 In this video I want to talk about how to solve equations involving radicals of some kind and there's two important cases you have to pay attention to. There is when the index of the radical is odd versus even, the two will behave very different in important ways. So some things to know about like if you take y equals the cube root of x or any odd power, some of the important things to remember here is that the domain of this number is going to be all real numbers, the domain of this function. There is no number we're forbidden to take the cube root of. We can take the cube root of negatives. There's not a problem with that whatsoever. So for example, the cube root of negative eight is equal to negative two because negative two cubed is equal to negative eight, right? So the domain is going to be all real numbers. The range is also going to be all real numbers. Every number can come out of a cube root. So if you end up with something like the cube root of x equals, let's say negative two, you don't have to disqualify like up no solutions, like no, actually that's completely kosher. Take the cube of both sides, you end up with x equals negative eight. Okay, so the domain of range has no restrictions. That makes solving this thing a lot easier. Another thing important about the cube root of x is that it's a one to one function. So it's a truly invertible function. So in order to solve this equation, you're just going to start undoing operations in reverse order. So we do first what's outside of the cube root here, we're going to add two to both sides and we end up with the cube root of two x minus four is equal to two. So it's equal to a positive number, but even if it equal to negative number, that wouldn't be a problem whatsoever. How are you going to get rid of the cube root? Well, you have to do its inverse operation. The opposite of the cube root would be the third power. So we're going to cube both sides and we end up with two x minus four equals two cube, which is eight. And now it's a linear equation, which we're probably very comfortable at this point. We're going to add four to both sides. We get two x equals eight plus four, which is 12. Then we're going to divide both sides by two. And we end up with x equals six as our solution for which we can double check to make sure if we take the cube root of two times six minus four. And we subtract two from that here, you'll notice two times six is equal to 12 like we saw before. 12 minus four is eight. The cube root of eight is two and two minus two is zero. And so we do see that the left-hand side agrees with the right-hand side. We have a solution. And so when you're working with odd radicals, there's really not a lot of difficulties because the function, its domain is all real numbers, its range is all real numbers and it's one to one. There's really no, there's no obstacles. There's no, no traps you have to look out for. So let's focus the remainder of this video then on even radicals. So we're talking about like square roots or fourth roots or sixth roots. What happens when you try to solve things like that? Well, some things to be aware of, like if you take y equals the square root of x, first of all, its domain is only going to be zero to infinity if we want to think of it as like a real valued function. And we've dealt with this issue before as we've worked with quadratic functions back in chapter three in this series. That if you take the square root of a negative, you're going to end up with some imaginary number, right? And so if you're willing to consider all complex solutions, that kind of relaxes the domain problem. If we're trying to graph this thing, we definitely need to restrict the domain just so we only get real numbers. So whether we include imaginary complex solutions or not will depend on whether we want to relax this condition on the domain. It is an issue, but it's not too much of an issue. We've dealt with it before. A bigger concern for us will be the issue with the range. The range of a square root function, and this would also be true for like the fourth root, right? If you remember the graph of these functions, the square root function would look something like the following. This would also be the basic graph for the fourth root. Looks something like this. You only see things above the x-axis, nothing below. So its range is going to be only the non-negative real numbers. This is important because if you run across something like the sixth root of x equals negative one, right? We cannot go past this. This would actually mean no solution because a sixth root could never equal a negative. A fourth root of square root, they can never equal negatives. So if you ever get an even root to equal a negative, there's no solution. Much like if you get the absolute value of x to equal negative one, right? That's not possible. So you get no solution. That's a concern we have to deal with as well. So we have to always be cautious. We always have to be cautious that when we're solving an equation involving square roots or any even radical, we have to make sure we have legit numbers from the range there. So when you look at like the square root of x-1, well, this is only solvable of x-7 is positive, or zero, right? If we're negative, you're not going to get any solutions. But that depends on x, right? There's some values of x that will be positive, some won't be, right? It would only be acceptable like when x is greater than or equal to seven, right? But on the other hand, when you look inside the radical, right, it has a restriction on its domain as well. So that has to be x is greater than or equal to one, which isn't, you know, that's compatible, right? We take the more restrictive thing right here. But this can get very confusing, paying attention to domains and ranges. So what I want to suggest is a much simpler approach. When you take your equation, the square root of x-1 equals x-7, my suggestion is this stuff about domains and ranges, I'm giving you permission to procrastinate. It's like, eh, I'll clean up my room later, mom. We'll do that later. So when you see the square root of x-1 right here, to get rid of the square root, we have to do the inverse operation, which is going to be squaring it. We have to square both sides. On the left-hand side, you're going to get x-1 because the square of the square root is an x-1. On the right-hand side, we do have to foil. x-7 squared means x-7 times x-7. And so when you have things like this x-7 to any power, the binomial theorem can come into hand, it can be very handy in doing these expansions here. The left-hand side will look like, excuse me, the right-hand side will look like x squared minus 14x plus 49. Now we've turned our equation into a quadratic equation. I'm going to move everything to the right-hand side. We're going to minus x and add 1 to both sides, minus x and add 1. This then gives us x squared minus 15x plus 50 is equal to 0. We could solve this by the quadratic formula, or we could just factor it, it turns out. Factors of 50 that add to be 15, we could do 10 and 5, make both of them negative. So we're going to get x minus 10, x minus 5 is equal to 0, and that would then tell us our solutions look like x equals 10 or 5. So we end up with two solutions here, but one has to be very cautious, right? Remember what I said earlier that the square root of x-1 is only going to work if the right-hand side was positive, or 0, 0 is okay there too. It can't be negative, but the right-hand side will only be positive when you're greater than 7. So that observation right there was like, wait a second, 5, what do you think you're doing here? You're not invited, you're not greater than 7, right? So actually x equals 10 is going to be the solution to this radical equation right here. x equals 5 is what one calls an extraneous solution, which I'm probably mispronouncing here. It's a fancy word that's kind of hard to say if you don't know how to pronounce words with more than five letters in it or anything like that. But anyways, this idea that 5, it looked like it was a solution, but it's really not. This is actually what I like to call a party crusher. The idea is they were not invited to the party, they weren't invited to the wedding, but they snuck in there anyways. So how did 5 sneak into this? How did it sneak into this process? Well, the issue really comes down to when we were dealing with the square root, we had to deal with a square at one moment. Consider the very simple equation x equals 2. I think we all can solve this equation right here. The solution to x equals 2 is x equals 2, great. But what happens if you square both sides of the equation? You're going to square the left, you'll get x squared, you'll square the right, you get 4. And then as you solve this equation, you end up with 2 solutions, x equals plus or minus 2, or the plus or minus 2. So you get this party crusher, negative 2, that sneaks into the game, it sneaks into the party, even though it wasn't invited at all. And this has to do with the fact that these even radicals are not 1 to 1 functions. This also had the issue with its range, right? These things are not 1 to 1. But really, it's this issue with the range here, we can't accept a negative number. But when you square things, you're kind of forgetting that. The inverse of the fourth root is actually a not 1 to 1 function, y equals y to the fourth. And so these issues here means that party crashes can sneak into the party here. But that's something that's easy to deal with. If you take the original equation, the equation that was not defiled by the filthy hands of man, if you take the square root of x minus 1 equals x minus 7, well, let's actually try my two possible solutions here. Let's take the square root of 10 minus 1 and compare it to 10 minus 7. What happens? Well, 10 minus 1 is 9. 10 minus 7 is 3. And on the right-hand side, the square root of 9 does turn out to be 3. Oh, so that one actually does work out. The left-hand side agrees with the right-hand side. So we see that 10 worked. On the other hand, though, if we take the square root of 5 minus 1, does this equal 5 minus 7? Well, you can see the problem here. The square root of 5, take 5 minus 1, you get a 4. The square root of 4 is going to equal 2. On the right-hand side, though, you're going to get 5 minus 7, which is negative 2. And negative 2 does not equal 2. And so that's that issue about a square root can't equal a negative. So this party crashers snuck into the game because of that. So when you're working with radical equations, it's imperative that you check for party crashers. Even if you did all of your algebra correct, we did all of our algebra correct, we can have party crashers. And so we have to check for them. And that's because of this range issue going on right here. Now, you can try to identify the party crasher by making one statement about, oh, the right-hand side has to be positive. But as the more complicated these functions get, the more difficult it can be to determine what is the appropriate domain to make all of this well-defined. So I just say, don't even worry about this stuff about domain and range. Just solve the equation and then check your solutions. Honestly, you should be doing that anyways. Check your solution. But it's necessary that with radical functions, radical equations, you need to check your solutions. Otherwise, you might not, you might have these extra numbers in there. And that's the thing that, like with this one, we have a square of things. We got two solutions from the quadratic. Turned out one of them worked, the other one didn't. Sometimes it could be that both of them work. And sometimes it could be that none of them work. It's imperative that you check your solutions. So this is the best way to know who's a party crasher and who's a genuine, authentic solution. Now imagine we have two radicals in our equation. The square root of 2x plus 3 minus the square root of x plus 2 equals 2. We can talk about the domain of this problem, right? Because we have to check our solution just like with rational functions. We have to check our solutions to make sure they live inside the domain here, right? This radical right here is only defined when x is greater than or equal to negative 2. And a similar statement can be said for this radical as well. So if we can solve the equation, find a number that's outside of their common domain, we have to throw it out. But there's also issues about the range, right? We have to make sure that these things combine in such a way that the range is correct as well. And key contract of all of this can be difficult. So we're just going to be like, pfft, I'm not going to worry about it right now. I'll deal with that at the end of the problem. I'm just going to check my solutions. Okay. And just if we get a negative value for x, that does not necessarily mean that's a problem, right? Everything is defined when x equals negative 1. Same thing right here. This is defined when x equals negative 1. So we don't just throw it x because it's negative. It has to make sure that the left-hand side and the right-hand side are both real numbers that are equal to each other. We have to agree with the domain and range. So how do you deal with two radicals? What if you have one? Well, the way you deal with radicals in algebra is the same way you deal with them if you are, you know, in a government, right? You want to separate the radicals and take them out one by one. I know it's kind of like a very violent solution, but, you know, it paints a picture in your mind that you'll hopefully remember when you have to solve such an equation on an algebra test, right? So we want to separate the radicals, right? So move one of the radicals to the right-hand side. We end up with the square root of 2x plus 3 is equal to 2 plus the square root of x plus 2. And now, because we have a square root on the left-hand side isolated, I want to get rid of it by squaring. But what's good for the goose is good for the gander. We have to do it to the right-hand side as well. When you square a square root, you will just get back the radicand 2x plus 3. On the right-hand side, we do have to foil this thing out when you square something. 2 plus the square root of x plus 2 right there. You have to do it twice. You can't just distribute the exponent over addition. You have to foil it or use the binomial theorem. And so when you look at the foil there, you're going to get 2 times 2, which is 4. You're going to get 2 times the square root. So I'm just going to write that out. You'll get 2 times the square root again. So I'm actually just going to write that as a coefficient of 4 right there. And then you're going to get the square root times the square root, which when you multiply those together, you're actually getting the square root squared. So the square root of x plus 2 squared, which, like we saw before, they cancel out. So you actually just get x plus 2, like so. And so now you'll see that we started off with 2 square roots. We isolated 1 and squared it away. Now we have just 1 square root. And so what we're going to do is repeat this process. We want to isolate the square root. So we're going to move things to the other side. So we're going to subtract 4, goose and gander. We're going to subtract x, goose and gander. And then we're going to subtract the 2. What's good for the goose is good for the gander. Separate all those things there. So you're going to get 2x minus x, which is an x. You're going to get 3 minus 4 minus 2, which is a negative 3. And then on the right-hand side, you have 4 times the square root of x plus 2. Now, if you want to get rid of the coefficient of 4 in front of the square root, you can do that by division, but I don't like needlessly introducing fractions into my equation if I can avoid it. So I'm actually going to leave it as a coefficient. We're going to square the left-hand side and the right-hand side to get rid of the square root on the right. So on the right-hand side, I'm going to start with that. You're squaring. Now you have to square the 4 and the square root. You're going to get a 16. And if you square the square root, you'll just get the radicand, x plus 2. On the left-hand side, you do have the foil again. So you get x squared minus 6x plus 9. We now have a quadratic equation. And so we want to combine some like terms. I would first distribute the 16, so we get 16x plus 32. And then combining like terms, we'll just put everything equal to 0 on the right. So minus 16x minus 32 minus 16x minus 32. We end up with x squared minus 22x minus 23 equals 0. We could solve this by completing the square of the quadratic formula. But I mean factors of 23 that add up to be 22 here, I take negative 23 and then plus 1. So the factorization is simple enough. So we end up with x equals 23 and negative 1. So those are what we think are the solutions. But remember, party crashers could be present. We need to come back to the original equation. Not any of the intermediate equations. We must go back to the original one. Check. Do those numbers work there? So if we try x equals 23, what happens there? Well, the left-hand side is going to look like the square root of 2 times 23 plus 3. Oops, that looks like a 13 there. Plus 3 minus the square root of 23 plus 2. For which as this simplifies, you're going to get 2 times 23, which is 46 plus 3, which is 49. And then you subtract from that 23 plus 2, which is 25. You end up with the square root of 49, which is 7 minus the square root of 25, which is 5. And that's equal to 2. So that's the right-hand side. So that works out. So 23 is a solution. If we try x equals negative 1, the left-hand side will look like the square root of 2 times negative 1 plus 3. And then we subtract from that the square root of negative 1 plus 2. Now be aware, there's nothing wrong with the domains of these functions. 2 times negative 1 is negative 2 plus 3 is 1. So you get the square root of 1. That's inside the domain of the square root. And negative 1 plus 2 is in fact 1, right? So you get the square root of 1 and 1, which is just 1 minus 1, which is 0, which is not the same thing as 2. So it turns out negative 1 didn't work. Right now is the party crasher that we anticipated. The only solution here is going to be x equals 23. So when you work with radical equations, there are problems with the domain, which our potential solutions turned out to be okay there. We didn't have to take the square root of a negative ever. But you have to also worry about the problems with the range that this party crasher that although negative 1 lives inside the domain of these two radicals, it does not make the left-hand side equal to the right-hand side. And it's snuck in there because of these issues with range and square roots, and it's squaring functions not being one-to-one functions. So if you remember to check your solutions to look for party crashers, then you can solve these radical equations. You take the inverse power to get rid of a square root, you take a square root, you get rid of a cube root, you take a cube root, you get rid of a fourth root, you take a fourth power. And that will then just have to make sure you check for party crashers and that will solve these radical equations.