 In this final video for lecture 35, I want to talk about how we can solve radical inequalities. What if we have an inequality that involves a radical of some kind? Take the cube root of 2x squared minus x is greater than x. So what I want us to do is I want us to solve the appropriate equation, right? If we replace the inequality with an equality, right? Whoops. If we take the cube root of 2x squared minus x equals x. Let's solve the equation. This will help us find the markers here, in which case if we then cube both sides to get rid of the cube root on the left, so we cube both sides, we end up with the polynomial equation 2x squared minus x equals x cubed. This is the polynomial equation. I want to set one side equal to zero, so we're going to take x cubed minus 2x squared plus x equals zero. So what I did is I moved all the terms to the right-hand side and then I flipped the equation around. If you don't like that, we can put the zero on the left-hand side. Not such a big deal there. So then we need to factor the polynomial there. I'm looking at x cubed minus 2x squared plus x. You can factor out the common divisor of x, leaving behind x squared minus 2x plus 1, for which then x squared minus 2x plus 1, I see is a perfect square trinomial. You're going to get x times x minus 1 squared, like so. This is going to give us our markers, in which case we then have these two numbers to pay attention to. We have x equals zero and we have x equals one. These are the numbers to be concerned about. Now, many ways as I've described this in the past is we could try to graph this thing. But the issue is what would you be graphing right here? We'd have to graph the function. We have to graph the function f of x equals the cube root of 2x squared minus x minus x, right? In which case then we're trying to figure out what is this thing greater than zero, right? So we want to above the x-axis. That can be a little bit of a challenge, right? This is a way more complicated thing. Now, this function can be very difficult to graph, right? Exactly without a graphing calculator. There's so much going on there. What can we do instead? Well, actually, I think it's going to be a lot easier if we pay attention to our equation over here. What if we actually did make it an inequality, right? We had this greater than x going on right here. Well, when you cube both sides, this is an increasing function, so it doesn't change the inequality. So it's still going to be pointing to the left. 2x squared minus x is greater than x cubed. And so if you follow through on all of these things, right, we then have to graph the function f of x equals x times x minus 1 squared when that's less than zero. So we're looking for things that are below the x-axis. Now, remember, the end behavior of this thing is going to be approximately x cubed. So it's going to point up on the right and down on the left. And then it'll cross the x-axis at x and it'll touch at x equals 1. So if we get a picture that looks something like the following, and if we want to be less than zero, we're talking about those portions where we're below the x-axis there. And so the solution to this thing would then look like negative infinity to zero. So graphing saves the day again, but we didn't graph the rational the radical function. We actually graphed the polynomial that turned out to be equivalent to it. Let's look at another example of this. So take the inequality 1 minus the square root of 5x minus 9 is greater than or equal to x minus 8. So when it comes to solving the inequalities, again, my recommendation is always you worry about the equation first. 1 minus the square root of 5x minus 9 is equal to x minus 8. On the previous example, we were able to keep track of the inequality sign, but that's because we had an odd radical. If you have an even one, things are a little bit more difficult, because at some point, we're going to have something like 5x minus 9 inside the square root. And this is going to be maybe greater than or equal to something like what it is. I'm not exactly sure. But then we square both sides. But what does the square do to the inequality? Well, the problem is when you take y equals x squared, this function is increasing. So over here, you would keep the inequality, but over here, it's decreasing actually would flip the inequality around. So because it both flips and preserves it, you have to break up the domain. It gets really, really complicated. So it's best just to kind of procrastinate that decision until the end. Just worry about the equation where those subtleties don't even come into play here. So if we solve this, we're going to subtract one from both sides. We end up with the square root of 5x minus 9 is equal to sorry, the negative square root of 5 minus 9 is equal to x minus 9 right here. We're going to times both sides by negative one. So we get the square root of 5x minus 9 is equal to 9 minus x. And now we have to square both sides. On the left hand side, we do have the on the right hand side, we'll have to foil it out. We get 5x minus 9 is equal to 81 minus 18x plus x squared. So we're going to add nine to both sides. We're actually going to subtract 5x as well. So add nine minus 5x because now it's a quadratic equation. So we get x squared minus 23x plus 90x equals zero. So since it's quadratic, we want to set the right hand side equal to zero. So we can factor it or use the quadratic formula. We'll get notice that factors of 90, we can take 5 and 18, that adds up to be 23. So we get 5 minus x and 5 minus 18 is equal to zero. And so that then tells us the solutions to the equation to be 5 and 18, right? And so now we're tempted. Okay, the next step is then we use these markers to start graphing this thing. But remember, we solved a radical equation. And it's a radical equation involving the square roots that there might be party crashers. Does 5 belong here? Does 18 belong here? It's hard to say. We have to check these answers here. And so if we check them, if you take the left hand side and you plug in 5, you're going to get 1 minus the square root of 5 times 5 minus 9. That should look like 1 minus, you're going to get 25 minus 9, which is going to give you 1 minus the square root of 16, which that gives you 1 minus 4, which is negative 3. Okay, the right hand side, that will look like we're supposed to take 5 minus 8, which is equal to negative 3, that passes. So 5 is in fact a solution to the equation. What about 18? Could there be 2? In that situation, you take the left hand side, which is equal to 1 minus the square root of 5 times 18 minus 9, in which case, like we saw before, 5 times 18 is 90. So you get 90 minus 9. We then get 1 minus the square root of 81, which gives us 1 minus 9, which gives us a negative 8. So no domain issues going on there. The right hand side should look like x minus 8, so we're supposed to take 18 minus 8, which gives us 10, which is not equal to negative 8. So this one didn't work at all. So we have to remove 18 from consideration. Great. So that's an important step. You don't want to forget that one. So now in terms of the markers, we just have one marker, which is going to happen at, it's going to happen at 5. Now, that's not exactly true. In terms of the graph here, we also do need to consider the domain, because if we come back to the original inequality, 1 minus the square root of 5x minus 9, this is supposed to be greater than or equal to x minus 8. You'll notice, for example, if I plug in x equals 0, then you're going to get the square root of negative 9, which is not a real number. And so the left hand side is not a real number. How can it be greater than a real number? That's a problem there, right? So there are some domain issues that we have to consider. So if we take 5x minus 9 and set that grade equal to 0, which would be the domain of the left hand side, we get 5x is greater than or equal to 9 or x is greater than or equal to 9 fifths. So we do also have to include this left boundary point, 9 fifths, like so. And so in terms of graphing it, we do want to, that 9 fifths does have to be included, right? The left hand side would be defined at 9 fifths. It would just equal 1, which is a possibility. And you'll notice that if I take 9 fifths and I subtract 8 from it and you actually get a negative number, that actually would pass off. So like this over here is actually part of the solution set. So I'm actually very tempted to include this picture right here. If I chose something bigger than 5, you'll see that actually it doesn't work out, the test value doesn't work. You could also try to approach this by graphing. If you thought of this function right here, f of x and this function right here is g of x. We're trying to see when f of x is greater than g of x. Now what's going on right here? We have a radical function, the square root of function, which we've done some type of horizontal stretch, some type of horizontal shift. It's been reflected across the x-axis and we've also shifted up by 1. So your picture is going to look something like the form. We have this decreasing square root function kind of like this. On the other hand, if you take y equals x minus 8, well that's just going to be a line that's increasing. And so it's going to look something like this. They'll intersect at x equals 5. And so you can see where is the radical greater than the line that's going to happen right here. So you can still solve this one by graphing or you can use some test points. That's always an option. So in the end, we see that our solution would look like 9 fifths comma 5. So 9 fifths will be included. That is part of the domain and that does make the left-hand side greater than the right-hand side. And if you plug in 5, that'll make the left and the right both equal to each other and equality is acceptable in this situation. So we do pass off that both 5 and 9 fifths should be part of the solution. And so yeah, solving inequality involving radicals can be a lot more challenging, especially when it's an even radical because there's issues with the domain, with the range, they're not one-to-one necessarily, at least the squares, the squares and the fourth powers are not one-to-one. So it's a little bit more advanced here. You can really understand how to solve this inequality. You were really set when it comes to inequality. You're going to be hard-fetched to find a more challenging inequality in a college address setting than the one we just saw here.