Problem

Solve each inequality, and graph the solution set…

04:09

Problem 11 Medium Difficulty

Solve each inequality, and graph the solution set.
$$
10 x^{2}+9 x \geq 9
$$

Answer

$$
\left[-\frac{3}{2}, \frac{3}{2}\right]+\underbrace{\left[_{-\frac{3}{2}}\right.}{1}+\frac{7}{-\frac{3}{2}}+
$$

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 have the inequality 10 X squared plus nine. Axe is greater than or equal tonight. The first thing we have to do to find the solution set to this is toe have it so that we're making a comparison to zero. In other words, I have to get zero on one side. So I'm going to do that. But subtracting nine from both sides here, that's the first thing we're gonna dio and we will get. 10 X squared plus nine X minus nine is greater than or equal to zero. And now what we have to do next. His factor. So when we go to fact, we know this train Oh mia will be the product of two by no meals. 10 X squared is five x times to exit. Could also have been 10 x times one X. But I think five and two is gonna work for us nine. There's gonna be three times three. Now, what we have to do is figure out our signs and I want a positive nine in the middle. So if I make this a plus three and this one of minus three, this works because you're gonna get five x times. Three. That'll be 15 X and then you have negative three times two exits and minus six acts. He's at up to get us deny next that we needed for our middle terms, we know we have the right factors here. Now we treat this like it's in equation. We let each factor equal zero and we solve. So with five x ministries equals zero and three double sides. Five x will equal three viable sides by five and we get X equals 3/5 and we have two X plus three equals zero. Subtract three, you get two X equals negative three. And then when we divide by two, you get X equals negative three halves. So those are our two routes now we haven't inequality, so those are not our solutions or solution is a solution set that represents all of the values that make that sentence true. So we know. And when we go to graft thes the solution sat on a number line, right? We're either going to be going to the left and right from these two values, or we're talking about all the values in between them, So I'm gonna pick a test point. In this case, zero is a convenient one. So I'm gonna test X equals zero and see if a checks if it checks, then I know my solution said is the value between negative three halves and three fests have a dozen check. Then I know my solution sets going the other way. So when I plug zero in, I'm gonna get five times zero minus three times two time zero plus three. And this has to be greater than or equal to zero. So this is going to be negative three times three, which is negative nine, which is not bigger than zero. So this test fails. That means the values between these routes are not part of the solution set, which means our solutions sat There's going to be Now I need brackets here because we said greater than or equal to zero. So I'm gonna have everything to the left of negative three halves, including the three negative three hands and everything to the right of 3/5 including the three fits. In other words, our solution set is going to be interval from negative infinity to negative three halves, including the negative rehabs or the interval from five halves? No. Excuse me. Its 3/5 interval from 3/5 to infinity. That's our solution set.

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