problem is the expression b squared minus 4ac, now have you got this right, awesome job and if not, seriously don't worry about it all, this required a tone of critical thinking. Now the reason that this expression determines the number of roots in our problem Is that, first off, we're taking the square root of it. And secondly, we're both adding and subtracting this number from negative b. Let's continue talking about this, since it's pretty subtle and seems to be a little complex.

Now that we know what the discriminant is. For a quadratic it's b squared minus determining how many real roots we'll have. Is it its absolute value? Whether or not it's a perfect square? Whether it's rational or irrational? Whether it's even or odd? Or whether it's positive, negative or zero?

The sign, or lack of sign, of a discriminant. So whether it's positive, negative or zero, is what determines how many real roots we have. I think we need to talk a little bit more about this.

Let's talk about signs for a second. For real numbers, we have 3 different cases. Every real number is either negative, 0, or positive. Now since discriminants are made of real numbers, assuming that the coefficients of our quadratic equation are real, then they had to fall into these 3 categories. Each of these types of discriminants matches a certain number of roots Of a parabola, or in other words, a certain number of solutions to our quadratic equation, which we can of course find by our quadratic formula, using the disciminate. So here are 3 types of discriminates, and which type do you think produces 2 real roots, which type do you think produces 1 real root, and which type do you think produces no real roots?

if the discriminate is positive, we end up with 2 real roots. If it's negative, we have no real roots and if it's 0 we have 1 real root. Let's look at why.

Here's our quadratic formula written out once again, and I've written the discriminant b squared minus 4ac in teal. It's important to note that we're taking the square root of this number, and as we've said many times before, we only know how to take the square root of non-negative numbers, non-negative real numbers to be exact. So if we have a negative number under here, we're not going to get any real solutions. However, we need to make another distinction among the numbers that do give us real solutions. If we have a 0 under here, then we only get 1 real solution. Since we'll have negative b plus or minus 0, and either adding or subtracting zero from negative b gives us the same number, just negative b. So we would just get negative b over 2a. If the discriminant is 0. That's only 1 answer. This is only 1 number. It's our plus or minus, which is what split this into 2 solutions originally, is no longer doing that for us. If b squared minus 4ac is positive, then we'll get some real number here and negative b plus that number will be different from negative b minus that number. So in that case we will wind up with 2 roots, two positions where we know that our parabola hits the x axis.

So, we're interested in exploring the situation where a determinant, b squared minus 4ac, is negative or less than 0. The reason this is an issue is because we know we need to take the square root of the determinant, and right now, we have no idea what taking the square root of a negative number means. None of the types of numbers we've dealt with so far can express the answer to this equation. So, I guess that means we're going to need a new kind of number. Let's look back then at all the kinds of numbers we have dealt with. Here's a diagram depicting our world of numbers that Julie introduced in the very first lesson of the course. I know this looks a little bit different because I decided to use rectangles this time instead of circles, but all the sets and subsets have the same relationship to one another. You can see that we have the natural numbers, which are inside of the whole numbers, which are inside of the integers, which are inside of the rationals. And together, the rationals and the irrational numbers make up the real numbers. The question I have for you then is where this new kind of number we're talking about belongs. We see that the number n is some negative real number. Where should we put the square root of n? Does it belong in the natural numbers? The whole numbers? The integers? The rationals? The irrational numbers? Or outside of the real numbers?

We know that we don't know how to get a solution within the real numbers, for expressing the square root of a negative number. So that means it needs to go outside of the real numbers, in what right now looks like a sort of no man's land. I think that means that we actually need to define a new part of our diagram.

They said that the square root of a negative number belonged outside of the portion of diagram that contains the real numbers. So in this area over here. And since we don't have a name for the type of numbers right now, we need to come up with one. And since these numbers are real numbers. You might as well call these numbers imaginary numbers, since they're not real. I think this is a pretty cool term. It's hard to even think about what the number that's imaginary means, or what it really is. We'll get a handle on this soon enough. For now though, let's talk about the simplest imaginary number we have. Well in my mind the simplest negative number I can think of is negative 1. The number related to this that we're interested in then, is the square root of negative 1. Now because this is the simplest of that new type of numbers that we're talking about; the imaginary numbers, it gets a special name. It gets its own letter, i, because it's the simplest imaginary number. Note that even though this is a letter representing a number, this is not a variable. It's a constant. Just like we have numbers like pi and e that also have letters going along with, with them. I is just representing the constant, the square root of -1.. So let's play around with I a little bit. Since we know that i is just a number, albeit, a new kind of number, but still just a number. We should be able to take it to different powers. For each of these, I'd like you to write in what number you think it's equal to. So what is i to the 0, what is i to the 1, what is i to the of these answers at least, will be imaginary numbers as well. So I itself will probably appear somewhere in here. Remember though, that because this is just another number exponents work with it in exactly the same way. I to the five is just equal to I times itself five. Times. Or 5 i's multiplied together. I think you might see something interesting happen when you do these. Also 1 last note. If you ever feel tempted to write the square root of 1, just write i instead.

Now since i is just another number, like all other numbers i to the 0 power is just equal to 1. This may seem a little bit funny, because 1 is a real number. But I think this is more a testament to the fact that i and all other imaginary numbers are in fact numbers, i to the first power is just i, i squared is i times itself. Which means we're multiplying the square root of negative 1 times the square root of negative 1, that should just give us negative 1. Now i cubed is going to be i squared times i. So we have negative 1 times i which is negative i. Interesting. So we only have ones and i's, the positive and negative versions over here. Let's see what happens in this column, i to the fourth should be i to the third times i again. That means we're multiplying the square root of negative 1 by itself and then taking the negative of that number. So this is negative, negative 1, which is just equal to 1. But that's funny, that's the same as i to the 0, i to the 5th is i to the 4th times i. So 1 times i which is i again. That's the same as i to the first. And we can see the pattern continues for i to the sixth and i to the seventh, so it seems like when we take powers of i, the numbers that they're equal to alternate in a cyclic way. We go from 1 to i to negative 1 to negative i back to 1 and so on. I think this pattern is really interesting, and keeping this in mind is going to help us remember the relationship between real numbers and imaginary numbers.

Since we're thinking about powers, I'd like to backtrack for a second and deal with an equation involving a real number. Can you tell me what the solution to X squared equals 3 is? I'd like you to write this as an exact answer, not as a decimal. And remember to be careful about what happens when you want to get rid of an exponent. Make sure that you cover all of the possible solutions for X.

X needs to be equal to plus or minus the square root of 3. Since we know that either the negative square root of 3 multiplied by itself equals 3, and the positive square root of 3 multiplied by itself equals 3.

So let's use a similar notation for a slightly different problem. What is the solution to x squared equals negative 1? We've already talked about one number this could be, so think about, consideirng the last quiz, if there are any others. And try and write this in compact a way as possible.

Just like we knew that x squared equals 3 had the solutions plus or minus the square root of 3, here x is equal to plus or minus the square root of negative It's just i, the number we've been talking about. So either positive i, or negative i can multiply by itself to equal negative 1. However, we said that the positive version of the square root of negative 1 is equal to I. And logically then, the negative version is equal to negative I.

Continuing along the same vein, let's play around with another similar problem. This time, can you tell me what the solutions to x to the fourth equals 1 are? I'll remind you that we saw earlier that i to the fourth equals 1. So that gives you one answer. And also tell you that there are several that belong in this box. Think really carefully about what these could be, and seperate your answers with commas once you've figured them out.

The numbers one, negative one, i and negative i all, when taken to the fourth power, equal one. In case you're curious about how exactly this works, I've written them all out for you here. I think probably taking negative i to the fourth power is the least intuitive of these, but we know that when we multiply two negatives together, the negative signs cancel one another our basically. So negative i times negative i is just i squared which is negative 1. So negative i to the fourth is really just negative 1 squared or 1. This is another way where imaginary and real numbers interact in a super fascinating way in my opinion at least. Who would have thought that there are all these different numbers that are multiplied by themselves 4 times equal to 1?

We've already seen the number negative i, so presumably we can put different real coefficients in front of i. 3i for example, is equal to just i plus i plus i, or 3 i added together. So again imaginary numbers operate in pretty much the same way as real numbers. Thinking about this, what do you think we get if we square 3i, apply the same rules that you know worked for real numbers. Real numbers, and simplify this answer as much as you can.

We know that the exponent needs to distribute to both factors that are multiplied together to make up this term. So this equals 3 squared times i squared. We know that 3 squared is just 9, and we know that i squared is just negative 1. So 3i squared equals negative 9.

We've already seen that we can solve the equation x squared equals negative 1. Using our new kind of numbers, imaginary numbers. And we got x equals plus or minus i as the solution to this equation. We should be able to then, solve equations like this. X squared equals negative 9. Please write your solutions in this box and simplify them as much as you can.

To get x by itself, and come up with our solutions, we need to get rid of our exponent. So, we need to take the square root of both sides. Taking the square root of x squared just gives us x, and on the right side, we need to take the positive and negative versions of the square root of negative 9. Now, let's think about what to do with this radical and the number underneath it. We can rewrite negative 9 as negative 1 times 9 and as we learned before, if we have two factors both written under a radical sign, we can apply the radical to both of them individually, just like we can with exponents. So now, we have plus or minus the square root of negative 1 times the square root of 9. Now, the square root of negative 1 is just i, and the square root of 9 is just 3. So, our answer is plus or minus 3i. Remember, that this means that if we substitutes in either

What are the solutions to x squared equals negative 25?

Our answer is similar to the question before, we get plus or minus 5i. I'll show you the work that I did to get this. So again, I went through the same process as in the question before. Separating the factors under the radical, into the positive and negative parts, which then gave me a real factor, and an imaginary factor. And of course, we can't forget that plus or minus sign, since we're taking the square root of both sides.

Let's use our new-found knowledge of imaginary numbers to deal with a situation we were interested in initially, when we have a parabola that doesn't touch the x axis. Here I'm showing you the graph of y minus 16 equals the quantity x minus know we won't get any real solutions. However, we also have new equipment to help deal with this situation. Maybe imaginary numbers will help us out. Let's see what we can do using the number i. I'd like you to solve this equation then either by completing the square or by using the quadratic formula.

Completing the square here is easy since, well the square is already completed for us, so I am not even going to bother using the quadratic formula all we need to do is modify the equations that it reads x minus 3 the quantity squared equals negative 16 then we just take the square root of both sides And of course we can't forget our plus or minus over here, on the right side. We can go through the same steps as we did in the past couple of quizzes, to end up with x minus 3 equals plus or minus 4i. Now here, to get x by itself, we just need to add 3 to both sides. And we end up with a final answer of x equals 3, plus or minus 4i.

Let's check our new answer. So we're going to plug in 3 plus or minus 4 i in the spot of x in our equation. So remember this is either 3 plus 4 i or 3 minus 4 i. So for now let's just pick one. Let's pick the plus. So if we let x equal 3 plus

It simplifies to give us negative 16. Which is exactly what we need since negative 16 plus 16 equals 0.

Let's think a little bit more about one of our solutions, 3 plus 4i, the one we were working with in the last problem. Now this is an interesting number, because it has one part of it that looks real and one part of it that looks imaginary. So my question for you is where on our world of numbers diagram it belongs. Please select the spot where you think it should be positioned.

spot outside of those two boxes. It looks like yet again we're going to need a new kind of number. At least, a new name for a type of number.

So once again we need a new kind of number classification, one that goes beyond just imaginary numbers and just real numbers. You can see I've drawn a new giant rectangle around both of those families of numbers and labeled it with the name complex numbers. So examining our new diagram, can you tell me which of the following statements are true? All real numbers are complex. All complex numbers are real. All imaginary numbers are complex. All complex numbers are imaginary. All real numbers are imaginary, or all imaginary numbers are real. Please check off as many of these as you think are true. Be careful in remembering how our diagram works. In particular, how do you tell if one set of numbers is a subset of another set?

Only two of these are true. All real numbers are complex and all imaginary numbers are complex. Let's look at our diagram. We can see that the complex number rectangle contains all of the other rectangles inside it. So that means that every rectangle that's inside the complex number rectangle Contains numbers all of which are complex numbers. Since the imaginary numbers and the real numbers are all inside of the complex numbers they are all complex. However the complex number rectangle is not contained inside any of these other rectangles. All of those categories are subsets of it. So there is no other category that we can group all complex numbers into on this diagram. Since the real numbers and the imaginary numbers are completely separate rectangles, they are separate groups. Neither one is a subset of the other.

We just came to the conclusion that all real numbers are complex and all imaginary numbers are also complex. However, we also saw a different number earlier the one that motivated our entire discussion of complex numbers, but seemed to not fit into the category of real numbers, or the category of imaginary numbers perfectly. So, I think what we need is a general statement for the form of a complex number, which of these do you think is a way to write that general form. Please note here that in all these answer choices a and b are real numbers.

A complex number is any number that can be written in the form a plus bi, where a and b are real numbers. And, of course, i is just the square root of negative complex numbers in the first place. Remember that number was 3 plus 4i. So here, a is 3 and b is 4. 3 and 4 are real numbers. So, this is perfect. That fits. 3 plus 4i is a complex number.

Since real numbers and imaginary numbers are both subsets of the complex numbers then we need to be able to write all the numbers that are in both of those subsets in the form a plus b i. Something special happens though for these numbers a and b for the real numbers and something special for the imaginary numbers. For all the real numbers what needs to be set equal to zero in this form, and for all imaginary numbers what needs to be set equal to zero?

For all real numbers we want the coefficient in front of the i to be 0 so b needs to equal 0. That way we'll get rid of the second term and only have the constant a, which is a real number. For imaginary numbers we need a to be equal to 0. That way all we'll have left is b times i the imaginary part of a complex number.

We can see from the general form for a complex number, a plus bi, that complex numbers have both a real part, which comes from the number a, and imaginary part, which comes from this part bi. Remember a and a are just real numbers here, but this part bi is an imaginary part, because of the i. And we can talk about the size of a real number, by plotting it on a simple number line, because there's just a linear progression of numbers, higher and lower than zero. Because complex numbers have these two parts though, the real and the imaginary, we can't plot them in a one dimensional way. Instead, we need to plot them on a two dimensional axis, similar to the Cartesian Coordinate Plane we've been working with. This fulfills a very different purpose though. We're not graphing equations on here, we're not relating variables x and y to one another, we're just graphing numbers that are written in this form. What we do is we label this horizontal axis on our plane, the Real axis. And we say that the vertical axis, is the Imaginary axis. So for any complex number, written in the form a plus bi, we can instead write it as the point, a comma b, on this coordinate plane. So the horizontal coordinate, comes from the value of the real part, and the vertical coordinate comes from the coefficient of the imaginary part. Let's look at an example. So i plot the point, 4 comma 6, on our set of axis, this translates into the complex number 4 plus 6i, that's what's being depicted by this point on this plane. Let's say we want to express negative 7 plus 2i, we go to negative 7 horizontally, and then up 2. So on this plane, we can graph any number written in this form.

Now that we've talked about how points are graphed on the complex plane, I'd like you to tell me what each of the points labeled on the graph represents. Be sure that you write your answers in the form a plus bi, where and, and b are just real numbers.

All we need to do here are figure out the coordinates of each of our points, and then recognize that the horizontal one is a and the vertical one is b. So for example, we know that point a has the coordinates 3,5, so the complex point it represents is 3 plus 5i. Similarly b is 5+3i. Let's look at a point in a different quadrant, maybe point f over here. The coordinates of f are negative

Once again, I've labeled some points on our complex plane. And I've labeled them with the letters a through h. For each of these three questions, I'd like you to fill in the letter that corresponds to the point that the question's asking about. So which point represents the number 7? Which one represents the number negative 7i? And which one represents the number negative 7 plus 7i?

Remember that the complex number a plus bi can be graphed on this coordinate plane as the point a,b. So, we need to figure out how each of these points is written in this form, or what a and b are each equal to. The number 7 is equal to 7 plus 0i, if you want to write it like this. So it should be the point 7,0, which is c over here. This makes sense. This is a real number, so it's written strictly along the real axis. A similar thing happens for negative 7i, which is just an imaginary number. Here, a is equal to 0 since there's no real part to this number, so its coordinates should be 0, negative 7, which is this point e. Now, negative 7 plus 7i is already written in a plus bi form or it's already recognizably written that way. So, its coordinates are negative 7, 7, which is point h.

So you've seen that complex numbers are just another kind of number, a type of number that subsumes both the real numbers and the imaginary numbers, and also contains some numbers that are part of each. Part imaginary and part real, so presumably we can manipulate complex numbers then in the same way that we manipulate real, Real numbers. Let's try this out and see if it works. What if I'd like you to add 2 complex numbers tegether, let's say 3 plus 4I and 5 minus done before together so, just try it out. Think back though, to adding like terms and how we practiced that pretty early on in the course.

The sum is just 8 plus 2i. We need to recognize here what things you can add together and what things we can't. 3 and 5 are both the real parts of these nubers so they can combine and 4i and negative 2i are the two imaginary parts so they can combine to give us 2i.

Let's try a subtraction problem now. What is 11 minus 6i minus the quantity 4 plus 2i?

The answer is 7 minus 8i. 11 minus 4 is 7. And negative 6i minus 2i is negative

Now let's try some multiplication of Complex Numbers. What do you think 4 plus polynomials together before.

The final answer we get when multiplying these complex numbers together is negative 7 plus 22i. So how do we get this? Well, as before, we need to make sure each term in the first set of parenthesis is multiplied by each term in the second set of parenthesis. So we start with 4 and multiply it by 2. And then also by 3i. And then we take 5i and multiply it by 2 and then by 3i as well. Then we just need to simplify it within each of these terms over here, combine like terms and come up with our final answer. I think the trickiest thing here is thinking about these middle terms and thinking about the final term. We can see here that multiplication involving imaginary numbers Works in pretty much the same that working with real numbers and variables does. We have 4 times 3i, the 4 multiplies the 3 to make 12. And then the i is still a factor,so it's multiplied in as well. 5i times 3i is just 5 times 3 times i times i or 15 times i squared, which is negative 1.

Here, I've taken the two solutions to the equation we solved before and I'd like you to multiply them together. So what is 3 plus 4i times 3 minus 4i? I'd like you to work all these out by hand but before you do, think about what pattern you see here or how these two factors relate to one another and try to predict what kind of answer you'll get over here.

Interestingly enough, we get 25. We know that when we multiply a real number by a real number, we get a real number. And we've seen before that if we multiply two imaginary numbers together, since we're multiplying something with an i to something else with an i, we end up with a real number for that term as well. In this case, we don't get an imaginary component for this final number because the middle terms cancel one another out. We have minus 12i and plus 12i. We can notice that the only difference between these two terms and the reason we get this answer is the sign of b. Keep this in mind as we move forward.

So these two numbers 3 plus 4i and 3 minus 4i. Both of which you remember were solutions to a quadratic equation we solved before, are called complex conjugates of one another. Everything about them is the same. Except for the sign between their real and imaginary components. And when we multiply them together, we get a real number. Considering this new term, what do you think the complex conjugate of the number 2 minus 3i is?

The complex conjugate of 2 minus 3i is just two plus 3i. Once again, we just look at the sign of the coefficient b over here, the coefficient of the imaginary part of the number and we flip it. It was negative and now it's positive. Let's look then at the graphical relationship between complex conjugates. The two orange points here are one pair of conjugates, 3 plus 4i and we notice is that the horizontal coordinate of either point in each pair is the same. But the vertical coordinate is just flipped so each point is just a reflection of the point it matches with across the real axis. If we have the number negative 8 minus 4i, then its conjugate must be the point that has the same horizontal coordinate and the opposite vertical coordinate, so it will be this negative 8 plus 4i which does seem to be correct.

Let's keep playing around with conjugates. First I'd like you to find the complex conjugate of 1 plus 9i and fill that into this orange box. Then I'd like you to find the product of 1 plus 9i in that conjugate. So write the conjugate again in this second set of parentheses and then multiply it by The original complex number. And fill the answer in this last box.

The complex conjugate of 1 plus 9 I is just 1 minus 9 I. Now let's find their product. Having a bunch of 1's in here actually makes our multiplication pretty easy. I like that. Now you just need to simplify this final term and combine like terms. Notice already we can see these 2 terms, negative 9 I and positive 9 I. Cancel one another out just like we had happen with our previous conjugate pair multiplied together. I would like to pause at this step for a second where we have 1 minus 81 i squared and note that this is a difference of 2 squares. This is actually what we would expect to get when we are multiplying things like this together. Remember that we learned quite early on in the course that when we multiply something in the form a plus b times a minus b. We get a squared minus b squared. This is exactly what's happening here. 1 is equal to 1 times 1 or a squared, and 81i squared is equal to 9i, the quantity squared. Now of course, we can simplify 81i squared because this is just a real number. Number. I squared is just negative 1. And negative 81 times negative 1 is positive 81, so our answer is 82.

We've been talking a lot about complex numbers together, but what if we have a situation like this, where we have one complex number divided by another complex number? Since right now, both the numerator and the denominator each have a real and an imaginary component and one is not directly a factor of the other one, we can't do anything with this expression. We often encounter fractions like this, and usually, what we do to manipulate them differently is to change the denominator in some way. What I'd like us to do then is to figure out how to change the denominator of this fraction so that it ends up being the smallest real number we can make it, other than 0, of course, since we don't want to divide by 0. So, think critically about the past few quizzes we've done and what you can multiply one complex number by to get a real number. Fill in whatever we need to multiply at the top and bottom of this fraction by, to come up with an equivalent fraction that has a real number in the denominator. I know this is new and it probably feels a little bit difficult right now but just give it a shot, and if you feel like you need to review a little bit to prepare for this question, you can go back and what the past couple of videos.

We saw earlier, that if we multiply a complex number by its conjugate, the middle terms cancel out and so we end up with a real number. This is exactly what we're looking for in this situation, to make our denominator what we want it to be. Since we don't want to change the value of this fraction that we had initially, we need to multiply by 1 in the form of the conjugate over itself. Now, we just need to multiply the numerators of these fractions and their denominators. At first start of multiplying out, gives us a fraction that I think is pretty clear we can reduce a little bit more. So, let's keep simplifying. Our final answer is negative 7 plus 22i all over 13. I think this definitely looks a lot more manageable than what we started out with. Awesome job.

We just simplified 4 plus 5 i, over 2 minus 3 i, to be written instead as negative 7 plus 22 i over 13. This is clearly a complex number, but it's not written in the form that we usually see complex numbers written in. So the last step, in modifying this initial complex number, I'd like you to write our answer in the form a plus bi. So put the real part of the number here and the imaginary component here. In fact I'll even write in the i for you. So you just need to tell me what a and b are here.

We just need to divide both of the terms in the enumerator by 13. So a is negative 7 over 13, and b is 22 over 13.

So here's another one for you to try out on your own. I've given you the expression 4 minus i divided by 1 plus 5 i. And I'd like you to go about changing the denominator to a real number, so that you can eventually write this number in the form a plus bi. Then fill your final answer into this box.

So our answer here isn't really very pretty, but it is something that we can tell right off the bat is a complex number, and we can tell how big the real part is, and how big the imaginary part is. Our answer is negative 1 over 26, minus 21 over 26 i.

I'd like to take a slight tangent here, very slight and very brief, and look at a number like this, 6 over the quantity 1 minus the root of 2. Now, at first sight, this may seem completely unrelated to what we were just talking about. But I'd like you to focus on the denominator for a second. The problem that I see with this quantity, and the reason that we can't really do anything with this fraction right now, is that there's a rational part and an irrational part in our denominator. I think in general, numbers are easier to work with the closer they get toward the center of that world of numbers diagram. So, let's see if we can make our denominator have only one component, probably just a rational component instead of an irrational one. So, what number would you need to multiply the numerator and denominator of this fraction by in order to come up with an equivalent fraction, where the denominator is a rational number and, of course, it's not equal to 0?

The same concept of a conjugate applies in this case as it did when we were working with complex numbers. Both cases make use of this reverse property of the difference of squares. Just as if we squared an imaginary number, we end up with a real number. If we square A number that has a square root sign over it. We'll just end up with that number that's under the root, which in this case is a rational number. So multiplying 1 minus the root of 2 times 1 plus the root of the denominator. Let's try it out. The numerator is quite simple to multiply out. You just multiply each of these terms by 6, and the denominator ends up being quite simple too. Since 1 times the root of 2 and - root of 2 times 1 cancel each other out. The square root of 2 squared is just 2. So we end up with dividing each of the terms in the numerator Are the 3 that's in the denominator. This gives us a final answer of 2 plus 2 times the root of 2. So many 2s everywhere.

Here's another one for you to try out on your own. I've given you the number, the square root of 2 divided by the quantity negative 3 plus square root of 5. And I'd like you to rewrite this by first changing the denominator that is a rational number, instead of having both rational and irrational components. And then I'd like you to simplify the fraction as much as you can.

After first multiplying both the top and the bottom by the conjugate of the denominator, negative 3 minus the square root of 5, we end up finally with an answer of negative 3 square root 2 plus 10 over 4. Now, of course, we could also write this as splitting the answer into two totally separate terms. Simplifying, that would give us negative 3 quarters root 2 plus 5 halves. I personally don't think this is quite as pretty as the first one we had before it, but some people may prefer to write things this way.

Now that we've had a lot of practice dealing with complex numbers, let's go back to that original problem we were trying to solve or at least a problem of a similar form. I want you to find the roots of this equation for a parabola Y equals X squared plus 4. Remember that a root is an X value when Y equals 0. If you get more than one answer, I'd like you to note that you should separate them by commas.

To find the roots we need to replace y with 0 and then we just isolate x. After a slight bit of arranging we've x squared equals negative 4. We can already recognize at this point that is going to need to be an imaginary number. Otherwise when we squared we wouldn't end up with negative. When we take the square root of both sides here, we need to make sure that 1 of the sides has a plus or minus before the radical sign. And simplifying the right side, gives us that x is equal to plus or minus 2i.

Now let's backtrack to an equation we've seen before. Y equals x squared plus 2x plus 10. I'd like you to do the same thing that we did in the last question, or at least look for the same thing. I want you to find the roots of this equation. Once again, if you get more than one answer, separate them by commas.

Using the quadratic formula we come up with the final answer of -1 plus or minus numbers, and we knew that you were going to have to do that because the number that's underneath the square root sign is negative.

Let's look at another equation. This time I'd like you to find the roots of y plus 10 equals negative 2 times the quantity x minus 3, squared. When you write the solutions to this equation, I'd like you to write them as compactly as you can, and also to use the standard form for complex numbers, a plus bi.

Since we're looking for roots, once again we set y equal to zero. Which is just going to leave a 10, on the left side of the equation, and change nothing about the right side. Now we need to move toward getting x by itself. You can notice that I actually didn't use the quadratic formula, in solving this equation. Instead, I just undid all the operations that had been performed on x on the right side of the equation. Eventually this led me to the answer of x equals 3 plus or minus i times the square root of 5. Now one thing to note about the way that I've written this answer. You'll see that I wrote i in front of the radical sign, whereas we'd normally write in the form a plus bi, where the real coefficient of the imaginary portion comes in front of the number i. However, to avoid confusion as to whether or not the i belongs under the radical sign, it's often written in front of the square root sign instead. This is similar to if we had something like 3 times the root of five, writing the number this way is usually the convention. Rather than this way, just to make perfectly clear what doesn't belong under the radical sign. Either way though, with the i as the first factor, or the i coming after the square root of 5, is mathematically correct.

As one last problem for now, I'd like you to find the roots of this equation. Y equals X squared plus 8X plus 16.

I decided that I'd like to switch what was on either side of the equation just because I prefer to have x on the left side of the equation. This is just a personal preference. It doesn't change the math that we're doing at all. I also set y equal to 0 since we are, after all, looking for the roots of this equation. Now, what we notice here is that you can actually factor this expression on the left. X squared plus 8x plus 16 Is just equal to the quantity x plus 4 times the quantity x plus 4 well these two factors are clearly both going to give us the same root since they are equal to each other, so from this we just get one equation x plus 4 equals 0 that needs to be satisfied in order for this equation to be satisfied Fine. So we get a root of x equals negative 4. Now even though we only got one answer here for x, we must notice that the factor that this came from appears twice in the equation. So it's not so much that there's just one solution, it's that the two roots of this parabola happen to be identical to one another. In fact this root is so special that it gets a special name. It's called a double root.

Let's talk a bit more generally then about all these different problems we've discussed and their roots. So reflecting on the huge number of quadratic equations we've worked with, how many solutions are there to any given quadratic equation? Please count repeated roots twice, and also remember that complex numbers and imaginary numbers, are in fact numbers. So answers that contain them do still count as solutions.

There are two solutions for any given quadratic equation. Let's think back to how this emerges from the quadratic formula. It comes out of the presence of this plus or minus here in the numerator. This is going to give us two answers, negative b plus the square root of b squared minus four a c all over two a, and then the same thing except for a minus sign in between the negative b and the rest of the numerator. This also makes sense if we look at the general form for a quadratic equation as well. We have ax squared plus bx plus c equals 0 and in order to end up with a term involving an x squared in it, because we know that the coefficient a is not allowed to equal 0. We need to be able to multiply 2 things together that have a factor of x in them, since that's the only way you can get an x squared down here. That means that either of these factors needs to be able to be set equal to 0 and give us both solutions. So there are 2 solutions.

Now here's a bit of a complex question. If a quadratic equation does not have 0 solutions with an imaginary component, how many solutions with an imaginary component does it have? Think really carefully about this. In fact, you might want to think about the quadratic formula and what kinds of solutions it gives us, and how those solutions relate to on another.

If we know that a quadratic equation has some number of solutions with an imaginary component, then it has to have two of those kinds of solutions. So, in other words, it has to have two complex solutions neither of which has a coefficient of b equal to 0. Let's look at the quadratic formula to think about why this is true. And the quadratic formula gives us solutions that are in this form. Let's write out the two solutions separately for the time being. Written out like this, we can see what may have seen hidden to you little bit before. The only difference between this solution and this one is that there is a plus here between the negative b and this part, and there is a minus here between the negative b and this part under the radical sign. Now, since we know the imaginary part of our solution has to come from this part with the radical, and what will happen, of course, only at the discriminate under here is negative. That means, that these signs that are opposite one another are separating the real component of either solution from its imaginary component. That means, that any time we find imaginary solutions, or complex solutions with nonzero imaginary components for a quadratic equation, those 2 complex numbers are going to be complex conjugates of one another. This is super powerful, super potent. We're going to continue to think about this concept through many of the coming lessons.

Finally let's return to working with Grant. He had just shown us this new equation that he'd come up with, that he thinks properly describes the trajectory of his fountain. The equation is y equals negative 7 x squared plus Fill in the coordinates right here.

To get to the vertex form for the equation for this problem, we need to go through the process of completing the square, just like you've done many, many times before. The equation in vertex form is y plus 3 equals negative 7 times x minus 1 squared, which means that the vertex is at 1 comma negative 3.

Now that you've worked so hard to figure out the vertex form of our equation, I've decided to show you the graph. Now I'll actually write the equation for the parabola in both forms, vertex form and standard form. And I'd like you to tell me what the roots of this parabola are. Remember simplify and write the answers in as compact a way as you can.

The roots of this parabola are 1 plus or minus i times the square root of 21 over 7. This is a simple case of using the quadratic formula to solve, and actually I had to do this problem way earlier on in the lesson. You may not have even remembered it. And this time you got to do it totally on your own. And you've also got to write the answer in the complex notation that you know about now.

Well, as we expected from our picture and from our work earlier on in the lesson, we got complex numbers as our roots. Now, looking at where the vertex of our parabola is below the x-axis, and the fact that the parabola is pointing downward, this curve doesn't really fit the situation that we're interested in. Remember, that y here in this equation and in this graph, is the vertical displacement from the ground of the water and x is the horizontal displacement of the water. Right now, the water is underground and that's not really what Grant's looking for in terms of decorations. So, what do you think is wrong with Grant's equation and his graph right now? There's one single thing that we can do to this graph that would fix the entire problem that we're having. The mislocation of the horizontal and veritcal coordinates of the water in Grant's fountain. So, what one step should we do to this graph to make it better describe that fountain's shape? Should we flip this parabola upside down? Should we translate it up? Translate it down? Make it wider? Make it skinnier? Translate it to the right? Translate it to the left? Or just leave it alone altogether?

We just need to move this parabola up. After all, falling objects on Earth do obey parabolic motion because of gravity. And we do have then the proper shape. It just happens to be positioned in slightly the wrong place.

So we know that we need to move this parabola up in order to make it fit our real world situation, but by how much do we need to move it up? If I'd like to have one of the roots of our parabola after we've moved it upward, equal to 0, then by how much do we need to translate this parabola upward?

If we want 0 to be a root, that means that we want one of our x-intercepts and also our y-intercept to be point 0,0. Right now, the y-intercept is at negative there to being up here. That means, we need to move the graph up 10 units.

After we translate the parabola up by 10 units, as we decided we need to do, what is going to be the equation of that new parabola? Please write your answer in vertex form first, and then also in standard form.

All we need to do to get our equation for a new parabola in vertex form is to look at the old equation in vertex form, and then shift the vertex up by 10. That means, the right side of the equation, which contains the horizontal component of the vertex is not going to change at all. The only thing that will change is the constant added to y on the left-hand side. We need to subtract 10 from here since we will be up by 10, and this gives us an equation of y minus 7 equals negative 7 times x minus 1 squared. To get this in standard form, we just need to multiply everything out and simplify. That gives us an answer of y equals negative 7x squared plus 14x.

Once again I have updated the graph class. So now it shows the old graph that we have this original problem, the orange one. And then above that, the new graph that we have of the equation, y equals negative 7x squared plus 14x. Thinking back to how we settle the problem at first, we say that we should let the base of the fountain be situated at the origin. This is why we wanted one of the roots of our equation to be 0. So you can see that we've got the fountain lined up right here. This is where the water is spouting out from. Now, my question is how far from the base of the fountain will the water land? Assume that everything here is measured in meters. So figure out what point on a curve corresponds to the place where the water from the fountain hits the ground, and then, figure out how far away from the base of the fountain that is?

What we're looking for here is just the other x intercept of our new graph. That's this point right here and its coordinates are 2, 0. So the water will land 2 meters from the base of the fountain. Alternatively, we could have of course figured this out from the equation alone. We can set up a quadratic equation, 0 equals negative 7x squared plus 14 x, then I could just factor the right side, which gives us 0 equals negative 7x times the quantity x minus 2. This equation will be satisfied if either negative 7x equals 0 or if the quantity x minus 2 equals 0. Of course, you get x equals 0 for the first solution, which is the one we already knew ahead of time. And then, we get x equals 2 for the second one, which is the one we just figured out.