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Write the solution set of each inequality if x is an element of the set of integers.$x^{2}-x<6$

$\{x |-2<x<3\},\{-1,0,1,2\}$

Algebra

Chapter 1

THE INTEGERS

Section 8

Quadratic Inequalities

The Integers

Equations and Inequalities

Polynomials

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

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so we have X squared minus X is less than six. So because we know you want a factor of China mule, we need to move this six to the other side. So I'm going to subtract six from both sides, so we'll have X squared minus X minus six is less than zero. So the factor this we need to split apart or middle term here. So needles apart coefficient times our last term and not give us negative six. And we need them to add two neg newborn. So when a list of factors of 616 and two and three and we know that we need one of the numbers and prepare to be negative and also needs to add some negative boy. So after testing a lunch combinations will get positive, too, and negative. Three is our appeared. So now we just need to split apart of middle turn. So have X squared minus two x no plus two X minus three x and then one a six. Listen, zero throwing a group the first tooth and group the 2nd 2 So we're gonna factor out the greatest common factor it from the person we get back here on exile, have X plus two you from the second we get back to what a negative three. So have explosives too, is gonna than zero and is listen zero Sorry. So because this is the same as this, we know that exposed to Times X minus three is less than zero. So solving inequalities takes a further step. Then just when the trying a mule's equal to zero because we're looking for a range of values. So we need to find arranged that meets all the qualifications. So because we know that this product is basically negative, less than zero means negative. The only way to get a negative product is if this one is positive and this one is negative, that would give us a negative, or this one is negative. And this one is positive, and I will also give us a negative. So I'm gonna say X plus two is positive. An X minus three is negative, and we also need to say X plus two isn't a new an X minus three is positive. So when I say like, for instance, this one here, greater than zero just means it's positive. So I just followed the signs. It really helps to do right down the signs under each factor. So that way you can make sure that she tests for all the possibilities. So no one is gonna sell critics. So subtract it from both sides. Will have X is greater than negative too. We added about sides here who have X is less than three. So we need to draw them. Relying it really hopes. No. A plot in ated to they will quite positive three. So know that X is greater than negative too. So to go this way and X is less than positive. Three. So this will go this way and we shade the combined area. It will have negative to this less than X is less than positive. Three for this one here we also need to solve for the other possibility. So over here you gotta subtract two from both sides. Say X is less than negative too. And here we're at through to both sides will have X is greater than positive three. So when I draw another number light, you have negative too. And three. So it's less than native to, and it's greater than positive three. Now, there's no common area to shade there, or this possibility does not exist. So we're only gonna focus on this answer. We also need to find a solution set. That just means we're gonna find about three or four numbers. The ex congee they're always go to above and two below. So we're gonna have brackets. I'm gonna start with some dots to show that it keeps going on. So two numbers below negative to that will be negative for negative three. And because we do not have a less standard equal to we cannot include negative, too. But we can start it positive three and two above three is or and five. And we also need to put some more dots to show that the range keeps going.

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