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


Welcome to the presentation
on quadratic inequalities. I know that sounds very
complicated, but hopefully you’ll see it’s actually
not that difficult. Or at least, maybe the problems
we’re going to work on aren’t that difficult. Well, let’s get started with
some problems and hopefully you’ll see where this is kind
of slightly different than solving regular
quadratic equations. So let’s say I had the
inequality x squared plus 3x is greater than 10. And remember, whenever you
solve a quadratic or I guess you would call it a second
degree equation– I guess this is an inequality. I shouldn’t use the
word equation. It’s tempting to sometimes do
it the same way you’d do a linear equation, kind of
getting all the x terms on one side and all the
constants on the other. But it never works because you
actually have an x term and then you have an
x squared term. So you actually want to get it
in kind of what I would call the– I don’t know if it’s
actually called this– the standard form where you
actually have all of the terms on one side and then a
0 on the other side. And then you can either
factor it or use the quadratic equation. So let’s do that. Well this is pretty easy. We just have to subtract 10
from both sides and we get x squared plus 3x minus
10 is greater than 0. Now let’s see if
we can factor it. Are there two numbers that when
you multiply it become negative 10 and when you add it
become positive 3? Well, yeah. Positive 5 and negative 2. And once again, at this point I
think you already know how to do factoring, so this should
be hopefully, obvious to you. So it’s x plus 5 times x
minus 2 is greater than 0. Now this is the part where it’s
going to become a little bit more difficult than just your
traditional factoring problem. We have two numbers, I
guess you could view it. We have x plus 5. I view that as one number. Or I guess we have
two expressions. We have x plus 5 and
we have x minus 2. And when we’re multiplying
them we’re getting something greater than 0. Now let’s think about
what happens when you multiply numbers. If they’re both positive and
you multiply them, then you get a positive number. And if they’re both negative
and you multiply them, then you also get a positive number. So we know that either both of
these expressions are the same sign, that they’re both greater
than 0, they’re both positive. Or we know that they’re
both negative. And I know this might be a
little confusing, but just think of it as– if I told you
that– I’ll do something slightly separate out here. If I told you that a times b is
greater than 0 we know that either a is greater than 0
and b is greater than 0. Which just means that
they’re both positive. Or a is less than 0
and b is less than 0. Which means that
they’re both negative. All we know is that they both
have to be the same sign in order for their product
to be greater than 0. Now we just do the
same thing here. So we know that either both of
these are positive, so x plus 5 is greater than 0 and x
minus 2 is greater than 0. Or– now this is a little
confusing, but if you work through these problems it
actually makes a lot of sense. Or they’re both negative. Or x plus 5 is less than 0 and
x minus 2 is less than 0. I know that’s confusing, but
just think of it in terms of we have two expressions: they’re
either both positive or they’re either both negative. Because when you multiple
them you get something larger than 0. Well, let’s solve this side. So this says that x is
greater than negative 5 and x is greater than 2. We just 2 both sides
of this equation. Or, and if we solve this side–
x is less than negative 5 and x is less than 2. I just solved both of these
inequalities right here. Now we can actually simplify
this because here we say that x is greater than negative
5 and x is greater than 2. So in order for x ti be greater
than negative 5 and for x to be greater than 2, this just
simplifies as saying, well, x is just greater than 2. Because if x is greater
than 2, it’s definitely greater than negative 5. So it just simplifies to this. And we’d say or– and here we
said x is less than negative 5 or x is less than 2. Well, we know if x is less
than negative 5, then x is definitely less than 2. So we could just simplify it to
or x is less than negative 5. So the solutions to this
problem is x could be greater than 2 or x could be
less than negative 5. And so let’s just think
about how that looks on the number line. So if 2 is here, x could
be greater than 2. So it’s all of these numbers. And if this is negative 5–
I shouldn’t have done it so close to the bottom. x is less the negative 5. So these are the numbers
that satisfy this equation. And I’ll leave it up to you
to try out to see that they actually work. Let’s try another one
and hopefully, I can confuse you even more. Let’s say I have minus x
times 2x minus 14 is greater than or equal to 24. Well, the first thing we want
to do is just manipulate this so it looks in the
standard form. So we get negative 2x squared
plus 14x– I’m just distributing the minus x– is
greater than or equal to 24. I don’t like any coefficient it
front of my x squared term, so let’s divide both sides of
this equation by negative 2. So we get x squared–
we divided by negative 2– minus 7x. And remember, when you divide
by a negative number you switch the sign on the inequality, or
you switch the direction of the inequality. So we’re dividing by negative
2, so we switched it. We went from greater than
or equal to, to less than or equal to. And then 24 divided by
negative 2 is minus 12. And now we can just bring this
minus 12 onto the left-hand side of the equation. Add 12 to both sides. We get x squared minus 7x plus
12 is less than or equal to 0. And then we can just factor
that and we get, what is that? It’s x minus 3 times x minus 4
is less than or equal to 0. So now we know that when we
multiply these two terms we get a negative number. So that means that these
expressions have to be of different signs. Does that make sense? If I tell you I have two
number and I multiply them, I get a negative number. You know that they have to
be of different signs. So we know that either x minus
3 is less than or equal to 0 and x minus 4 is greater
than or equal to 0. So that’s one case. And the other case is x minus 3
is greater than or equal to 0, which means x minus
3 is positive. And x minus 4 is less
than or equal to 0– oh, I went to the edge. So let’s solve this and
hopefully it’ll simplify more. So this just says that x is
less than or equal to 3. And this says x is greater
than or equal to 4. So both of these things have to
be true. x has to be less than or equal to 3 and x has to be
greater than or equal to 4. Well, let me ask
you a question. Can something be both less than
or equal to 3 and greater than or equal to 4? Well, no. So we know that this
situation can’t happen. There’s no numbers that’s
less than or equal to 3 and greater than or equal to 4. So let’s look at
this situation. This says x is greater than
or equal to 3 and x is less than or equal to 4. Can this happen? Sure. That just means that x is
some number between 3 and 4. If we were to draw this on the
number line, we would get– if this is 3, this is 4. And it’s greater than or
equal to so we fill it in. And less than or equal
to so we’d fill it in. And it would be any number
between 3 and 4 would satisfy this equation. And I’ll leave it up
to you to try it out. I know this is confusing at
first, and this is actually something that they normally
don’t teach really well, I think, in most high schools
until 10th or 11th grade. But just think about you’re
multiplying two expressions. If the answer is negative
then they must be of different signs. If the answer is positive
they must be the same sign. And then you just work
through the logic. And you say, well, no number
can be less than 3 and greater than 4, so this doesn’t apply. And then you do this side
and you’re like, oh, this situation does work. It’s any number
between 3 and 4. Hopefully that gives you
a sense of how to do these type of problems. I’ll let you do the
exercises now. Have fun.

About James Carlton

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31 thoughts on “Quadratic inequalities | Polynomial and rational functions | Algebra II | Khan Academy

  1. Sir I am trying the same concept to find domain of sqrt of 9-x^2 I am not getting the correct result

  2. U guys your teachers spent their college just to teach us. Dont you ever say that a 9 min. Vid is more simplier than waht your teacher teaches for 2,3,4 lessons.
    Sometimes u gotta ask and Listen carefully.

    There is no use of being shy

  3. Ohh, this just explained 2-3 years of confusion when doing this. I never understood how the answer got there and always just skipped to it, but after that explanation it got rid of all of that confusion. Thanks so much!

  4. My teacher just made me mug up the factorising inequalities, he never actually told what they meant and that made me so mad. But thank you so much sal.

  5. Mom : All you do is looks at your phone all day and never study

    me: * actually finished studying calculus at age 13 *

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