 This video will talk about solving radical equations. Again, solving radical equations are similar to solving quadratic equations that have only a square in them. So just like when we had x squared equal k, we took the square root of both sides. Again, this is what I'm talking about. We have x squared equal 2k, and we said we could take the square root of both sides, and we had x on this side. And the reason why we could do that is because we had a square root which is inverses that cancel each other out. Well, now we're going to be starting with a radical, and we want to get rid of the radical. So to do that, we would square, and the square will cancel the square root, and again we'll get x. So to do that, that's the concept. So we need to isolate the radical by moving things around, taking the radical isolated. And that includes coefficients. And then once we get that isolated, then we could square both sides. To do it to both sides, then we have to do the whole side on each side. And then we're going to either have a linear or quadratic equation to solve. And then the key thing is you cannot omit this. Unless you graph, you cannot omit this step, and that is to check. Because sometimes you get it linear, and quadratics have answers that solve them, but wouldn't fit with the radical. So don't forget, you have to solve radical equations graphically as well. You can either set them equal to zero and look for your zeroes or x-intercepts. Or you can find the intersection of the two sides. We're going to do some of those that way. The only problem with graphs is they require you to change your window often. So sometimes that can be tricky. Alright, so this first problem, the square root of x equals five. That's a pretty simple one. In fact, it's so easy, I want to just show us how to graph it. And one way we said you could do it is if you set it equal to zero. So if I set it equal to zero, I'm going to keep the square root here, and I'm just going to subtract the five to the other side. And that'll be equal to zero. So now this is going to be my y1, and the zero is going to be my y2. So I have a second x squared for the square root of x, close my parentheses minus my five, and then y2 is going to be zero. And this one is pretty nice. It should work in a simple standard window. And it doesn't quite fit in a standard window, but I can see that I just need to add some positive x's. So go to my window, and I want to make sure I don't have to come in here real often. So I'm going to try 30. I think that should be nice and big enough for me to look at. And sure enough, I can see that there's an intersection. So I'm going to write about there. So second trace intersect, because that's going to be my x-intercept. Enter, enter, enter. And it takes me over there and tells me that x is equal to 25. So I would just want to sketch my little graph here. I've already told you this is y1 and this is y2, and then this tells me that x is equal to 25. I'm going to solve this one by squaring both sides. So if I square both sides, then we have the square, cancel the square root on this side, and we have 4x plus 5, and then it's equal to x squared, which means I have a quadratic equation, so I need to set it equal to 0. So I'm going to take the x squared to the other side, because I'll use the quadratic formula in my calculator so it doesn't matter what x squared is positive or negative. So negative x squared plus 4x plus 5 is equal to 0, and over in my calculator, using the quadratic formula, a is negative 1, b is 4, c is 5, and I find out that x is equal to negative 1 and x is equal to 5. Well, I have to check both of those. So if I check negative 1, I have the square root of 4 times my negative 1 plus the 5, and that's supposed to be equal to my negative 1. And when I do the work to it, this gives me the square root of 4, negative 4 plus 5 equal to negative 1, or the square root of 1 equal to negative 1, but the square root of 1 is positive 1, which is not negative 1. So this is not one of my answers. But if I go and check the other one, I have the square root of 4, remember we're plugging back into the original equation, times my x, which is 5, plus the 5 all under the radical, equal to my x, which is the 5, and I work on this side, I get the square root of 20 plus 5 equal to 5, and the square root of 25 we know is 5. So that one works, and my answer is x equal 5. So let me have a problem like this one. I want to show you how to graph it. So I put in the left hand side is y1 and the right hand side is y2. That's what I have right here. And I'm set up for a standard window, and we look at the graph and we can see this part, but I can't see the negative 27. But if you think about it, it's going to be down here. But let's change our window so that we can see it. So I need my y minimum to be, let's say, negative 35. So if I graph it, there's my radical and there's my negative 27. And they are never going to cross. So there is no solution to this one. And again, what are we going to do here? Well, we could graph it, or we could square both sides. I have an isolated thing right here. You can decide whether you want to graph it or you want to square both sides. I'm going to square both sides just for teaching purposes. We square and we get 13x plus 4 on this side. And we square the square root over here and we get 5x plus 20. And now it's a nice, simple linear equation. I'll subtract my 5x from both sides and that'll give me 8x plus 4 equal 20. And if I subtract the 4 from both sides, that'll give me 8x equal to 16. And if I divide by 8, x should be equal to 2. But is it really equal to 2? Well, let's see. The square root of 13 times 2 plus 4 is supposed to be equal to the square root of 5 times 2 plus 20. The square root of, this is 26 plus 4, which isn't that being the square root of 30. And this one gives me 10 plus 20, the square root of all that, which is the square root of 30. And they are exactly equal, so we can say that x is equal to 2. Okay, this is an interesting-looking problem. Let's just peel away the layers so we have to subtract the 6. The 6 is being added to this square root thing, so I have to subtract 6 from both sides. The negative 4 square root of x minus 1 is going to be equal to a negative 24. And if we divide by negative 4, everything, both sides, we get the square root of x plus, or x minus 1, equal to positive 6, negative divided by negative. Now we're ready to square both sides. You have x minus 1 here, equal to 36, and x is equal to 37 when I add one to both sides. Oh, we go and we check. 6 minus 4 square root of 37, and then minus my 1 underneath the radical, and all of that is supposed to be equal to negative 18. Well, 6 minus 4 times the square root of 36, is still supposed to be equal to negative 18. And 6 minus 4 times the square root of 36 is 6, and that's supposed to be equal to negative 18. And now we have 6 minus 24, and guess what? Now we have negative 18 equal to negative 18, so we have a real solution. And some of you may be saying, wait a minute, you had a radical equal to a negative. But I had a radical equal to a negative until I had the radical isolated, but I ended up dividing off a negative, so it ended up being a 6. And over here, I had the 6 minus a product, and that's how I got my negative. It wasn't because the radical by itself was equal to a negative. This one might have been really nice to graph. Our last problem. When we look at this one, I want to do this one by hand, and I want you to be very careful. A lot of people like to do these by hand, which is totally fine, but you've got to remember to do this side correctly. So when I square my square root here, remember I'll just get x plus 10. But when I square that side, it's really like saying x minus 2, I have two factors of that. So you need to do the foil on it. You should get x squared minus 4x plus 4, not x squared plus 4, which is what I often get when I'm grading. So I need to subtract x from both sides. So 10 will be equal to x squared minus 5x plus 4. And when I bring the 10 to the other side by subtracting it, I get x squared minus my 5x minus 6. And if I go to my quadratic formula, just to get this done quickly, find out that a is 1, b is negative 5, c is negative 6. And that tells me then that x is equal to 6 and negative 1. And I have to go check. So I'm going to say that the square root of 6, that's the first x I see, plus 10 would be equal to 6 minus 2. This is the square root of 16. And 6 minus 2 would be 4. And the square root of 16 is 4. So 6 works. And if I try and do check the other one, I get the square root of negative 1 and then plus my 10. And that's supposed to be equal to my x, which is negative 1 and minus 3. I have a square root of 9, negative 1 plus 10, equal to negative 1 minus 3 would be negative 4. And the square root of 9 is 3, which is definitely not negative 4. So x equals 6 is my only solution.