Problem

Solve each inequality, and graph the solution set…

04:00

Problem 30 Hard Difficulty

Solve each inequality, and graph the solution set. See Example 4.
$$
(x+2)(4 x-3)(2 x+7) \geq 0
$$

Answer

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Precalculus

Algebra

Beginning and Intermediate Algebra

Chapter 11

Quadratic Equations, Inequalities, and Functions

Section 8

Polynomial and Rational Inequalities

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Video Transcript

Okay, All right. We have inequality X plus two times four x minus three times two x plus seven, which is greater than or equal to zero. And we want to find the solution set for this inequality. It's already factored for us. So the first thing we have to do right now is take each of these factors, let them equal zero so that we we confined our roots. So our 1st 1 were exposed to his equals. Zero subtract suitable from both sides. And you get X equals negative, too. The four x minus three is equal to zero. Had three double sides. You get four X equals three, so x will be three Ford's. And if two X plus seven equals zero, subject someone from each sign. Two x will equal negative seven. Divide each side by two x his negative seven halves. So those are only part of the answers, right? We want to know where we are greater than or equal to zero. Well, we have found where we are equal to zero, right. But what intervals are we greater than zero for? Well, this is where we have to think about some test values. if we have the endpoints of our intervals and negative seven halves negative, too. And 3/4 that I need to know what's going on between these intervals, right? I would have points at these values. Those are marking where the ends of our intervals are. So I need to pick values that are around these negative seven halves. That's negative 3.5. So negative four would be a good point to pick something between negative seven halves and negative too negative. Three would be a good test value between negative two and 3/4. Zero would be a convenient one to check in something to the right of 3/4. One would be an ex easy energy to use. So these air the test values I'm gonna use to determine which intervals are part of my solution set. So if I test X equals negative four first, that's going to give us negative four plus two. When I plug this in for acts times four times negative four minus three times, two times negative four plus seven. And if this interval is part of our solution set than this will check, this will be negative. Two times negative 19 times negative. A plus seven is going to be negative one. So this is going to be positive 30 times. Negative one is negative. 38 which is definitely not greater than or equal to zero. So we're going to say no. That interval will not be part of our solution set. It doesn't check out for us. So now let's see what happens. The nexus Negative three. That will give us negative. Three plus two times, four times negative. Three minus three times, two times negative. Three plus seven. I want that to be greater than or equal to zero. That's going to negative one times. That's gonna be negative. 12 Ministry is negative. 15 times one, and that is gonna work out to be positive 15 which is in fact, greater than zero. So that tells me that the values on that interval will be part of my solution set. All right, so that means the values between negative seven house and negative to work the next value on when a test is gonna be X equals zero. So when I put zero in forex, I'm gonna get zero plus two times four time zero minus three times, two times zero plus seven and again, is that greater than or equal to zero? Well, this will give us too times negative. Three write zeros were always easy to use their convenient times. Seven. That's going to be negative. Six times seven is negative 42 which is not bigger than zero. So that interval does not work from the last one we need to check then, is when X is equal to one that will give us one plus two times four minus three times two plus seven. So it's gonna be three times one Tim's nine, which is 27 which is greater than zero. So that indicates that this interval is also part of our solution set. So on the number line, it would be that interval right there. So what is our solution? Set our solution sad. Our solution will be the interval from negative seven halves too negative, too, or from 3/4 to infinity. And that's how we would write that

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