Lesson Video: One-Variable Quadratic Inequalities Mathematics • First Year of Secondary School

Video Transcript

In this video, we’re gonna look at solving quadratic inequalities. First, we’ll take a simple approach of looking at the graphs, but then we’ll take an algebraic approach. If you can find roots of quadratics or solve quadratic equations already, then you can do this. But you do need to present your working out carefully to avoid making a common mistake near the end. Let’s take a look at some questions now.

So question one then.

Using the graph, find the values of 𝑥 for which 𝑓 of 𝑥 is greater than or equal to zero. And we’re given that 𝑓 of 𝑥 is equal to minus 𝑥 squared plus five 𝑥. And we’re also given the graph which shows 𝑦 equals 𝑓 of 𝑥, which actually maps out that quadratic. So this is a symmetrical parabola, cuts the 𝑥-axis at zero and five, and also cuts the 𝑦-axis at zero. So we’ve gotta find out when 𝑓 of 𝑥 is greater than zero; in other words, when the 𝑦-coordinate is greater than or equal to zero.

So with 𝑦 equals 𝑓 of 𝑥, and we’re looking for when 𝑓 of 𝑥 is greater than or equal to zero, we’re really- we’re looking for are all the points on that curve where the 𝑦-coordinate is greater than or equal to zero. Well here at zero and here at five on the 𝑥-axis, the 𝑦-coordinate is zero. And on all these points on the curve in between, the 𝑦-coordinate is greater than zero. So that’s the region that we’re interested in. So we wanna know the 𝑥-coordinates that generate those. Well zero, five, everything in between in terms of the 𝑥-coordinate, so from zero to five, and everything outside that region. So bigger than five onwards or less than zero onwards down to negative infinity, that’s not included in our region because these parts of the curve are not greater than or equal to zero. So we could write that like this, zero is less than or equal to 𝑥 is less than or equal to five.

But we could also write it in interval format. So the critical values are zero and five. That’s either end of the interval. Now zero is included, so we need to clu- include a square bracket at the end. And five is included, so we put the square brackets around that end. So that’s in interval format. So we could also put it in set notation. So we’ve got the set of 𝑥 such that 𝑥 is a real number between zero and five. So the process then was to have a look at the graph and identify all the points on the graph which match the criteria that we’re looking for. In this case, it was 𝑓 of 𝑥 is greater than or equal to zero. So it’s then a question of finding which 𝑥-coordinates match the criteria and which 𝑥-coordinates don’t match that criteria, and then just kind of summarising that in one of these formats: the appropriate format whether it’s the inequality format, the interval format, or the set format depending on what the question asks for.

Okay, let’s move on to the next question then.

And we’ve got to use the graph to find the values of 𝑥 for which 𝑓 of 𝑥 is greater than zero, so strictly greater than, not equal to zero. Well in this case, at 𝑥 equals one, we’ve got a 𝑦-coordinate of zero, so 𝑓 of 𝑥 is equal to zero. So that’s not gonna be in the region we’re looking for. And when 𝑥 is four, we’ve got a 𝑦-coordinate of zero or 𝑓 of 𝑥 equals zero. So that again is not gonna be in the region that we’re looking for. However, when 𝑓 of 𝑥 is greater than zero; that’s all of these points up here on the graph and going off into infinity in that direction, and then all of these points up here on the graph and going off into infinity in that direction.

So thinking about the corresponding 𝑥-values, well we’re not including four, but everything to the right of that is included. We’re not including one, but everything to the left of that is included. So the bits that we don’t want when the 𝑦-coordinate is less than or equal to zero down here is everything from 𝑥 is one up to 𝑥 is four inclusive.

Okay, so how do we go about writing this out? Well we’ve got a noncontinuous region. So we’ve got everything to the left of one and everything to the right of four. So we have to put that as two separate inequalities. So 𝑥 is less than one or 𝑥 is greater than four. So thinking about the equivalent in interval format, we’re going all the way from negative infinity all the way up to but not including one. And we’re going from not including four all the way up to positive infinity.

So the critical values there are negative infinity and one and four and infinity. So infinity always has the round parenthesis after it. And we’re not including the one in that region, so we put around parenthesis around it. We’re not including four, so that’s got a round parenthesis. And then we’re going out to positive infinity, which again has a round parenthesis. And both of those two regions are valid, but nothing in between. So it’s the union of those two regions. So in interval format, we’d write it like that. And we can write our answer in set notation, so the set of 𝑥 such that 𝑥 is real, where 𝑥 is less than one or 𝑥 is greater than four.

But another way of writing it is that basically, you know, the set of real numbers is all of these numbers here, along the 𝑥-axis, the number line if you like. But we want to exclude this region here from 𝑥 is one up to 𝑥 is four, so this region here. So we could say it’s all of the real numbers minus this region. So we just put it in this format, so the real numbers, and then we’re subtracting this interval here from one to four inclusive because we don’t want one and we don’t want four to be our region. We want to exclude that from our region of solutions, so lots of different ways of presenting our solution sets there.

So with number three, we’ve gone to a purely algebraic approach.

Find the values of 𝑥 that satisfy 𝑥 squared minus three 𝑥 minus ten is less than or equal to zero. So what we’re gonna do is, we’ll say let’s consider the equation 𝑦 equals 𝑥 squared minus three 𝑥 minus ten. So we’re basically putting all of this lot equal to our 𝑦-coordinate. Now this is a quadratic, and the coefficient of 𝑥 squared is one, so that’s positive. So we know that this would be a positive happy curve. We also know that the constant term on the end is negative ten, so 𝐶 is equal to negative ten. And that’s where it cuts the 𝑦-axis. And it cuts the 𝑥-axis when the 𝑦-coordinate is equal to zero. So since 𝑦 is equal to 𝑥 squared minus three 𝑥 minus ten, what we’re saying is it cuts the 𝑥-axis when 𝑥 squared minus three 𝑥 minus ten equals zero.

And that factors, so 𝑥 squared minus three 𝑥 minus ten factors to 𝑥 plus two times 𝑥 minus five. And now we’ve got it in the format. We’ve got something times something is equal to zero, so one of those things must be equal to zero in order to get a-a result of that product of zero. So either 𝑥 plus two equals zero or 𝑥 minus five equals zero. And that means that 𝑥 must be the equal to negative two to make that equal to zero or 𝑥 must be equal to five to make that equal to zero.

So now we’ve got enough information for us to be able to sketch the curve of 𝑦 equals 𝑥 squared minus three 𝑥 minus ten. Well it cuts the 𝑦-axis at negative ten, and it cuts the 𝑥-axis at negative two and positive five. So that’s maybe about here, negative two; and positive five is over here. Now it’s a quadratic, so that’ll be a symmetric parabola. So the axis of symmetry, because it’s gonna be midway between negative two and positive five, is gonna be sort of here somewhere. And the curve is gonna look something like that. Now as we’ve said at the beginning, 𝑦 is equal to all of this stuff, and what we’re trying to find is the 𝑥-values for which that is less than or equal to zero. So we’re looking on this particular graph for where 𝑦 is less than or equal to zero.

Well 𝑦 is equal to zero here and 𝑦 is equal to zero here, so negative two and negative five are the 𝑥-values that generate a 𝑦-coordinate of zero. And we’re also looking for the region for which 𝑦 is less than zero, so that’ll be everything in between. So that’s all the way round here. So in terms of the 𝑥-values that generate those 𝑦 coordinates, well 𝑥 is negative two, 𝑥 is five, and everything in between. They are the valid 𝑥-coordinates. And for the 𝑥-coordinates we’re not interested in, well look up here, you can see that the 𝑦-coordinate is greater than zero so we’re not interested in that. So in terms of the region we’re not interested in, it’s this region out to infinity here; and it’s not including negative two, but it’s this region out to a negative infinity over here.

So the 𝑥-values we’re looking for is to generate that 𝑦-coordinate of less than or equal to zero are negative two is less than or equal to 𝑥 is less than or equal to five. So that’s in inequality format. In interval format, the ends of the interval we’re looking for are negative two and five, and they are both included. So we need to put the square brackets around those. So that’s in interval format. And using set notation, we can say that we’ve got the set of 𝑥 such that 𝑥 is real where negative two is less than or equal to 𝑥 is less than or equal to five.

So the process that we went through there was we, first of all, we came up with an equation for 𝑦 equals some combination of 𝑥, some function of 𝑥, and then we worked out where that generated a value of zero. And then we were trying to think of, you know, okay we were looking for the function to be less than or equal to zero in this case, or it might be equal to zero or greater than zero in other cases. So you’re then doing those comparisons. Now the bit that’s really important that I was talking about at beginning in terms of your working out is to do this sketch. If you do the sketch, it’s really clear to see whether you’re looking for points above the 𝑥-axis or points below the 𝑥-axis. If you don’t do that, lots of people go through these questions and they-they find out these critical 𝑥-values, but then they just kinda guess at whether we’re going between the 𝑥-values or outside of the 𝑥-values. So this final sketch here is just really helpful in getting it nice and clear in your mind whether you’re looking for points for 𝑦-coordinates above the line or below that line, the 𝑥-axis.

Okay for our last example then, let’s find the values of 𝑥 that satisfy minus 𝑥 squared minus 𝑥 is less than negative twelve. Now I’m gonna rearrange this problem. So we’re actually gonna solve an equivalent problem that comes up with the same answers, but it’ll be slightly easier to do. I don’t like working with these negative 𝑥 squareds and I don’t like having you know somethings on one side of the inequality and somethings on the other. It’s much easier to have an equality- an inequality based around zero, so we can look above and below the 𝑥-axis. So what I’m gonna do is I’m gonna add 𝑥 squared to both sides and I’m gonna add 𝑥 to both sides of this inequality. So I’ve got something is greater than zero.

So first of all, let’s add 𝑥 squared to both sides. And on the left-hand side, if I had 𝑥 squared, I’m just left with negative 𝑥 cause negative 𝑥 squared plus 𝑥 squared is zero. That cancels itself out. And then on the right-hand side, I’ve got 𝑥 squared, so plus 𝑥 squared minus twelve. You could say minus twelve plus 𝑥 squared, but I think it’s easier to write it that way around. So now we’re gonna add 𝑥 to both sides, and that gives us zero is less than 𝑥 squared plus 𝑥 minus twelve.

So this is the problem we’re gonna solve; zero is less than 𝑥 squared plus 𝑥 minus twelve. And it generates an entirely equivalent set of 𝑥-values- of solutions as the original problem. So let’s consider the quadratic equation 𝑦 equals 𝑥 squared plus 𝑥 minus twelve. And then effectually, what we’re trying to do is that means that this lot here is our 𝑦-coordinate and we’re trying to say when is that 𝑦-coordinate greater than zero. So if this is 𝑦, remember we’ve got the greater than; that’s on the bigger side of the greater tha- of the inequality to zero. So we are talking about when 𝑦 is greater than zero. It’s important to get that the right way round at this stage.

So looking at that quadratic then, that’s obviously one 𝑥 squared, so our 𝑎 value is positive. So this is another one of those happy curves. And the constant term on the end is negative twelve, so that means that it cuts the 𝑦-axis at negative twelve. And then the graph of that quadratic would cut the 𝑥-axis when 𝑦 is equal to zero; definition of the 𝑥-axis: the 𝑦-coordinate is zero. So that’s when 𝑥 squared plus 𝑥 minus twelve is equal to zero. Now again, I can factor 𝑥 squared plus 𝑥 minus twelve, 𝑥 plus four times 𝑥 minus three, and now again we’ve got two things multiplied together giving us zero. The only way you can get an answer of zero when you multiply two things together is if one of them is in fact zero.

So either 𝑥 plus four is equal to zero or 𝑥 minus three must be equal to zero. So when 𝑦 is zero, when it cuts the 𝑥-axis, that must be when 𝑥 is equal to negative four or 𝑥 equals three. So now we’ve got enough information to do a bit of a sketch. We can see that it cuts the 𝑦-axis at negative twelve and it cuts the 𝑥-axis at negative four and positive three. And the line of symmetry, remember quadratics are always these symmetric parabolas, is gonna be slightly to the left of the 𝑦-axis cause it’s gonna be midway between those two points where it cuts the 𝑥-axis; negative four is further away from zero than three is, so the line of symmetry’s slightly to the left of the 𝑦-axis.

So the graph looks something like that, a little bit wobbly, but it doesn’t have to be one hundred percent accurate. And we can see that the 𝑦-coordinate on each point on that graph is equal to the square of the 𝑥-coordinate plus the 𝑥-coordinate take away twelve. So in this problem, we’re looking for when the 𝑦-coordinate on that graph is greater than zero, which 𝑥 coordinates generate a 𝑦-coordinate of greater than zero.

Well when 𝑥 is equal to negative four, the 𝑦-coordinate is equal to zero, so that’s not greater than zero. So that’s not in the region we’re looking for. And when 𝑥 is three, that generates a 𝑦-coordinate equal to zero. So that’s not in the region we’re looking for. And in between these points down here, we can see that the 𝑦-coordinate is less than zero, so that’s no good. So the bits that we’re looking for are up here and out to infinity in that direction, negative infinity. And up here and out to positive infinity in that direction, that’s where 𝑦 is greater than zero.

So let’s consider the corresponding 𝑥 coordinates. We said three isn’t included, because that generates a 𝑦-coordinate of zero. But everything to the right of three up to positive infinity is included in our region because that’s generating 𝑦-coordinates which are above zero. Negative four is not included cause that’s a 𝑦-coordinate of zero. But everything to the left of that is included because again our corresponding 𝑦 coordinates would be greater than zero.

For the region that we’re not interested in; remember we said that all these points here are less than zero, they have a 𝑦-coordinate of less than zero, so not included is negative four, not included is positive three, and everything in between the two — whoops — everything in between the two across here is not included in the region that we’re looking for. So again, our green region has been split into two. It’s a noncontinuous region. So in terms of inequalities, 𝑥 is less than negative four or 𝑥 is greater than three. Remember, at negative four and three, we’re generating 𝑦-coordinates equal to zero. So that’s not the region we’re interested in.

In terms of intervals, we’re going all the way from negative infinity up to negative four. And we’re going all the way from three up to positive infinity. So we just need to think about the brackets or parentheses around these; well the infinities always have the round parentheses. And remember, four is not included in the region, so we use the round parenthesis; and three is not included in the region, so we use the round parenthesis. And we’re looking at the union of those two regions.

In set notation, we can say it’s the set of 𝑥 such that 𝑥 is real where 𝑥 is less than negative four or 𝑥 is greater than three. Or again, we can say look it’s all of the real values of- that 𝑥 could be apart from this region here, and so we could exclude that from our answer. So another way of writing it is the set of real numbers subtract the region from negative four to three. And the region that we’re excluding includes negative four and it includes three.

So in this case, we had to just rearrange the problem slightly to come up with an equivalent problem, which was relatively easy to solve. We did some basic analysis on 𝑥- and 𝑦-intercepts, did the graph again because that really helped us to understand the solutions, and then we’ve given you a range of different ways of presenting your answer.