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In each of the following exercises, solve the given inequality.$$x^{2}-2 x-1<0$$

$$1-\sqrt{2}<x<1+\sqrt{2}$$

Algebra

Chapter 0

Reviewing the Basics

Section 5

Solving Non-Linear Inequalities

Equations and Inequalities

Oregon State University

Harvey Mudd College

Baylor University

Lectures

02:45

In each of the following e…

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03:28

02:16

03:41

03:18

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All right, I've got a question here. X squared minus two X minus. One is less than zero. And first they want to do is factor this out. And we could see that it's not a standard factor. The more factor, even they are perfectly, I should say. So we want to do is you want to go ahead and use our quadratic formula. We say X is equal to negative B us minus, maybe swear. Minus for a C over two A. Okay, this law, you here is your A This is your being. This is your seat. When you go ahead and plug that in into your handy dandy calculator, you will get that your X is equal to a one minus root, too. And your ex is also equal to a one plus route to All right, And then from there, what we're gonna do is you're gonna take your ex values. You're gonna plot it on a number line. And you have that going from negative infinity too, Infinity. All right. Now what we're gonna do is you're going to try to determine what region of X values does. Do they were regions X value, satisfy your inequality, so we'll plug in a value between one plus route to and your infinity. So let's see. We got one plus route to that is approximately 2.41 So let's plug in three. We plug in three for X will get three squared minus two times three minus one. So that will be nine minus six, which is three minus. One is to we know, too is certainly not less than zero. Therefore, this'll region does not satisfy the inequality. Pick a value between one minus routes to let's see what one minus Ruutu comes out. It comes out to be a negative value, so we could just choose something in the middle. Here we could pick zero. So we set our X values equal to zero. We would have zero minus zero minus one, which would be a negative one, and negative one is certainly less than zero. Therefore, we could say, Well, this this region of values would satisfy the inequality. Finally, we'll pick a value between one minus route to negative infinity. We could choose negative three and we would get a negative. We get a nine minus a negative six, which is 15 plus minus, one would be 14. So that's still be a positive value. And we know that positive would not be less than zero. So we could say that our domain for X exists when X is greater than our domain for exes, when X is greater than one minus route to and less than one plus route to and thes values of X would satisfy the inequality. All right, well, I hope that clarifies the question. Thank you so much for watching.

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