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In each of the following exercises, solve the given inequality.$$(3 x-4)^{4}(2 x-3) \leq 0$$

$$x \leq 3 / 2$$

Algebra

Chapter 0

Reviewing the Basics

Section 5

Solving Non-Linear Inequalities

Equations and Inequalities

Oregon State University

McMaster University

University of Michigan - Ann Arbor

Lectures

03:16

In each of the following e…

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03:26

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00:54

Solve each inequality.…

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01:28

Solve each inequality alge…

All right, we've got a question here. That is three X minus four for the fourth multiplied by two X minus three. And that is all less than or equal to zero. Alright, so first thing we're gonna do is gonna solve for R X values by setting them equal to setting each a factor equal to zero. And we'll say, Well, then X must be equal to a 4/3. Because we add for divided by three and then once again, the reason why I'm not including I just want to clarify the fourth here because if you took the route fourth root of this, you would get three x minus four. And then if you took the fourth of zero, which is zero. All right. And then finally we take the second parentheses, which is two x minus three would set that equal to zero. So for X and we get X is equal to three over to All right. And then what we do is we trial in number one and we say that well, x, this inequality is true when we have three halves rehabs. And this inequality is also true when we have when we have 4/3. Okay, so this is valid. But what we want to do is determine the entire domain of X. We want to determine what is a valid in this region, which is when, uh, which is between three halfs and infinity. It's a valid between this region, and then, finally, is a valid from four thirds to negative infinity. So we're gonna do is we're gonna start off by this region, just plug in any value into our inequality and see whether or not becomes true. So we'll take a value from three halfs and affinity for example to. So I have three times two minus a four for the fourth power two times two minus three. And we're gonna try to solve that answer. So we get a four minus three here, which is one. And then here we have a six minus four, which is to to to the fourth as a 16, and we know that that will be a positive value there. Therefore, this inequality would not be true in this region. All right, now, if we pick a value from four thirds and three halfs, for example, we could do 14 All right? We'll do. We'll do a 1.4. So 7/5 for example. 14 So three times. Yeah, he 1.4 minus minus. Force to the fourth power. And then two times 1.4 minus tary. All right, here. Just looking at it. We could see two times. 1.4 is 2.8 minus, uh, three would give us a negative value. Should be negative point to and then a three times 14 multiplied by 1.4 is 4.2 minus four is point to the fourth. Gives you a very small value, but regardless, it's being multiplied by a negative, so we know that any negative value is gonna be less than zero. Therefore, we can confirm that X the region within four thirds and three half does exist. But that was a big okay. So that region of X value does exist now. Finally, a region of values from four thirds to negative. Infinity will pick the numbers, too. We'll have three times to and we'll subtract four fourth power and a two times two and we'll subtract three. So we'll get six minus four is to once again a 16. Oh, I'm sorry. I'm picking two. I meant to pick. Let's pick one to would, two would fall in the other region to would fall in this region. We already saw for that here. So a number between four thirds a negative thing. He would be one. Okay, so we'll do three times one minus four to the fourth. And then two times one minus three. So we have three minus negative four, which is negative. One to the fourth. Power would just be a positive one. And then two times one Which one? One minus three is negative too. So we'll end up getting a and you see here you see writing this rice or two times one is two to minus two times two minus three is negative one and then a three times one minus four to the fourth would be one And yes, so that will be a negative value. Okay, sorry. Sorry about the confusion there. So we take a one and we multiply. A negative one will get a negative one. Therefore, the region here also exists. Okay, so then we could just confirm that the region within our where inequality is true. It's when all values of X are less than or equal to three. Halfs. Okay, so for X values or less than or equal to three, half this inequality here would be true. All right, well, I hope that clarifies the question there. Thank you so much watching.

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