This Square Root Is Hiding Something

Ajouté : 2026-05-05

[00:00:00]Hello everyone.

[00:00:02]In this video, we're going to be solving a radical equation.

[00:00:07]Kind of like an easy one, maybe. I don't know. We'll see. And I'll be presenting two methods.

[00:00:11]Uh we have the square root of x plus two times the square root of x minus one is equal to five, and we're supposed to solve for x values. So, one thing that we can talk about before we get into solution methods is the domain.

[00:00:26]As you know, for this on this channel, we're looking for real solutions because I have another channel called a plus bi dedicated to complex numbers, the world of complex numbers. You can go ahead and check it out and let me know let me know what you think. Uh there are also a bunch of lecture videos you can check if you're new to complex numbers.

[00:00:44]And let me know what you think. So, we're looking for real solutions, and in order for the solutions to be real, we have to make sure that whatever is inside the radical, which is called a radicand, such a weird word, but what's inside the radical needs to be greater than or equal to zero. Obviously, you don't want to deal with something like the square root of negative one because that belongs to the complex world, not to the real world. Of course, when I say complex world, it kind of includes the real world, but we're talking about the complex complex world, okay?

[00:01:15]Whatever that means. So, to solve this problem, like I said earlier, we're going to look at the domain. So, I want x minus one to be greater than or equal to zero. That means x needs to be greater than or equal to one. Anything less than one, we're going to reject because it's not going to work. But that's not the only thing because this is kind of like a the nested something, which might be a clue for some people.

[00:01:39]Uh or sorry, I meant nested radicals.

[00:01:42]Sorry, I gave it away already. But uh we have another radical. So, radical wrapped inside another radical. So, we have to consider the outer radical, and that means this expression right here needs to be greater than or equal to zero as well. So, x plus two times the square root of x minus one needs to be greater than or equal to zero. But what is that supposed to mean, right? It's kind of like a weird inequality in and of itself. So, to be able to solve this, we could probably look at something like this.

[00:02:14]Subtract x from both sides, and then try to square both sides. But here's a problem. When you square both sides, are you going to keep the same sign or you're going to change the inequality, right? So, we could do both and then test it out uh with the findings, that is a safer method. So, let's go ahead and square both sides. Four times x minus one equals x squared, and I want to keep it as is. From here, we get uh 4x minus 4 is greater than or equal to x squared. Let's bring everything on the same side and kind of flip sides here. x squared minus 4x plus 4 is less than or equal to zero. By the way, this is a perfect square in case you haven't noticed. And what does that mean? How can a perfect square be less than equal to less than zero? It can't be less than zero, but it can be equal to zero, in which case x has to be a two.

[00:03:06]So, from this branch, the only valid value for x is x equals two. Does that work? When we did the squaring, uh we made an assumption uh that uh when we square the sides, we're not going to change sign. But anyways, let's do this.

[00:03:25]Try the other one.

[00:03:26]What if the sign changes?

[00:03:30]Uh I meant less than or equal to.

[00:03:33]Oops.

[00:03:35]Less than or equal to x squared. And obviously, when you bring everything over, it's going to look like this, which will make more sense, hopefully, because this is x minus two squared, and this is always always always true. So, we don't have an issue with the right-hand side. It's all good.

[00:03:52]Everything's satisfied, including two, by the way. You got to be careful here, though. If we had something like this, um oops, that's not what I meant. If you had something like this, then x equals two would have to be excluded, but that's not what we have. All right? So, it's all good. Uh this is nice, and the other branch gives us x equals two, which is already included in this because everything will work, which means this is probably always going to be uh greater than or equal to zero. And you can test it. Um it's kind of hard to It's not hard to see because we already know x needs to be greater than or equal to one. So, if you replace x with one, you're going to get one plus zero, which is already positive, and anything greater than one will give you a larger value. There's a plus sign. This is going to be positive.

[00:04:40]All good. You see? You can also look at it that way. But anyways, this was the domain, and we the only finding we have is x must be greater than or equal to one. Okay? Now, let's get to the solution. First method.

[00:04:51]First method is basically going to focus on uh like normal method. The second one is the abnormal method. That's why I saved it for the second, okay?

[00:05:00]Bear with me now.

[00:05:02]When you have a radical equation, how would you solve it? You would square both sides, wouldn't you?

[00:05:07]Okay, so let's do it. Square both sides, get rid of at least one of the radicals.

[00:05:12]This gives us something tiny bit simpler. And then, of course, your next step should be what? You don't want to square both sides again because that's going to keep giving you radicals. You don't want that. Subtract x from both sides first, and then square both sides.

[00:05:25]That's actually a smarter move, okay?

[00:05:27]It's something you need to learn if you're dealing with radical equations.

[00:05:31]And I did make a bunch of videos on radical equations. Go ahead and check out my what's it called? Playlist on radical equations. Okay. When you square both sides, it's going to be four times x minus one equals 625 minus 50x plus x squared. Let's bring the x squared over here. x squared minus 50x plus 625 uh equals 4x minus 4, and then we're going to bring everything over to the left-hand side. x squared minus 54x, and 625 plus 4 is 629.

[00:06:09]Okay? So far, so good.

[00:06:11]And that's equal to zero, of course.

[00:06:13]Now, you want to solve this quadratic, don't you? Come on.

[00:06:17]And when you solve it, what are you going to get? Well, looks like 629.

[00:06:23]And that could be a prime number, I don't know.

[00:06:25]Uh probably not. Hopefully not because then we would be stuck, right? Uh but um you can try factoring it. Uh looks like it's not divisible by three, for sure.

[00:06:36]How about how about uh seven? What am I thinking, right? Obviously not by five or um not six, not eight, nothing even, obviously.

[00:06:49]Uh if it's not divisible by three, it's not going to be divisible by nine, but maybe seven. Seven might work. And how do you check if a number is divisible by seven?

[00:06:58]You take the ones digit, double it, and then subtract it from the rest of the number as if this was a two-digit number. You'll get 40 uh I was hoping to get 42, but it didn't work. We're getting 44, which is not a multiple of seven, which means 629 is not divisible by seven, either. It could be divisible by hmm.

[00:07:18]11 doesn't work. Maybe 13.

[00:07:23]Uh let me think. I don't think 13 is going to work, either. So, we're kind of stuck here. Such a bad number. I don't know why.

[00:07:30]Uh I hope I didn't make any mistakes. But anyways, if by solving this quadratic equation, and you can set it up like this, uh negative b plus minus the square root of b squared minus 4ac, maybe this will give us a clue divided by two. Okay. Let's go ahead and square 54, and then we're going to multiply four times 629, and then we will, hopefully, get a nice solution from here. Okay?

[00:08:02]So, we have like two options here.

[00:08:04]Either work this out or try to solve it if there are good solutions, obviously.

[00:08:10]Uh maybe try completing the square.

[00:08:12]Let's go ahead and try completing the square. I'm going to go ahead and subtract 629 from both sides, and then we'll we'll get lucky, right?

[00:08:20]We'll get lucky, maybe. Now, to complete the square, I I need I need to consider half of 54, and then square it, which is going to be 27 squared. That's the number I need to add to both sides. And what is 27 squared? It is 729.

[00:08:36]And when you add 729 on the right-hand side, boom, you you get 100, which is a perfect square. And that's perfect. Now, we have x minus 27 squared on the left, and on the right-hand side, we do have 100. And obviously, you could also go off of this, uh but this will probably be a tiny bit more painful. Actually, it's not going to be that painful because this is going to give you something like 26, 2700, and this is going to be close to that, so the difference is probably going to be a two, three-digit number, which you can easily uh square root. Okay, but this is really good. Uh it's the idea behind the quadratic formula, by the way. So, from here, you can tell x minus 27 is either 10 or x minus 27 is equal to negative 10. If you add 27 to both sides, x equals 37 or x equals 17. So, those were the two numbers, actually, if you think about it. Those were the two numbers that made up 629, and when I told you this might be a prime number, uh it kind of makes sense because look at this. I mean, I was going to test all the numbers up to 17, and that's when I found out, okay, there you go, it's divisible by 17, and then you find the other one. You see? It's really hard to figure out unless you kind of know this.

[00:09:54]Hopefully, from now on, you're going to know. But basically, those two solutions. But, million-dollar question is do they both work and do they fit the domain?

[00:10:04]You already know the answer, right? But, let's just, for fun, test them out.

[00:10:09]So, let's go ahead and look at the original problem. If x is 37, let's see what happens. If x is 37, uh you're going to get 37 plus two times the square root of 37 minus one, which is 36. As you know, this is six. Two times six is equal to 12. 12 plus four 37 is 49.

[00:10:28]Wait a minute. Did I get that right?

[00:10:30]Okay. Uh 49 and the square root of 49 is seven. Uh-oh. Houston, we have a problem. 37 did not work. Okay, let's test the other number, which was 17. If x is 17, then we're going to be getting the square root of 17 plus two times the square root of 16. Square root of 16 is four. Two times four is eight. 17 plus eight is 25 and the square root of 25 is five. Yay. x equals 17 works. Uh nay, x equals 37 does not work. Therefore, the only solution to this equation seems to be x equals 17. But, let's go ahead and look at this problem from a different angle, which I hope you'll like and let me know what you think. But, this problem is actually was designed uh with that in mind. So, that But, why am I giving you the first super painful method? It's not super painful, but uh tiny bit. Uh first of all, it's good to know multiple methods and also no pain, no gain, okay? Now, here's how it works.

[00:11:29]This expression should be very familiar to you and once I show you, hopefully, you'll remember for the rest of your life, okay?

[00:11:35]Maybe not too long, I don't know. Uh x minus one. So, we're going to subtract one and add one. As you know, if you subtract one and add one, it's zero.

[00:11:43]Nothing is going to change. Actually, a lot of things are going to change because if you think about it, this is the square root of x minus one squared.

[00:11:51]This is two times the square root of x minus one times one and this is one squared. What does that look like to you?

[00:11:58]Doesn't that look like a perfect square?

[00:11:59]Absolutely. This is the square root of x minus one plus one quantity squared.

[00:12:04]Then, I'm square rooting it and setting it equal to five, which means the square root and the square will cancel out. Of course, this quantity is not never negative, is not ever negative.

[00:12:16]So, we end up with the single result, which gives us the following.

[00:12:23]And then, we square both sides and, finally, we arrive at the single solution as before. But, before, it was a little bit more painful, wasn't it?

[00:12:33]And this brings us to the end of this video. Thanks for watching. I hope you enjoyed it. Please let me know. Don't forget to comment, like, and subscribe.

[00:12:39]I'll see you next time with another video. Until then, be safe, take care. Don't forget to check out CyberMath shorts channel and A plus B I and until next time. Bye-bye.