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Write the solution set of each inequality if x is an element of the set of integers.$x^{2}-7 x+10>0$

$\{x | x<2 \text { or } x>5\},\{\ldots ., 1,2,6,7, \ldots\}$

Algebra

Chapter 1

THE INTEGERS

Section 8

Quadratic Inequalities

The Integers

Equations and Inequalities

Polynomials

Yoon L.

March 11, 2022

list all the integers that satisfy -4<x<2

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So we have X squared minus seven X plus two. He's got it in his ear. So 1st 1 back to our left hand side. So you're gonna want to part our coefficient times there last time when we get positive soon and then our middle term, which we want them to add to his native seven. So now a list of factors of 10 1 intend in two and five. And because it's positive, they both either beat need to be positive or they both need to be negative. But because our seven it's negative, you know, they both need to be negative. So in this case, we're choosing negative to a negative. But because I asked the negative seven so, uh, split the middle turn. So X squared minus two x minus five picks been plus tune and they were good. The 1st 2 and group the second to know we factor out the greatest common factor. So bloody X, we have X minus two and pull out a negative five. Here we have X minus two on the inside. And we know that because this is the same as this we have are two factors so excellent is to an expert in a spot of is greater than zero. So because this isn't any quality, it takes an extra step to solve for X because you have to find a range of X values and because a positive times a positive gives us a positive and we want we want our back there's to be multiply and give us a positive number. But because a negative times negative also gives us a positive number we need to solve. For Wayne, the factors are less than zero, so I also need to saw or X minus two times X minus pi. It's less than zero because when I vote, if you multiply them both and it gives you a negative number, it turns to a positive because negative times negative give us a positive so we'll have X minus two is greater than zero over here, and X minus five is greater than zero. And we acted about. Sides here still have excess grated into, and we add five to both sides here, so we'll have exits guarded in pots. So okay, he's a dreidel, A little number line. To combine. These two qualifications will have to hear and it's greater dance. It goes this way, and we have five here and it's great. And so it goes this way and the only one the only way to combine this is it. X is greater than by. So now we need to software or the wind still have X minus two is less than zero an X minus. Pi is less than zero and we add to the both sides. So accept less than two. And we had Bob to both sides and exits list in five. So I gave the door a little number line. We'll have to and five fixes listen to So it goes this way and excess lesson five. So it goes this way and it means both qualifications when exist, list into. So that's how we come by on that one. So because our shading goes to different directions we're gonna use or because it's not in between so we're gonna use or so X is gonna get by or X is less than two. We also want to find a solution set. So I just do two numbers and I have been below so huh ex congee three. No, not through work it has to be greater than five so two numbers guarded by this 67 and then two numbers below to ISS one and zero, so that could be a solution said.

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