Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

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

$$x \leq-3 \text { or } 0 \leq x \leq 2$$

Algebra

Chapter 0

Reviewing the Basics

Section 5

Solving Non-Linear Inequalities

Equations and Inequalities

McMaster University

Baylor University

University of Michigan - Ann Arbor

Idaho State University

Lectures

05:39

In each of the following e…

01:50

03:40

04:15

02:42

05:55

04:07

02:36

02:31

02:55

03:14

03:16

05:26

02:30

03:06

03:28

06:15

06:36

all right, We've got an inequality here, which is X times X minus two, multiplied by X plus three. And all of that must be less than or equal to zero, right? We're going to solve for our X values by setting it equal to 01 X is equal to zero X minus two, which is zero. Pull that back a little bit, give ourselves more room. And therefore we could then say, Well, X is equal to positive, too. On the X Plus three is equal to zero for X is equal to negative. All right, now, if that's the case, we would draw out a number line. We would say, uh, this inequality exists when excess native three act is equal to zero and what actually equal to two. All right, but we're looking to find the domain of X value. So we want to figure out whether it exists in this entire region here to infinity, whether exists from 0 to 2 03 and finally from negative three to include, All right, we do that, we'll start off by this region. Here, we'll plug in a random value between two and infinity. For example, three and we'll do. Three multiplied by three minus two, three plus three, and we'll look to see if it's less than or equal to as you will. So we know three minus two is one, and the three minus 13 times one is three, multiplied by six would give us 18. We could see that That's clearly not great, less than or equal to zero. Therefore, this region does not exist. All right, let's plug in a value from 0 to 2. So we have one. So you have one minus two and a one plus three. So would be negative. One multiplied by a positive four, which would be negative for. And we know that's less than or equal to zero. Therefore, this region does exist. Okay, we'll pick a value from zero to negative three, for example, Negative one. You have negative one minus two negative. One plus three. And so we have. Excuse me. Negative one multiple minus A two is negative. Three negative three multiplied by negative one is a positive three and in positive. Three multiplied by here, which is to give us six. So that's a positive value. And we know that would make sure that this inequality is not true. Therefore, it doesn't exist in that region. Now, finally, we're plugging a value from negative infinity, so we'll pick negative four. And so I have negative for minus two negative four plus three. And we'll get a negative six here and a negative one here in the negative four here. And we know that would be a negative 24. So therefore we could say, Well, it does exist in this rich. All right? And then, if we were to write this out, we would say, Well, if your ex is greater than or equal to zero and less than or equal to the region does inequality is satisfied. And if your ex is less than or equal to negative three, you're inequality is also satisfied. All right, well, I hope that clarifies the question there. Thank you so much for watching

View More Answers From This Book

Find Another Textbook

02:09

Solve each of the quadratics by first completing the square. When the roots …

02:00

Find the unknown.$$(2 r-3)^{2}-24=0$$

01:24

Find the unknown.$$(x+3)^{2}=25$$

02:43

Solve for the unknown, and then check your solution.$$x+11=5$$

00:36

Find the unknown.$$y^{2}-49=0$$

01:52

Rewrite the equation in the form $a x^{2}+b x+c=0,$ with $a > 0,$ and the…

06:11

Solve for the unknown, and then check your solution.$$0.35+0.24 x=0.2(5-…

02:25

In each of the following exercises, solve the given inequality.$$x^{2}+4…

01:57

Complete the square in each of the following by putting in the form $(x+B)^{…