 Welcome back to our lecture series Math 1050, College Algebra for Students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Missildine. In this video, I want to talk about quadratic inequalities. Much of Chapter 3 has been involving quadratic functions and quadratic equations, but as often as necessary, we switch over to inequalities. Now, when we're trying to solve inequalities, the technique I always suggest is that we actually begin by solving the equation. If you want to solve a quadratic inequality like x squared minus 4x minus 12 is less than zero, I would first begin by solving the equation x squared minus 4x minus 12 equals zero. Then we solve the equation like we would in the other quadratic equation. We could complete the square, we could use the quadratic formula, or in this case, a very simple factorization is possible here because we need factors of negative 12 that add to be negative 4. We can get away with having x minus 6 and x plus 2. Notice that negative 6 times 2 is negative 12, but negative 6 plus 2 is negative 4. This gives us two markers, x is equal to 6 or negative 2. Now, these markers are not necessarily the solutions to inequalities, but they do help us understand what the solution looks like. These are the markers here. What I would then do is I would draw a number line like you see illustrated here to the right. I would then mark off 6x equals 6 and then x equals negative 2. I would mark these off. Now, looking at the linear inequality right here, we could pick a test point because we have different intervals. We want something less than 2, like maybe negative 3, something between negative 2 and 0, or negative 2 and 6 like 0, or then you can have something bigger than 6, like maybe 10, and then you could try plugging these numbers into the original inequality and see what happens. If you get a true, like it passes, then that's part of the solution set. If not, then no. Now, personally though, I don't like to do the test points. I actually just like to think of the graph. Like when I look at this quadratic function, I look at the leading coefficient, it's a positive one, one's your a value. And since it's positive one, it means it's going to concave upward. So if I were to draw this thing really quickly, I would get something like the following. I get a mark at negative 2, I get a mark at 6, and since the graph is concave upward, I'm going to get a picture, look something like the following. Oh no, that's a hideously drawn parabola. That's okay, it doesn't matter. Because if we're comparing our function to 0, if we want to be less than 0, that means we're looking for those points which are below the x-axis. If you're below the x-axis, you're talking about this region right here. And so then we record that the solution set is going to be from negative 2 to 6 because we want those things below the x-axis. Because it's strictly less than 0, we do not include the markers. The x-intercepts will not belong to the solution set because negative 2 for x or x being 6, that would show that you're equal to 0, but we have to be less than 0 so we get those things between them. Let's look at another example. Take 2x squared is greater than or equal to x plus 10. If I was solving an equation, I would set the right-hand side equal to 0. So we're gonna do that for inequality as well. 2x squared minus x minus 10 is greater than or equal to 0. So I want to kind of stop us right here. What I already see is since this is greater than or equal to 0, I'm gonna be looking for those points which are above the x-axis when I'm done. And so also because of the leading coefficient right here, your A value is positive. Again, our graph is gonna be concave upward. So if I were to graph this thing like so, I know that my function would look something like the following. It's concave up and we're gonna get these two different x-intercepts. What are these x-intercepts? I don't know. That's why I have to solve the equation. So you solve the equation 2x squared minus x minus 10 equals 0. Again, you could solve this by the quadratic formula. You could solve this by completing the square factoring that does work in this situation. Notice that if I take 2 times 10, 2 times negative 10, that's equal to negative 20. And I need factors of negative 20 that have to be negative one. You could do negative five and four. So with these numbers in hand, we can then try to factor by groups or you could try to do a guess and check approach. You get like 2x times x something. You could try that. I'm just gonna factor by groups. So you're gonna get 2x squared minus 5x. That's your first group. And then for your second group, you're gonna get 4x minus 10. This is still equal to zero. From the first group, you can pull out just an x, I guess, leaving behind 2x minus five. From the second group, you can pull out a two leaving 2x minus five. Those do match up. And so our factorization is gonna look like 2x minus five and x plus two equals zero. That then tells us that our solutions are gonna be x equals five halves and x equals negative two, which we then could label on our graph right here, negative two and positive five halves or 2.5 if you prefer. And so since we're looking for those things above the x-axis, we'll be grabbing this portion right here. This part, the y-coordinate is above the x-axis and over here is also above the x-axis as illustrated in our graph right here. So our solution set would look like the following. We get negative infinity up to negative two. Now, do I include the marker or not? Since I'm gonna be greater than or equal to zero, negative two will be included in that. So we get negative infinity up to negative two, union bracket five halves to infinity, which again, we're gonna include the end points negative two and five halves because we're allowed equal to zero. And since we want greater than zero, that means we want those things above the x-axis. Well, how about this example, x-square plus 2x plus one? If I try to solve this one by factoring, I'm actually gonna discover that x-square plus 2x plus one actually factors as x plus one quantity square, which should be greater than zero. And so as you see right here, this has a single x-intercept at negative one and because you're leading coefficient is one and it's positive or my graph is concave upward. And so you see, you get a picture that looks something like the following. You get a point on the x-axis and then everything else is above the x-axis. If we're looking for those points, which are greater than zero, that would be grabbing this sector right here and this sector right here. So the y-coordinate is greater than zero whenever x is not equal to one. So your solution would actually look like negative infinity to negative one union negative one to infinity. And so you get the solution, you get the solution to be everything except for negative one. Negative one feels really left out right now. It didn't get to be on the dodgeball team. Everything but negative one is a solution here. I do wanna, and I wanna show you that this in this situation, if one were to calculate the discriminant here, you'll notice that the discriminant of this quadratic is actually equal to zero. And if your discriminant is equal to zero, you basically have only one of two pictures. So the first picture is kind of like we saw a moment ago. I'm gonna put a little bit lower here. Your first picture, your parabola, its vertex is the x-intercept. So it looks something like this. This would be because your A value is positive and your concave up. And therefore you get something like that. The other option is that your vertex is still on the x-axis, but then maybe the graph is concave down. So your A value is negative and then your concave down. Personally, when I'm working with a quadratic inequality, I always want the leading coefficient to be positive. So if my leading coefficient was negative, I'd probably just times it by negative one and then we'd just be done with it. So that is, you can always turn it into this type of problem, but if it is, it's concave downward. If you were then asking yourselves, what's the solution look like if I'm greater than zero? Well, in this situation, since we'll say that the vertex happens at h, if you're looking for your quadratic to be greater than zero, then your solution would look like negative infinity to h union h to infinity, like so. And if you also modified the problem so that you want f of x to be greater than or... I wrote, I didn't write what I meant. We want f of x to be greater than zero. Sorry about that. If f of x was greater than zero, then you'd want everything above the x-axis. You get everything except for h. Let me actually write my h a little bit better here. Now, if you wanted, if you want this thing to be f of x is greater than or equal to zero, in that situation, you'd want everything to the left of the vertex, everything to the right, but since you could equal zero, then you could get the vertex itself. In that situation, your solution so it would actually be all real numbers. Or you might write that with the r right there. So all real numbers. On the other hand, if your graph was going downward, like I said, we don't even need the green picture. That's an option, but you can always flip it above. And so I'm gonna focus just on this upward picture right here in yellow. What happens if you looked at f of x as less than zero like I accidentally wrote at the beginning? If you want the stuff that's less than zero, then you're gonna be grabbing this stuff down here, which there's nothing there. So we actually find out that the solution would be the empty set, a.k.a. this thing right here. That doesn't seem so good. And I mean, it's possible, you might get no solution if you want the things that are below. On the other hand, if you wanted like say f of x is less than or equal to zero, that way you would want everything that's below the x-axis, which is nothing. And everything that's on the x-axis, that would give you the interval of just h to h, which of course is just a single number, h itself. So the discriminant equals zero case is sort of like an interesting duck here. It basically comes down to we get everything but the vertex, everything, nothing or nothing but the vertex. These are the possibilities you get. Well, let's look at one last example here. Take x squared plus x plus one is greater than zero. I wanna mention that in this situation, the discriminant b squared minus four ac, if we look into that, we're gonna get one minus four, which is negative three, which as negative, this would mean that there's gonna be no x-intercepts on the graph. And since the leading coefficient is positive, we actually see, like you can see on the screen above, we would see that our graph would actually live entirely above the x-axis. Like so, so it never hits the x-axis. So what would our solution look like? Well, if you want the things that are greater than zero, you want things above the x-axis, in that situation f of x is greater than zero, what we would get, like we see here, is we want everything that's above the x-axis, that's everything, that's everything, that's everything. Our solution would be all real numbers, like so. Well, what if we ask the question, what if we want f of x to be greater than or equal to zero? So we would include the x-intercepts. Well, there are no x-intercepts, so it doesn't change the solution at all. It's still gonna be all real numbers. Now, on the other hand, if we wanted to ask the question, well, when is f of x less than zero? Like we see here, less than zero would be down here, but there's no part of the graph that lives there. We would again get nothing. This is the empty set. And if we switch to include the x-intercepts, so when f of x is less than or equal to zero, that is still the empty set, because there's no x-intercepts to include. So the negative discriminant case is even kind of weirder than the discriminant zero case that we saw before, because when your discriminant is negative, it means that your function will be either entirely above the x-axis or entirely below, and that means their solutions could be all real numbers or no real numbers. You only get those two options, just flip them with coin basically. Now, because of these two kind of anomalous cases, the discriminant equals zero, discriminant equals negative, for the most part, I would probably have you practice with positive ones, because that's where some interesting things happens, but be aware that these special cases are possibilities, and that's how one solves the quadratic inequality. If you can solve the quadratic equation, and you have a basic idea of how to graph a quadratic equation, then you can put those principles together and solve the quadratic inequality.