Common Core Algebra II.Unit 9.Lesson 3.Solving Quadratic Equations with Complex Solutions
Jan 4, 2017
Hello, I'm Kirk Weiler, and this is common core algebra two. By E math instruction. Today, we'll be working on unit 9 lesson three quadratic equations with complex solutions. Now, you've been working with imaginary and complex numbers for the last couple of lessons. And we've been working with the quadratic formula for quite some time now. Today, we're going to put those two together and see how some quadratic equations have solutions that include imaginary components. But first, let's take a look at the quadratic Formula One more time. All right. We all know the quadratic formula we've been seeing it since common core algebra one. But there's this portion of the quadratic formula. This portion that lies underneath the square root. And up until this course, whatever was underneath the square root could not be negative. But remember now that negative one is defined as the number I and that allows us to take the square roots of all sorts of negative numbers, like the square root of negative 49 would be the square root of 49 times the square root of negative one, which would be 7 times I so now, because we can allow the discriminant, that's what we're going to call it. We're going to give this thing a name, and we're going to give it a name immediately. The discriminant, because the discriminant can be negative, we can have quadratic equations that have solutions that involve the number I. And let's jump right into exercise one and see how that happens. All right? I'm going to clear out this text. And let's do it. All right. Exercise number one. Use the quadratic formula to find all solutions to the following equation, express your answers in simplest a plus BI form. Now, this is what I call a red flag. All right? In other words, if I look at this and I think, oh, I should use the zero product law and I should factor in factor and factor. And factor, I never going to be able to find anything that makes those factor nicely. And I'm not going to be able to because the problem is going to end up having complex solutions. So in fact, let's do it. Let's jump right in and let's use our formula negative B plus or minus the square root of B squared minus four AC all divided by two a, let's use that formula to solve this equation. All right? As always, we want to figure out what a, B and C are, a is one. B is negative four and C is positive 29. And let's throw those numbers in. All right? Negative negative four plus or minus the square root of negative four squared minus four times one times 29. All divided by two times one. Now, when students make mistakes with the quadratic formula, almost always, it's because they get that discriminant. Let me write that word down again. We'll see it later on. They get that discriminant portion wrong. So I always recommend that you take some time and make sure to do that correctly. So pause the video now and work out what's underneath the square root. Don't worry about taking the square root yet. Just work on what that number is that I've circled in red. All right, let's go through it. So in this problem, when we work all of this out, we get four plus or minus the square root of negative 100. Now it used to be that when we got a negative underneath the square root, we knew we had made a mistake because, hey, you can't take the square root of a negative number. Until we introduce the number I, right? Now remember, when we have something like the square root of negative 100, that's easy enough to deal with. That's the square root of a hundred. Times the square root of negative one, so that's just going to be ten times I many of you won't break it up that way, you'll just immediately look at this and go, all right, square root of negative 100. Well, that's the square root of a hundred, but I'll just tack an eye on. Now finally, it says express your answer in simplest a plus BI form. Don't forget that division distributes, so I have to divide the four by two, and I have to divide the ten by two. But once I get that, I get two plus or minus 5 I and that's not too bad at all, right? I mean, it's really the same thing that we've done with the quadratic formula all along. The only difference is now we have to bring in the notion of the I so now we have a quadratic equation that has complex solutions with imaginary portions. Pause the video now because in a moment you're going to work on some of these on your own. And write down anything you need to. All right. I'm going to clear out the text. And let's go on to exercise two. We're going to get some more practice with this. Now an exercise too, I've created some of them where you have to do a little bit of manipulation first. Remember, that in order to use the quadratic formula, you always have to have it rearranged. So that you've got the quadratic expression equal to zero. Right now, we've got an equal to 7 X minus ten. So what we need to do is we need to subtract a 7 X from both sides. And then we need to add a ten. All right. And that's going to give us a new quadratic equation. X squared -12 X plus 40 equal to zero. Don't turn that thing into an expression. Make sure you've still got the equals zero there. All right? But now what I'd like you to do is pause the video and work through this problem just like we work through the last one. All right, let's go through the numbers. So a is one again. B is equal to negative 12. And C is equal to 40. So when we use the quadratic formula, we'll have negative negative 12. Many of you will automatically write that down as positive 12, and I would encourage you to do that. Negative 12 squared minus four times one times 40, all divided by two times one. So that's going to be positive 12 plus or minus and again, most important thing that you're going to be working with here is going to be that discriminant, that quantity underneath the square root, make sure you get that right. If for some reason it turns out to be a positive number right now, now you're in trouble because this question is really trying to assess. It's really trying to assess if it were let's say on a standardized test, can you use the quadratic formula? Can you get an equation equal to zero? And can you deal with the idea of an imaginary number due to a negative under a square root. So if you get a positive under that square root, you got a little bit of a problem. Anyway, it all turns out to be negative 16. Which gives us 12 plus or minus four I, all divided by two, and then I'll distribute that division by two by dividing the 12 by two. Whoops, almost lost my minus there. And the four by two, and that gives 6 plus or minus two Y and there we have it. Simplest a plus BI form. All right? Pause the video now and write down anything you need to. All right, let's clear out the text. And keep moving along. Exercise two letter B, very similar to the one before, so I'm going to have you work through the entire problem on your own. Let me just warn you, though. The final simplification will be somewhat more challenging than the ones that we had before, because the number under the square roots not going to be nearly as nice as negative 100 nor negative 16. So pause the video now and take 5 minutes or so or longer if you need it to work through this problem. All right, let's go through it. So just like before, we need to start the problem by rearranging the quadratic equation so that it's equal to zero. In order to do that, I'm going to subtract a ten X from both sides and subtract four. And that's going to leave me with X squared plus 6 X plus 11 equals zero. Great. So a is equal to one, B is equal to 6. And C is equal to 11. Substituting those into the quadratic formula. We're going to have negative B plus or minus the square root of B squared minus four times a times C, all divided by two times a again, be very, very careful as always when you're evaluating the number under the square root, that discriminant, and in this case, it'll end up being the square root of negative 8. Now as I mentioned before, that negative 8 isn't nearly as nice as negative 100 or negative 16. You know, I see a negative 100 underneath there, ten I see a negative 16 under there, four I but 8 is not a perfect square. So we do a little bit more work with it. Now, how much more work that's kind of up to you? Remember, I can break it all was up as a perfect square. A not perfect square. And the negative one. Because remember, multiplication is commutative, I just broke up negative 8 as four times two times negative one. That'll give me negative 6 plus or minus the square root of four is two. The square root of two I really can't do anything with. It's an irreducible, non perfect square. And then the square root of negative one is I some teachers like to take that I and write it here instead just because they don't want it to be confused and have people think it's under the square root. I'm going to leave it right there for a minute. But now I'm going to distribute that division by two, notice I'm going to do the division always as the last thing I do. It's going to be the last thing I do. I'm not going to do it up front or anything like that. Therefore, I'm going to get a negative three, here these twos cancel plus or minus the square root of two. I that's a good way to write it. You'll find that some math teachers will prefer that it be written with the I before the square root of two. Which I find interesting. I do it sometimes myself. It's a little inconsistent with what we had been doing in other problems, but they're all good, both those two answers. All right. So pause the video now. And write down anything you need to. All right, let's clear out the text. And then take a look at the discriminant and the quadratic formula in a little bit of a different light. All right. Exercise number three. What we're going to be doing here is we're going to be taking a look at a graph of a parabola. So make sure you have your TI 84 plus or some other graphing calculator out. All right, and we're going to see what the connection is between a parabola, all right? And it's zeros, and it's X intercepts. So letter a. Algebraically find the X intercepts of this parabola, express your answers and simplest a plus BI form. Now X intercepts are also known as zeros, right? And they're known as zero specifically because they are the values of X that solve for where the function is equal to zero, where the Y value is equal to zero. Now, of course, to do this, we can use the quadratic formula. Why don't we do that together? We're going to get X equals negative B, which is going to be positive 6. Plus or minus the square root of negative 6 squared B squared. Minus four times a times C stand that square root symbol far enough. I didn't give a lot of room here. So we're going to have 6 plus or minus the square root. I forgot to flip my page over. 6 plus or minus the square root of negative 16 divided by two. And that'll be 6 plus or minus four I divided by two. Which is three plus or minus two I all right, so there are the zeros of this quadratic. All right? Now, what I want to see is what does that mean in terms of the X intercepts? I mean, there they are, three plus or minus two I so where does this thing actually cross the X axis? Well, I've given you a grid there. Well, I've given you a set of axes. We're not going to bring out our TI 84 plus right now. What I'd like you to do is use your graphing calculator in the window indicated. To sketch a graph of that parabola. It'll pause the video now and take a minute to do that. All right. So simple enough, right? We throw that in Y one. We set up our window, et cetera, we get a parabola that roughly looks like that. Nice Y intercept at 13. But notice what it doesn't appear to do. It doesn't cross the X axis. And that's the whole point to letter C it says from your answers to a and B what can be said about parabolas whose zeros are complex roots with nonzero imaginary parts. Wait, what? Let's read that again. What can be said about parabolas who's zeros? These things. Are complex roots with nonzero imaginary parts. Now, why didn't I just say, you know, what can be said about parabola that parabolas have zeros that are complex? Well, the reason that I can't ask it that way is that all numbers are complex numbers. All numbers are complex numbers. All of them. Right? The key is, do they have an imaginary component that isn't just zero I? Do they have some non trivial imaginary component? And the plain fact is what you can say about parabolas whose zeros are complex roots with nonzero imaginary parts, is that they do not cross the X axis. I'm going to put an exclamation point there just for emphasis, right? And the reason why is very simple. The X axis, this guy right here, right? That is a real number line. A real number line, right? You've been dealing with real number lines for a lot that is a real number line. You've been dealing with real number lines for a long time. Okay? And the plain fact is, if our answers include an imaginary component that isn't zero, then it can't cross that axis because that axis only contains purely purely real numbers. So that's really kind of cool. And of course, we all know that some parabolas cross the X axis, and some parabolas don't. And ones that have nonzero imaginary components don't cross the X axis. All right. Pause the video now and think about this hard before we discuss it a little bit more on the next slide. Okay. And clear that out. So we've been talking about this. This word, this idea, the discriminant, right? The quantity that is underneath the square root, and what we just saw is that that turns out to be a negative number. Right? Then the parabola will miss the X axis, so it'll either miss it like that, or it'll miss it like that. Okay? And we'll work with this a lot in the next lesson. But this is really kind of cool. The discriminant will tell you whether or not a parabola will ever hit the X axis or not. Okay? So let's get a little bit of a workout on this in the next problem. And maybe we should erase some of that text so that we can actually see the next problem. There we go. Exercise force says, use the discriminant of each of the following quadratics to determine whether it has X intercepts. All right? Well, it's this simple, right? We could, of course, graph the thing. Or we could solve the entire quadratic formula. But at the end of the day, all we really need to know is what the value of B squared minus four AC is. So let's do that. What would it be? It would be negative three squared minus four times a times C and I'm going to kind of go through this calculation a little bit. That's 9. And then negative and negative becomes a positive, so that's 40. And that's 49. Now what does that really tell us? Well, that tells us if we were going through the entire quadratic formula, there wouldn't be any I involved. There'd be no I there would be no imaginary component that wasn't zero. So yes, yes. It does. Because B squared minus four AC is greater than or equal to zero. It just can't be negative. All right? Why don't you try letter B? All right, let's take a look. Hopefully what you found in letter B was that this will not have X intercepts. So I want to make sure that looks like a negative and not like a times. In this case, we'd have 6 squared minus four times one times ten. And again, just kind of working out all the math. We'd have 36 minus 40. And that would be negative four. Now remember, this is that quantity that's sitting under the square root in the quadratic formula. And because now that it's negative because it's less than zero, no. It won't. Because B squared minus four AC is less than zero. It's negative. Try one more. Try letter C. All right. Let's go through it. B squared minus four AC. That's going to be three squared. Minus four times two times 5. That's going to be 9 -40. Just negative 31. So again, no. Because B squared minus four AC is less than zero. All right? And again, it's really kind of neat. With parabolas, unlike lines. I mean, some lines can not have X intercepts if they're just horizontal. But generally lines have X intercepts, but you can have parabolas that open upwards that miss the X axis. You can have parabolas that open downwards that miss the X axis. And of course, many parabolas hit the X axis twice, some even real rare ones hit the X axis only once. But we'll look at that more in the next lesson. Pause the video now and think about this before we move on. Okay. Let's do it. One last problem. Exercise 5, which of the following quadratic functions when graphed would not cross the X axis. Now, there's many different ways to do this problem. I'd like you to pause the video and do it any way you so choose. But just make sure to work on it hard until you're absolutely certain that you know which one would not cross the X axis. All right, great. Well, again, one of the clear ways that you can do this problem is to simply evaluate or sorry to graph each one of these on your calculator. And you'd see that all of them cross the X axis except for number three. Now, if we want to use the discriminant on this sorry my writing pad got a little bit away from me. If we want to take a look at the discriminant, which is B squared minus four AC, what we would find is we'd have negative four squared minus four times four times 5, which would be 16 minus 80. Which would be negative 64. And it's because of that negative on the discriminant, that it does not cross the X axis. Now, what's kind of cool is that a lot of students, even without a graphing calculator, could tell that these two must cross the X axis. Right, and think about why that is, like in number one. Number one, we have a negative, as beautiful. Negative Y intercept, right? One, one, two, three. And yet, it opens upwards. Right? So it's got to have X intercepts. Number two, it's kind of the opposite of that. It's got a positive X intercept. Y intercept. But it opens downwards. So it also has to have X intercepts. Three and four, you can't really use that type of reasoning. So you have to either use your graph and calculator to graph the two parabolas, or you have to use the discriminant and whether or not it's negative or positive. All right. Well, pause the video now. And then we'll wrap up the lesson. Okay. Let's do it. So in this lesson, we finally, for one, soft quadratic equations. That had solutions that involved imaginary components. In other words, they had solutions that had the number I in them. All right? And that wasn't so bad. I mean, it was just using the quadratic formula again. But then we took that idea and we connected it to whether or not a parabola has X intercepts. In other words, whether or not it touches the X axis and what we saw is if the quantity underneath the square root, known as the discriminate, was negative, then the parabola would completely miss the X axis. We'll explore that a little bit more in the next lesson. For now, I'd like to thank you for joining me for another common core algebra two lesson by E math instruction. My name is Kirk weiler, and until next time, keep thinking and keep solving problems.