Solve each rational inequality. Give the solution set in interval...

Question

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. {x^2 - 3x- 4} /{ x^2 + 6x + 9} ≤ 0

Accepted Answer

Hey, everyone in this problem, we're asked to provide the solution set for the following irrational inequality in interval notation. The inequality we're given is X squared minus four, X minus 12 divided by X squared plus 10, X plus 25 is less than or equal to zero. We're given four answer choices, options A through D, all of them have the solution set ranging from negative 2 to 6. And they just differ in which end points they include in the solution and exclude in the solution. Now, if this was a question on a test, since you already have the interval, OK. Each of them have the same interval, all you would really need to do is test the end points to see whether those are solutions to get the correct answer. OK? But here we're gonna walk through this entire problem so that you know how to approach this type of problem. And then if the answer choices were different, you'd be able to tackle it. Now we're given X squared minus four, X minus 12 divided by X squared plus 10 X plus 25. And that is a less than or equal to zero. Now, in this problem, we already have zero on one side, we have one rational function on the left hand side. OK? So that's a really great, what we wanna do is to simplify, to factor this rational expression. OK. We're gonna factor the numerator, we're gonna factor the denominator. Now, the numerator X squared minus four X minus 12, we can write this as X squared plus two, X minus six, X minus 12. OK? We're gonna write it like that so that we can factor and you can factor any way that you want. OK? Whichever way you like to factor best, you can go ahead and do that. This is gonna be divided by, well, we have X squared plus 10, X plus 25. We're gonna write that as X squared plus five, X plus five, X plus 25. And so we're just breaking up those middle terms and this is all less than or equal to zero. Now we're gonna go through and we're gonna factor in parts, OK? Now that we have four terms, we're gonna start by factoring the first two terms. OK? And I'm talking here both the numerator and denominator. We're gonna apply this. OK? We're gonna factor the first two terms, then we're gonna factor the second two terms and we're gonna see what we get. So we have X squared plus two X and we can factor an X out of there. So we get X multiplied by what's left over which is X plus two. Now, from our second two terms, we have negative six, X minus 12. So we can factor negative six and we're left with X plus two. OK. So we have X multiplied by X plus two minus six, multiplied by X plus two. That's our numerator. And in the denominator, we're gonna do the same thing. So we factor in X or left with X plus five. those fi in that first two terms, then we get plus five multiplied X plus five. And this is all less than or equal to zero. Moving to the next step, you can see that we have this common factor now of X plus two in the numerator. And that's why we splitt our term, our middle term negative for X like we did. OK? So that we could get this common factor. So we have that common factor of X plus two. If we were to bring that out, what's left is X minus six. So we get X plus two multiplied by X minus six. And now that numerator is factored, we do the same to the denominator. We have that factor of X plus five. When we take that out, we actually have another factor of X plus five. So we have X plus five multiplied by X plus five. We're gonna write that as X plus five squared and this is all less than, or equal to zero. All right. So Now we have our rational expression, we have a single rational function. Everything is factored. The other side, we have zero. So we want to think about our points of interest. OK? And our points of interest are gonna be where either the numerator or the denominator equals zero. OK? We're gonna have those points are where the function could switch from being positive to negative. Now, let's make our work a little bit easier here. OK? Let's save ourselves some intervals to check X plus five squared is always gonna be positive. OK? X plus five squared is always greater than or equal to zero. OK? So the denominator is always greater than or equal to zero. OK? All we need to do is write that X cannot be equal to five. OK? If X is equal to or sorry, not X equal to five, X equal to negative five. If X is equal to negative five, then the denominator will be zero will be dividing by zero. And we can't do that. The function will be undefined. So that won't be a solution. OK? So we have this restriction that X cannot be equal to negative five. But otherwise that denominator is gonna be positive. So we want the left-hand side to be negative K less than or equal to zero. The denominator is always gonna be positive. So all we really need to look at is whether X plus two multiplied by X minus six is less than or equal to zero. Hm Now our points of interest again are gonna be where the left hand side is zero. Oops. OK. So our points of interest if we have the factor X plus two equal to zero, OK. The entire left hand side is zero, X plus two equals zero. Tells us that X is equal to negative two and same thing for X minus six. If that is equal to zero, the entire left hand side is equal to zero. And that gives us X is equal to six. So we have these two points of interest and we want to think about what happens around these points. So we're gonna draw a number line so that we can evaluate our intervals. So we have negative two on our interval and we have six on our interval, we're gonna start from the left. Now on the left, we have X values less than negative two. If X is less than negative two, then X plus two is gonna be negative. It's gonna be less than zero, X minus six is also gonna be negative, it's gonna be less than zero. And so the left hand side of our inequality, which we're gonna call L H S OK. This simplified inequality which is X plus two multiplied by X minus six. It's going to be greater than zero. It's gonna be positive maybe because we have a negative multiplied by a negative. Now we want that left hand side to be less than or equal to zero to satisfy our inequality. And so that means that this interval is not going to be a part of our solution. All right now moving to the right, we have the point exactly at negative two at X equals negative two. The left hand side is gonna be equal to zero, we have zero less than or equal to zero. That's true. OK? And so the point negative two is going to be included in our solution. So we're gonna draw a round circle there to indicate that that point is gonna be included. OK? Because we have that less than or equal to. Now we're moving to our next interval and that's gonna be from negative 2 to 6. OK? So for X values between negative and two and six, what happens? Well, I can think about substituting in a test point like X equals zero, X plus two is gonna be greater than zero, X minus six is gonna be less than zero. OK? It's still gonna be negative. So the left hand side when we multiply these two factors together is going to be negative, it's gonna be less than zero, which is exactly what we want. OK? That this interval satisfies our inequality. And so this is gonna be part of our solution. So we're gonna give this interval, a green check mark. So we're gonna move further to the right and we get the point X equals six. And at X equals six, we have the same situation as when it's negative to OK. The left-hand side is gonna be equal to zero, which satisfies our inequality. So we're gonna draw a closed circle to indicate that we include that 00.6. And this is just a visual reference for us when we go to write down that interval of exactly what we found. Now, we have one more interval to test. That's when X is greater than six. That means that X plus two is gonna be greater than zero and X minus six is gonna be greater than zero. And so we multiply those two factors together, we are gonna get that the left hand side is greater than zero does not satisfy our inequality. So red X on that interval, it is not part of our solution set. And so as expected, we found that the interval for the solution is from negative 2 to 6. OK? And we found that both end points should be included. And so we use square brackets in our interval notation to indicate that we're including those end points. And if we go back up to our answer choices, and we can see that the correct answer is gonna be option C because that's the answer choice that has those square brackets to include the end points. Thanks everyone for watching. I hope this video helped see you in the next one.

Solve each rational inequality. Give the solution set in interval notation. See Examples 4 and 5. {x^2 - 3x- 4} /{ x^2 + 6x + 9} ≤ 0

Key Concepts

Rational Inequalities

Interval Notation

Critical Points and Test Intervals

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