Problem

Solve each inequality, and graph the solution set…

03:44

Problem 15 Easy Difficulty

Solve each inequality, and graph the solution set.
$$
6 x^{2}+x \geq 1
$$

Answer

$$
\left(-\infty,-\frac{1}{2}\right] \cup\left[\frac{1}{3}, \infty\right)
$$

Related Courses

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

All right. So we are solving the inequality where we have six six x squared plus X greater than or equal to one Now, first thing we have to do is having inequality where we are comparing everything to zero. So the first thing I'm gonna do is subtract one from both sides when we get six X squared plus X minus one greater than or equal to zero. Now we have this quadratic when we have to do next this factor it. So we're going to treat this like it's an equation. Six X squared Aiken factor three x times two x one can only be one times one, and I need a positive one in the middle. So if I put a plus one here and a minus one in the 1st 1 that will give me a positive when I need for my middle term. Now what comes next is I have to take huge factor. Let it equals zero. So we will get three x minus one equals zero. So to solve, I would add wonderful signs you get three X equals one, divide by three and we get X equals 1/3. That's our first route. Our other route will be two X plus one equals zero. Subtract one from each side. You get two X equals negative one. We would then divide both sides by two. And X will be equal to negative 1/2. Now we know where solution set can either be the values between negative 1/2 a positive 1/3 or to the left and right of those values. So what we have to do now is do a test point and a convenient one to check is X equals zero because zero is between negative 1/2 and positive 1/3. If this tracks that I know, that's where my solution set lies. If a dozen check, then I know it's going the other way. What we mean by that is this. When I plug zero in, I'm gonna have three times zero minus one times, two times zero plus one. And if this checks it'll be greater than or equal to zero. Well, this is gonna be negative. One times one Well, negative. One times one is negative one. And that is definitely not greater than zero. So it doesn't check that fails. What? That tells me. Then is when we go to look at our solution set, we have a value at negative 1/2. We have another value that it three halves. Excuse me. At 1/3 we tested X equals zero and found out that that interval did not work. So our solution said then has to be going to the left, starting a negative 1/2 and going to the right starting at 1/3. We also don't forget our inequality was greater than or equal to zero. So we want to include the endpoints as part of our solution sets. So in other words, the solution said here will be the interval from negative infinity to negative 1/2 or the interval from 1/3 the positive infinity and that's our solution of this problem.

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