### 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.