Solving rational inequalities by testing intervals in terms of a constant

0:00- [Instructor] What we wanna do in this video is find the solution set for this inequality right over here. And like always, if you are inspired and I encourage you to be inspired, try to find the solution set on your own before we work through this together. All right, now let's work through this together.

0:16So what I wanna first do is algebraically manipulate this. So I have a zero on one side, and I'll try to get the zero on the right-hand side. So let me just subtract two from both sides. So if I subtract two from both sides, I'm left with 16 over X squared minus 2X minus two is less than zero.

0:38And now what I could try to do is write it as just one big rational expression, so I could rewrite the two. So let me rewrite this as 16 over X squared minus 2X, and then I'm going to rewrite this right over there. Actually, lemme do this in a different color.

0:55I'm going to rewrite this right over here. Instead of subtracting just two, I'm gonna write the equivalent of two. That's the same thing as two times X squared minus 2X over X squared minus 2X. All I did is found a common denominator here. And so now I can rewrite this entire thing as, in the denominator I still have X squared minus 2X, and actually I'm going to have to factor that eventually.

1:24So why don't I just factor it right now? So I'll have X times X minus two in the denominator. And in the numerator, all of this business right over there, I have 16. And let me distribute, let me multiply things out. So minus 2X squared, and then plus 4X. Oh, let me, I lost the less than equal zero.

1:52Less than or equal to zero, less than or equal to zero. And then let me go up here and let me rewrite this numerator in the standard form where we write the highest degree first. So it is minus 2X squared plus 4X plus 16. All of that is over X times X minus two is less than zero.

2:16And I'm gonna try to factor this top out. Let's see, I can factor out a negative two. So I have negative two times X squared minus 2X minus eight. All of that over X times X minus two is less than zero. And let's see, what two numbers, if I were to multiply get to negative eight, and if I were to add it get to negative two?

2:44So let's see, negative four and positive two work. So I could rewrite the numerator as negative two, I'm just factoring this expression here. Negative two times X minus four times X plus two. Did I do that right? Yes. Negative four times two is negative eight.

3:01Yep, that looks good. All of that over X times X minus two is less than zero. So why did I do all of this? Well, I'm gonna figure out where does this rational function either have values, X values that make this numerator over here equal to zero, or that make this denominator undefined and will have some type of a discontinuity.

3:26Because those can be the areas of sign changes. So we can see that our function, the numerator here at least will become zero at X equals four and X equals negative two. This is where the numerator is equal to zero. And then we also have interesting things happening at X equals zero that makes the whole expression undefined, the zero, or we have zero in the denominator.

3:56And we also have X equals two. So now let's test this function right over here, over intervals to the left, in between, and to the right of these values. And I will- And see whether the function is less than zero, because the intervals where it is less than zero are the intervals that satisfy this inequality, which is equivalent to this original inequality.

4:21So let's start there. And let's see, the lowest value here is X is equal to negative two. So let's test the interval for X is less than negative two. And so let's try, say, with X equals negative three, X equals negative three. And I could have tried any value less than negative two here.

4:39But let's try it on this expression here. We're going to have negative two, I'm write a little bit smaller 'cause it's gonna take a little bit of space, times negative three minus four times negative three plus two. All of that over negative three times negative three minus two.

4:59And what we really care is just the sign. So we could see this, the denominator over here, negative three minus two, that's negative five. Negative three times negative five, that is positive 15. The numerator over here, see negative three minus four, that's negative seven.

5:15Negative three plus two is negative one. So we're gonna have a negative times, well, we could just see this is positive. 14 times a negative, that is negative 14. Negative 14. The important thing is that this is negative, so I know it's kind of messy here, but all of that right over there is negative, which means that over this interval for X is less than negative two, we're satisfying the inequality.

5:43So this is going to be part of our solution set. Let's keep going, let's go to the next interval, when we're between negative two and now we move up to zero. And we could try out the value X is equal to negative one. Same idea here. Oh, maybe I'll do it right over here.

6:01So it's we have negative two times negative one minus four, times negative one plus two, all of that over negative one, times negative one minus two. All right, in the numerator here, negative one minus four is negative five. Negative one plus two is positive one.

6:22So negative two times negative five times one is positive 10 in the numerator. And then in the denominator over here, negative one minus two is negative three. And negative one times negative three is positive three. So we have a positive divided by a positive.

6:40This is going to be positive over this value. So it's not less than zero, so this is not going to be part of our solution set. Now let's keep going in a new color. And maybe this pink color here would be nice. Let's now go between zero and our next value, which is two.

6:59Between zero and two. And we could try, let's just try X equals one. So same idea, negative two times one minus four times one plus two, all of that over one times one minus two. In the numerator over here, let's see, this is negative three, this is positive three.

7:25So our numerator, we have a negative times a negative times a positive. So the numerator is going to be positive. And then our denominator, we have a positive times, this is negative one, times a negative. So our denominator is gonna be negative. So this whole thing is going to be negative.

7:41So this whole thing, our function is negative here. Negative, which means we're less than zero, which means that this also is part of our solution set. And then last but not least, let's try that last interval for X is... Oh, actually we have two more intervals to test.

8:01We have the interval between two. X is between two and four, let's try that one out. And we could try, let's say, positive three here. So if we say positive three, we'll have negative two times three minus four times three plus two. All of that over three times three minus two.

8:25So let's see, this is going to be negative one, this is positive five. So our numerator, negative times negative times positive is positive. Our denominator is also positive, that's just a one there. Also positive, so not part of the solution set. Now last but not least, I'm running out of colors, we can try the interval for X is greater than four.

8:48And let's say we try X equals five there. And I'm really running out of space here. Lemme see, maybe I'll do it right over here. We have negative two times five minus four and once again I'm just trying some value in that interval, times five plus two, all of that over five times five minus two is going to be equal to, or I could just evaluate the sign.

9:13So this is positive one, this is positive seven. So we have a negative numerator. And then our denominator right over here is positive. Five minus two is three, positive. So this is gonna be negative, negative divided by positive is negative. So this is also part of our solution set.

9:33So our solution set is that X can be less than negative two or X could be between zero and two or X could be greater than four. Sometimes if you see it in set notation, it might look like this. Let me see where I could write it. We could write that our solution set goes from negative infinity to negative two, and we're not including negative two, we're not including the boundary here because it's strictly less than.

9:58Or, that means union, or we're between zero and two, or, or union, we are between four and positive infinity. That's our solution set right over there. Sorry for all of the jumble. I had to write a lot in limited screen space.