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## Algebra (all content)

### Course: Algebra (all content) > Unit 9

Lesson 12: Quadratic inequalities# Quadratic inequalities

Sal solves x^2+3x>10. Created by Sal Khan.

## Want to join the conversation?

- Could you use the quadratic formula to solve a question like this?(14 votes)
- You can use the quadratic equation to find the endpoints of the intervals that will be you solution, and would then need to test in which of those intervals the inequality is true. So in this case you could use it to find -5 and 2 [(-3 +- Sqrt(9+4(10)1))/2 = (-3 +- 7)/2 = -10/2 or 4/2]. This breaks up the number line into 3 intervals {x<-5, -5<x<2 and x>2} for a quadratic inequality, the solution will always either be the two outer intervals or the middle one (think about how a graph would look), so you can just check the middle. In this case you can check x =0, and since 0^2 + 3(0) is not greater than 10, you know it will be the outside intervals {x<-5 or x>2}(12 votes)

- What would happen if the greater than turns into less than?(6 votes)
- x² + 3x < 10

x² + 3x - 10 < 0

(x + 5)(x - 2) < 0

( x + 5 > 0 and x - 2 < 0 ) OR ( x + 5 < 0 and x - 2 > 0 )

( x > -5 and x < 2 ) OR ( x < -5 and x > 2 )

Because it is impossible that x < -5 and x > 2, our final answer is x > -5 and x < 2, which we can write as -5 < x < 2.(8 votes)

- how do we solve an inequation like x-3/x-5>0(1 vote)
- When you have an inequality like this, you can't multiply the denominator with 0 and make the simplification easier. This is because you are unsure of whether the value of "x" is positive or negative. And due to this , if x was negative you'd have to flip the inequality but like I said you do not know for sure. So this how you do it :

(x-3)(x+5) / (x-5)(x+5) > 0

(x-3)(x+5) / (x^2 - 25) > 0

Now you can multiply the denominator with zero since even if your "x" is negative is will become positive since it's been squared. So,

(x-3)(x+5) > 0

So , you'll have two values, x > 3 or x > -5 . Now the final part is figuring out the interval. When you have two values , the method I use to find the interval is using the wavy curve method ( It would be tough for me to explain it , better to understand it visually than by writing). According to the wavy curve method, when you have "x" is greater than ( >) something, then you leave the values between the two numbers and take all the other values.

Since our values are -5 and 3, the solution will be -∞ .....-8,-7,-6,-5 and 3,4,5,6....+∞. When you write this in interval notation , you'll have (-∞,-5)U(3,+∞)(8 votes)

- how will you solve this x² + 8x +15 < 0(1 vote)
- Problem: x^2+8x+15<0

First step: Factor out the inequality. (what times what equals 15 and when added together makes 8?)

(x+3)(x+5)<0

Step 2: Solve for x. This inequality has two answers.

X can either be -3 or -5, since both, when plugged in for x, will make the inequality equal to zero.

Step 3: Draw a number line with the points -3 and -5 plotted with hollow circles, since the inequality has a < sign.

Step 4: Plug in -4 (representing the space in between the points -3 and -5 on the number line) into the (x+3)(x+5) and solve.

(-4+3)(-4+5)= -1 The answer is negative.

Step 5: Look at x^2+8x+15<0. The answer is suppose to be LESS than 0, which means that the answer has to be a NEGATIVE number. Since when -4 is plugged into the inequality, the answer is negative, the space in between the points -3 and -5 on the line is the SOLUTION.

Step 6: Write the solutions to the inequality in the form of X<? or X>?. Since the answer is in between -3 and -5, the solutions should be X>-5 or X<-3.

Final solutions: x>-5 or x<-3

Hope this helped!!(6 votes)

- how would you solve this?:

find the set of values of x for which

3 - 5x - 2x^2 < 0(2 votes)- By factorising, and then solving for two different x values:

-2x^2 - 5x + 3 < 0

( 2x - 1 )( x + 3 ) < 0

2x - 1 < 0 or x + 3 < 0

x < 1/2 or x < -3(2 votes)

- Could the Quadratic formula be used to solve this?(2 votes)
- I thought that a quadratic inequality would generate an area like it happens in linear functions. is it possible?(1 vote)
- That can be the case. If you have a quadratic inequality then you need to factorise it to work out your values of x. Once you have done this, you can plot them on a graph. If the inequality is > 0, than the area the inequality is representing will be be above the x -axis. However, if the inequality is < 0 than the area represented is below the x - axis.(3 votes)

- x^2>1 and x^2<1

what are the values for x.(1 vote) - I'm new-ish to algebra and I have no idea how to approach this question: -2b^2+1000b-1000. Originally it's from these two equations which I merged: 2(a+b)=1000 and a+b+2ab<1500. I might've done a step wrong but I can't factor the first equation and when I apply the quadratic formula, it comes out with square roots.(1 vote)
- how will we solve (x-1)(x-2)(x-3) > 0(1 vote)

## Video transcript

Let's say that we want to
solve the inequality x squared plus 3x is greater than 10. We want to figure out
all of the x's that would satisfy this inequality. I encourage you to
pause this video now. And I'll give you a hint. Try to manipulate the
way that you would have if this was a
quadratic equation. But then as you
get to the end, try to reason through it, because
the reasoning might departure a little bit from
what you are used to. So I'm assuming you've
given a go at it. So the first thing
that we might want to do, just to get
into a form that we're more comfortable with, is
subtract 10 from both sides. If we subtract 10
from both sides, then on the left
hand side, we're going to have x squared
plus 3x minus 10 is still going to be greater than. If we add or subtract the
same thing to both sides, it won't change the inequality. But it's now going
to be greater than 0. 10 minus 10 is 0. Now, this gets us into a
form that we're more used to seeing quadratic
expressions in. If this was an equal
sign right over here, we'd want to factor this thing. So let's just try to factor
here too and see what happens. So we're going to factor it. We're going to
think of two numbers whose product is negative 10
and whose sum is positive 3. And we've had a lot of
practice doing this. If you think about the factors
of 10, it's 1, 2, 5, and 10. 2 and 5 seem tempting,
because their difference is 3. So if you have positive
5 and negative 2, that seems to work out. Positive 5 and negative 2. Their product is negative
10, their sum is positive 3. We could rewrite
this as x plus 5. Let me do that in
that yellow color so you see where this
5 is coming from. X plus 5 times x minus 2 is
going to be greater than 0. Now, if this was
an equality here, we would say well, how
do we get this equals 0? If either of these
things were equal to 0, then this entire expression
would be equal to 0, because 0 times anything is 0. But we don't have
an equality here. We have a greater than symbol. So let's think about how we
could reason through this. And I'll do a little
bit of an aside here. If I were to tell you
that numbers a and b, and if I were take
the product a times b, and if someone were to tell you
that product is greater than 0, what do we know about a and b? Well, we know that they
have to have the same sign. They're either both
positive-- a positive times a positive is going
to be a positive-- or they're both going to be
negative-- a negative times a negative is a positive. It's going to be greater than 0. So we know the same thing here. Let me write it down. So we know either a is greater
than 0 and b is greater than zero-- so either
both of them are positive or both of them
are negative-- or a is less than zero and
b is less than zero. So we apply that
same logic here. You could view
this x plus 5 is a, you could view this x minus
2 as our b of the product of two things. The product is greater than 0. That means that either both of
these expressions are positive or they're both negative. So let's write that down. I'll write it this way. So either both of these
expressions are positive. So either x plus 5 is
greater than 0 and x minus 2 is greater than 0-- let
me write it this way-- or they're both negative. x plus 5 is less than 0 and
x minus 2 is less than 0. So now let's think about
all of these inequalities independently. But let's maintain this logic
here of the and and the or. So let's look at this. Either they're both positive,
so if both of these expressions are positive, what
do we know about x? Well, if you subtract 5 from
both sides of this inequality, you get x is greater
than negative 5. And if you add 2 to this
inequality, both sides of that inequality, you're going
to get x is greater than 2. So if x is greater than negative
5 and x is greater than 2, what do we know about x? Well, any x that's
greater than two is going to be greater
than negative 5. So we could just simplify
this right over here to say that x is greater than 2. So all of this, this
is equivalent to saying x is greater than 2,
because clearly anything that is greater than
2 will satisfy that. And both of these
things have to be true. For example, x
equal negative 4, it would satisfy this
inequality but not this one. And so the and would break down. And negative 4 does not
satisfy both of these. In order to satisfy
both of these, you essentially have
to satisfy that one. So this expression
simplified to that. Now what about this? What about this statement
right over here? Well, x plus 5 less than 0,
subtract 5 from both sides. That is, x is less
than negative 5. And add 2 to both sides
of that inequality, you get x is less than 2. Now if x is less than negative
5 and x is less than two, what do we know about x? Well, that just means that x
has to be less than negative 5. If it's less than
negative 5, it's definitely going
to be less than 2. And we got to remind ourselves
that we have this or here. And that's essentially
describing the solution set for this quadratic
inequality here. x is going to be
greater than 2 or x is going to be less
than negative 5. And we could actually plot this
solution set on a number line. So if this is our number line
right over here, and let's say that this is 0. Let's say that's 1. 2 right over here. This is negative 1, negative
2, negative 3, negative 4, negative 5. So x could be greater than 2,
not greater than or equal to, so I'm going to put
an open circle here. So it could be greater than 2. Or it could be less
than negative 5, not less than or
equal to, so I'm going to put an
open circle here. And so it could
be less than that. So x could be any number. It could be negative 6,
negative 6 would satisfy this. You could verify that. Negative 6 squared is
36 plus negative 18, which is going to be 18,
which is greater than 10. Or you could have, say,
a positive 3 would work. 3 squared is 9 plus
another 9 is going to be 18, which once again,
it is greater than 10.