Rational inequalities | Polynomial and rational functions | Algebra II | Khan Academy
- Articles, Blog

Rational inequalities | Polynomial and rational functions | Algebra II | Khan Academy


In this video I want to do a
couple of inequality problems that are deceptively tricky. And you might be saying,
hey, aren’t all inequality problems deceptively tricky? And on some level
you’re probably right. But let’s start with
the first problem. We have x minus 1 over x
plus 2 is greater than 0. And I’m actually going to show
you two ways to do this. The first way is, I think, on
some level, the simpler way. But I’ll show you both methods
and whatever works for you, well, it works for you. So the first way you can think
about this, if I have just any number divided by any other
number and I say that they’re going to be greater than 0. Well, we just have to remember
our properties of multiplying and dividing negative numbers. In what situation is
this fraction going to be greater than 0? Well, this is going to be
greater than 0 only if both a and– so we could write both a
is greater than 0 and b is greater than 0. So this is one circumstance
where this’ll definitely be true. We have a positive divided
by a positive; it’ll definitely be a positive. It’ll definitely be
greater than 0. Or we could have the situation
where we have a negative divided by a negative. If we have the same sign
divided by the same sign we’re also going to be positive. So or– a is less than 0
and b is less than 0. So whenever you have any type
of rational expression like this being greater than 0,
there’s two situations in which it will be true. The numerator and the
denominator are both greater than 0, or
they’re both less than 0. So let’s remember that and
actually do this problem. So there’s two situations
to solve this problem. The first is where both of
them are greater than 0. If that and that are both
greater than 0, we’re cool. So we could say our first
solution– maybe I’ll draw a little tree like that– is x
minus 1 greater than 0 and x plus 2 greater than 0. That’s equivalent to this. The top and the bottom– if
they’re both greater than 0 then when you divide them
you’re going to get something greater than 0. The other option– we just
saw that– is if both of them were less than 0. So the other option is x
minus 1 less than 0 and x plus 2 less than 0. If both of these are less than
0 then you have a negative divided by a negative,
which will be positive. Which will be greater than 0. So let’s actually solve in
both of these circumstances. So x minus 1 greater than 0. If we add 1 to both
sides of that we get x is greater than 1. And if we do x plus 2 greater
than 0, if we subtract 2 from both sides of that equation–
remember I’m doing this right now– we get x is
greater than minus 2. So for both of these to hold
true– so in this little brown or red color, whatever you want
to think of it– in order for both of these to go hold true,
x has to be greater than 1 and x has to be greater
than minus 2. This statement we figured out
means that x has to be greater than 1 and this statement
tells us that x has to be greater than minus 2. Now, if x is greater than 1 and
x has to be greater than minus 2, x clearly has to
be greater than 1. You know, 0 would not satisfy
this because 0 is greater than minus 2, but it’s
not greater than 1. So for something to be greater
than 1 and minus 2 it has to be greater than 1. This whole chain of thought
where I’m saying the numerator and the denominator are greater
than 0, that’s only going to happen if x is greater than 1. Because if x is greater than 1,
then x is definitely going to be greater than negative 2. Any number greater than
1 is definitely greater than negative 2. So that’s one situation in
which this equation holds true, and we can even try it out. Let’s say x was 2. 2 minus 1 is 1 over 2 plus 2. It’s 1/4. It’s a positive number. Now let’s do the situation
where both of these are negative. If the x minus 1 is less than
0, if we add 1 to both sides of that equation that tells us
so x minus 1 is less than 0. That’s the same thing– if we
add 1 to both sides of that– as saying that x
is less than 1. So that constraint boils
down to that constraint. Now this constraint, x
plus 2 is less than 0. If we subtract 2 from
both sides we get x is less than minus 2. So this constraint boils
down to that constraint. So in order for both of these
guys to be negative, both the numerator and the denominator
to be negative– we know that x has to be less than 1 and x
has to be less than minus 2. Now if something has to be less
than 1 and it has to be less than minus 2, well, it just
has to be less than minus 2. Anything less than minus 2 is
going to satisfy both of these. So this boils down to
just x could also be less than minus 2. And remember, this is an or. Either both the numerator and
the denominator are positive, or they’re both negative. So both of them being positive
boil down to x could be greater than 1, or both of them being
negative boils down to x is less than minus 2. So our solution is x could be
greater than 1 or– that’s both of these positive–
or x is less than minus 2. That’s both of these negative. And if we wanted to draw it on
a number line– let me draw a number line just like that. That could be 0 and then
we have 1, so x could be greater than 1. Not greater than or equal to. So we put a little circle
there to show that we’re not including 1. And everything greater than 1
will satisfy this equation. Or anything less than minus 2. So we have minus 1, minus 2,
anything less than minus 2 will also satisfy this equation by
both making the numerator and denominator negative. And you could try it out. Minus 3. Minus 3 minus 1. I’ll just do minus 3 minus
1 is equal to minus 4. And then minus 3 plus 2. Minus 3 plus 2 is
equal to minus 1. Minus 4 divided by
minus 1 is positive 4. So all of these negative
numbers also work. Now, I told you that I
would show you two ways of doing this problem. So let me show you another way
if you found this one maybe a little bit confusing. So the other way– let
me rewrite the problem. You get x minus 1 over x
plus 2 is greater than 0. And actually, let’s mix it up
a little bit and you could apply the same logic. Let’s say it’s greater than or
equal– well, actually no. I’ll just keep it the same way
and maybe in the next video I’ll do the case where it’s
greater than or equal just because I really don’t want
to– maybe I want to incrementally step up the
level of difficulty. We’re just saying x minus
1 over x plus 2 is just straight up greater than 0. Now one thing you might say is
well, if I have a rational expression like this, maybe I
multiply both sides of this equation by x plus 2. So I get rid of it in the
denominator and I can multiply it times 0 and
get it out of the way. But the problem is when you
multiply both sides of an inequality by a number– if
you’re multiplying by a positive you can keep the
inequality the same. But if you’re multiplying by a
negative you have to switch the inequality, and we don’t
know whether x plus 2 is positive or negative. So let’s do both situations. Let’s do one situation
where x plus 2– let me write it this way. x plus 2 is greater than 0. And then another situation
where– let me do that in a different color. Where x plus 2 is less than 0. These are the two
possibilities for x plus 2. Actually, in those situations
can x plus 2 equal 0? If x plus 2 were to be equal to
0 than this whole expression would be undefined. And so that definitely won’t
be a situation that we want to deal with it. It would an undefined
situation. So these are our two situations
when we’re multiplying both sides. So if x plus 2 is greater
than 0 that means that x is greater than minus 2. We can just subtract 2 from
both sides of this equation. So if x is greater than
minus 2, then x plus 2 is greater than 0. And then we can multiply
both sides of this equation times x plus 2. So you have x minus 1 over
x plus 2 greater than 0. I’m going to multiply both
sides by x plus 2, which I’m assuming is positive because
x is greater than minus 2. Multiply both sides
by x plus 2. These cancel out. 0 times x plus 2 is it just 0. You’re just left with x minus
1 is greater than– this just simplified to 0. Solve for x, add 1 to
both sides, you get x is greater than 1. So we saw that if x plus 2 is
greater than 0, or we could say, if x is greater than
minus 2, then x also has to be greater than 1. Or you could say if x is
great– well, you could go both ways in that. But we say, look, both of
these things have to be true. If for x to satisfy both
of these, it just has to be greater than 1. Because if it’s greater
than 1 it’s definitely going to satisfy this
constraint over here. So for this branch we come
up with the solution x is greater than 1. So this is one situation where
x plus 2 is greater than 0. The other situation is x
plus 2 being less than 0. If x plus 2 is less than 0
that’s equivalent to saying that x is less than minus 2. You just subtract 2
from both sides. Now, if x plus 2 is less than
0, what we’ll have to do when we multiply both
sides– let’s do that. We have x minus 1
over x plus 2. We have some inequality
and then we have a 0. Now if we multiply both
sides by x plus 2, x plus 2 is a negative number. When you multiply both
sides of an equation by a negative number you have
to swap the inequality. So this greater than sign will
become a less than sign because we’re assuming that the
x plus 2 is negative. These cancel out. 0 times anything is 0. We get that x minus
1 is less than 0. Solving for x, adding 1 to
both sides, x is less than 1. So in the case that x plus 2 is
less than 0 or x is less than minus 2, x must be less than 1. Well, I mean we know– if you
say something has to be less than minus 2 and less than 1,
just saying that it’s less than minus 2 will do the job. Anything less than minus 2 is
going to satisfy this one. But not anything that satisfies
this one is going to satisfy that one. So this is the only constraint
we have to worry about. So in the event where x plus 2
is less than 0, we can just say that x has to be
less than minus 2. That’ll satisfy this equation. So our final result is x is
either going to be greater than 1 or x is going to
be less than minus 2. So once again, I can graph
it on the number line. x is greater than
1 right there. You have 0, minus 1, minus 2. And then you have x is less
than minus 2, you’re not including the minus 2. And just like that. And that is the exact same
result we got up here. So whichever version
you find to be easier. But you can see, they’re both a
little bit nuanced and you have to think a little bit about
what happens when you multiply or divide by positive
or negative numbers.

About James Carlton

Read All Posts By James Carlton

65 thoughts on “Rational inequalities | Polynomial and rational functions | Algebra II | Khan Academy

  1. Thanks this video was helpful but could you do one on geometry? I'm finiding that subject a little bit tricky and it would be highly appreciated if you did a video on the basics of geometry

  2. Can someone tell me the rules for a closed dot and an open dot on the number line when graphing a quadratic inequality? Actually, I mean when you have one closed and one open.

  3. your really good and i understand it could you make a part 3 if thats ok? and if u already have THANKS! i will check it out you are a really good taecher.

  4. What if the problem is greater than/less than or equal to? I remember that when you graph it on a number line either the variables in the numerator/denominator have to be open circles but I dont remember which.

  5. I'm so glad you make these videos. I get so confused with the explanations the math teacher I have gives, but when you explain how to solve them, it just make sense. Thank you so much!

  6. okay this is probably a very stupid question but I have this problem which is (3x-6)/(2x+1) + x > 0 what do I do with that +x? Like I'm completely lost .-.

  7. bruh all you had to do was split each equation and do X-1=0 and X+2=0 and you'll get your two answers. easy.

  8. You're taking too much time on explaining while you can just say that when you solve the x-1 and x+2 the signs of both -1 and +2 will changed into +1 and -2 lol

  9. WHAT ARE YOU TRYING TO IMPLY HERE? WHAT I WANT TO SEE HERE IS THAT YOUR FIRST STEP MUST BE EQUATING THE NUMERATOR AND DENOMINATOR TO ZERO. THANKS.

  10. two ways to solve it? you trying too hard come on theres no difference between them you just making the "second step" more complicated its the same thing as step one

  11. i understand how to solve rational functions but not solving rational inequalities. its a bit harder.

Leave a Reply

Your email address will not be published. Required fields are marked *