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Numerade Educator

Missouri State University

University of Michigan - Ann Arbor

University of Nottingham

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So this is really similar to the last problem. We're solving an inequality, but notice this time, the only difference is that we have and inclusive inequality. So I'm actually going to be including this boundary point of five. So x squared plus one actually could be equal to five. And then, of course, it could be greater than five as well. So, again, I'm going to start by subtracting one. So I have X squared needs to be greater than or equal to four. And then I'm gonna take the square root of both sides. And this is really important because we're going to see this time and time again during this course. But it's important to recognize this now. When I take the square root of X squared, I don't get X, I get absolute value of X. And when I take the square root of four, maybe you're in the habit of saying, Oh, that's plus or minus to know the square of four is, too. But when I take the square root of a variable squared, I need to consider the fact that the variable could be negative as well as positive. So actually, this is equivalent to just saying absolute value of X is greater than or equal to two. Now, if I would absolute value of X to be greater than or equal to two. Actually, two things can happen, sort of like in the last example I needed. Two things toe happen to be less than three here. One of two things could happen for absolute value of X to be greater than or equal to two. The first one is that X could be greater than or equal to two. But there's also something else that can happen. So I haven't Or here I could also have ex being so negative that when I take the absolute value, I end up with something greater than or equal to two. So in other words, X could be less than or equal to negative, too, because then when I take the absolute value, I will be greater than or equal to. And again. The key word here is going to be, or so I get to solution sets here from my two cases to consider. Okay, One of them is to your infinity, and the other one is negative. Infinity to to and the key word here is or so with. And I took an intersection of the two sets that I found with or I'm actually going to take the union. That's negative. So the solution set is going to be negative. Infinity too negative, too. Union to to infinity. And you see that any number in the set when I take the opposite value will be greater than or equal to. So here is my solution set.

Georgia Southern University