Hello wonderful mathematics people, This is Anna Cox from Kellogg Community College. Radicals in multiple terms. Add or subtract. Simplify by combining like radical terms. If possible. Assume that all variables in Radicans represent positive real numbers, meaning no absolute value brackets are really necessary. So in this first example, we're looking for like terms. And what we mean by like terms is we mean that the inside of the radical is the same. So the inside is 7 and the inside is 7 for these two. So if I have 5 of something and another one of something understood number one in front, if there's not a number, I've got six of those somethings. So those two terms are going to combine to give me 6 radical 7. Then I have 1/4 root of 11 and another fourth root of 11. So those are like terms. So if I have -8 of something and a +9, that gives me one of them or the 4th root of 11. I don't have to write the one. It's understood on the next example. These are not the same on the inside to start with. So we got to figure out how to make them the same. If we think about 54, we could think of 54 as 9 * 6, and nine is 3 * 3, and 6 is 2 * 3. So there's two, there's 1-2 and three threes. So the cube root of 2 * 3 ^3 * X, the cube root of 2, we don't know. But the cube root of 3 ^3, the exponent and the index are going to cancel, leaving us a three on the outside. And then the cube root of X, we don't know, so we're going to leave it on the inside also. So we get three cube roots of 2X. This next one, the cube root of 2, I don't know, but I do know that the cube root of X to the 4th, that three goes into four one time with one leftover. So now the insides of those radicals, the radicand portions are the same. So I'm going to realize that I have 3 -, X of the cube roots of two XI can't combine the 3 -, X because those aren't like terms. But the cube root of 2 X and cube root of 2 X we're going to factor out, leaving us the 3 -, X On this next one. We're going to be able to realize that the inside is not currently a monomial. So we're going to factor out a nine and get 9 times the quantity y + 3 plus the quantity y + 3. Then sqrt 9 is 3 and now those radicans are the same. So if I have 3 of something and one of something that's going to leave me four of those somethings. I'm going to let you do a few. On the next type of problem, we're going to do distribution. We're going to multiply. So we're just going to actually multiply things out. If the number is on the outside of the radical, we're going to multiply those, and the numbers on the inside we're going to multiply. So 5 square roots of 30 -, sqrt 18. Now 30 can't reduce because 30 is 5 * 6 or 5 * 2 * 3. There's not 2 of any of them because it's a square root in order to take one out. But 18 will reduce because 18 is 2 * 9 or 2 * 3 * 3. So if we thought about this as 2 * 3 ^2, then the index of two goes into the exponent of the 3 ^2 1 time 2 goes into two once. So A3 root 2. So we'd get 5 sqrt 30 - 3 sqrt 2 is our final answer. This next one we're going to foil. So first terms are going to give us 12 square roots of 5 ^2 - 16 square roots of 15 + 9 square roots of 15 - 12 square roots of 3 ^2 and the square root and the squares are going to cancel. So we get 12 * 5 minus the 16 and the 9 root Fifteens are like terms, so we can combine those and then 12 * 3 down here at the end. So 12 * 5 is 60 - 7 root 15 - 12 * 3 is 36. Our final answer, 24 - 7 root 15. So a few for you to try. We're going to rationalize the denominators next. And to rationalize a denominator, we need to multiply by what's called its conjugate. So if I have 7 plus root 6, the conjugate is 7 minus root 6. And if I do it to the bottom, I've got to do it to the top. So now the top turns into 3 * 7 minus root six. I'm not going to distribute that out quite yet because I'm going to see if once I foil and simplify, if there's anything in the top that would reduce with the bottom. In this case, the three possibly. So we're going to get 49 -, 7 root 6 + 7 root 6 minus root 6 ^2. That root 6 ^2 is going to just turn into 6 and we get the 3 * 7 minus root 6. Now the whole reason for the conjugate is to make the radicals in the denominator go away. 49 -, 6 is 43, so all the radicals, the negative and the positive cancelled and then the square root and the square cancel, so all the radicals in the denominator go away. So our final answer is going to be 21 -, 3 root 6 / 43. This next one we're going to multiply by the conjugate and we get 4 root 3 + 3 root 2. The conjugate is keeping everything exactly the same except for that middle term. It's based on the concept of the difference of squares. If I have a ^2 -, b ^2, a + b and a -, b, multiply, getting rid of those middle terms of those radicals. So here, when I foil this out, we're going to have to foil the top and the bottom. So we're going to get 28 root 6 + 21 root 2 ^2 + 16 root 3 ^2 + 12 root 6. That's all the numerator, the bottom 16 sqrt 3 ^2 + 12 square roots of 6 -, 12 square roots of 6 - 9 square roots of 2 ^2. So continuing with this one, the 28 and the 12 are going to combine to give me 40 square roots of six. The square root and the square are going to cancel, so 21 * 2 and 16 * 3. And the bottom we're going to get the square root and squares canceling again, so 16 * 3 - 9 * 2. So simplifying this up, we're going to get 40 square roots of 621 * 2 is 42 + 48, so 90 over 48 -, 18, so 30. And then we can see that all of those terms are evenly divisible by 10. So 4 square roots of 6 + 9 / 3 is going to be our final answer for that one. Let's look at some more. We're going to give you a few to try. This next one is to rationalize the numerator. And it's the exact same steps, except now we're going to multiply by the conjugate of the top instead of the bottom. I'm going to let you finish it up now that I've talked it through. What the rationalize the numerator is this last grouping. These are no longer the same indices and we have to get them the same indices before we can do anything. So I'm going to think about four and three have the lowest common multiple of what? And the answer is 12. So the lowest common multiple of four and three is 12. Now what we're going to do is we're going to think about four times what gave me 12, and the answer is 3. So I'm going to do the same thing to the exponent. So 3 * 3 is going to give me 9 over on this next one I took 3 and I * 4. So now I'm going to multiply the exponent by 4. Once the indices are the same, then we combine the insides. In this case the bases are the same, so we add the exponents. Now 12 is smaller than 17, so 12 goes into 17 one time with five leftover. That's our final answer. This next one, when we look here, we understand that if there's not a number, it's understood to be a 2SO2, and three is going to give us a common indicee of six. So multiplied this understood 2 by 3. So this is going to turn into 2 ^3, X to the 9th, Y to the 9th. Now this 4 is not prime, so we want to think of the four as 2 ^2 and then X and Y ^2 on the inside. So I took that index of three and multiplied by two. So now the exponents are all going to get multiplied by two. Once we have the same indices, then we're going to combine the inside. So we're going to get 2 to the 7th, X to the 11th, and Y to the 13th. 6 goes into seven one time with one leftover. Six goes into 11 one time with five leftover. Six goes into 13 twice with one leftover. That's our final answer. Division's going to work the exact same way. We're going to get common index, so 5 and understood 2 is going to give us 10. So we're going to have multiplying all the exponents by 2X to the 6th Y to the 8th X ^2 y ^2. Now with division, once the index indices are the same, we subtract the exponents on the inside. So 6 -, 2 is four, 8 - 2 is 6. So right here I had an error. This is an understood 2 that we multiply by 5 to get 10. So these should have been fives, and then 6 subtract 5 is 1 and eight subtract 5 is 3. So we would have gotten the 10th root of XY cubed as a correct answer. This next one we're going to do each term separately because we're going to do distribution properties. So we've got the cube root of X ^2 y times the square root of XY minus the cube root of X ^2 y, the 5th root of XY cube. So the common indices on this first two are 6, and we're doing the exact same skills we did a minute ago. So 3 * 2. So we get X to the 4th y ^2 and this is 2, and we're going to multiply it by 3. So X ^3, y ^3 minus in between 3:00 and 5:00 is going to be 15. So X to the 10th, Y to the 5th, 15th X ^3 y to the 9th. So once the indices are the same, 6th root of X to the 7th, Y to the fifth minus the 15th root of X to the 13th, Y to the 14th. 6 is smaller than the 7th. So six goes into seven one time with one leftover, and then six doesn't go into 5, and 15 is bigger than the other two. So those can't simplify down. So that's our final answer and there's a few for you to try. Thank you and have a wonderful day.