Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. Add or subtract. Simplify by combining like radical terms if possible. Assume all variables and radicands represent positive real numbers, meaning no absolute value brackets are necessary. So if I have 3 root 7 + 11 root 7, that's going to give us 14 root sevens. If you thought about having 3X's plus 11X's, we combine our like terms and that's 14X's. Same concept here. The root sevens are in common. You could always think about using distributive property to take out the root 7, which would leave us 3 + 11. Three plus 11 is 14 and we can multiply it on any order we want. We usually put the whole number first in the radical at the end. This one if I have root three y + 8 root 3 YS, we have an understood one in front. So 1 + 8 is 9 cube roots of Y. This one if we have one radical 6 + 8 radical 6 - 3, radical 6. Well, 1 + 8 is 9, Radical 6 -3 radical six 9 - 3 is 6 radical 6. If you're having a hard time seeing it, you could always think about pulling out the radical 6. Understood. One in front plus 8 - 3. One plus 8 minus three 9 -, 3 is 6, so radical 6 * 6 and we usually put the whole number in front. This next one we can only combine like terms and the radical and the radican portions have to be the same including the index. So if I have 5 root 7 plus root 7, we get 6 root sevens. If I have -8 fourth root of 11 and +9 fourth root of 11, negative 8 and +9 gives us 1/4 root of 11, 9 root 50 - 4, root 2. These are not exactly the same. I need to figure out how to make sqrt 50 simplified. Well, 50 is 25 * 2, and we know sqrt 25 is 5, so we get 9 * 5 root 2 - 4, root two, 9 * 5 is 45, root 2 - 4, root 2, or 41 square roots of two. For a final answer, this one they're not the same, so we need to simplify them. 54 We could think of 54 as 6 * 9. Six is 2 * 3, and 9 is 3 * 3. So 54 is 2 * 3 ^3 * X. We're going to simplify these up. We don't know the cube root of 2, but the cube root of 3 ^3 is 3, and we don't know the cube root of X. The second term, we don't know the cube root of 2, but the cube root of X to the fourth three goes into four one full time with one leftover. So now we have 3 -, X cube roots of 2X. I can't combine the three and the negative X because they're not like terms, but they both have that cube root of 2 X. So using distributive property, we can take that out. This next one, nine y + 27, they have a nine in common, so we're going to factor out a nine and get y + 3. Sqrt 9 is 3, so 3 square roots of y + 3 plus sqrt y + 3, the understood one in front. So 3 + 1 is 4 square roots of y + 3. We're going to multiply now, so we're going to actually distribute. We're going to get sqrt 5 times sqrt 5 or sqrt 5 ^2. We're going to get sqrt 5 times sqrt 2 or 5 * 2. Sqrt 5 ^2 is going to leave us 5, sqrt 5 we don't know, and sqrt 2 we don't know. So we're going to just multiply that to get sqrt 10. So 5 -, sqrt 10 is the final answer. This next one distributive property, again sqrt 3 has an understood one in front. So we're going to have the 1 * 5 and sqrt 3 times sqrt 10. We're going to have 1 * 1 square root of three times sqrt 6, so 5 square roots of 30 -, sqrt 18. And the question is, can that simplify up? Well, 35 * 6 five times 2 * 3. I need 2 because it's the square root to take one out. I don't have two of anything. I don't have two fives or two twos or two threes. If we look at 18, that's going to be 3 * 6 or 3 * 2 * 3. There we do have two threes, so we're going to take out a three. We can rewrite 18 as 2 * 3 ^2 sqrt 2. We don't know, but sqrt 3 ^2 is going to be 3 square roots of 2SO5 square roots of 30 -, 3 square roots of two. This next one, same thing even if the index changes. So we're going to get the cube root of 27 - 4, the cube root of 63. So 27 3 * 9 or 3 * 3 * 360 three 7 * 9 or 7 * 3 * 3. We need a cube root, so we need three of them. We do have 3 threes for the 27. We don't have 3 of anything in the 63. There were only two threes and 1/7. So our final answer, the cube root of 3 ^3, 3 -, 4, the cube root of 63. Now we're going to foil. We're going to do the first terms, the outer terms, the inner terms, and the last terms. If you squint really hard and have a good imagination, that looks like Charlie Brown when we foil. So first outer inner last 4 * 4 + 4 * -3 square roots of 2 + 3 square roots of 2 * 4 - 3 square roots of 2 * 3 square roots of 2. So 16 - 12 root 2 + 12 root 2 - 9 square roots of four. This was actually the difference of squares formula. So the positive term in the middle and the negative term in the middle are going to cancel. So we're going to get 16 - 9 * sqrt 4 is 2. So 16 - 18 or -2 going to foil again. So 4 root 5 * 3, root 5 -4 root 5 * 4, root 3 + 3, root 3 * 3, root 5 - 3, root 3 * 4, root 3. We're going to multiply the numbers that are in front, so the 4 * 3 and we're going to multiply the numbers inside. 5 * 5 is 5 ^2 4 * 4 five times 3, 3 * 3 three times five, 3 * 4 and 3 * 3. So simplifying this up, we get 12 * 5 - 16. I can't simplify sqrt 5 * 3, so we'll make it square to 15 + 9. Sqrt 15 - 12 * 3 12 * 5 is 60 - 16, root 15 + 9 root 15 - 36. So we're going to get 24 - 7 sqrt 15 next one. It's a quantity squared. So we're going to have 4 + sqrt X -, 3 * 4 plus square roots of X -, 3. And we're going to do our foil again. So our first terms are going to give us sixteen 4 * 4 outer terms, 4 square roots of X -, 3 inner terms 4 square roots of X -, 3, and last term sqrt X -, 3 times itself or quantity squared. So we'd get 16 + 8 sqrt X - 3. The square root and the square at the end are going to cancel. So we're going to combine. Unlike terms, we're going to have X + 13 + 8 sqrt X - 3. We want to rationalize the denominator. So if it's got a single term, a monomial, we're just going to multiply the top and the bottom by whatever that single term is. So we're going to get sqrt 66 / sqrt 6 ^2. The square root and the square are going to cancel sqrt 66 / 6. Now this is where it's going to get a little tricky. We're going to use our concept of our difference of squares a ^2 -, b ^2, so a + b * a - b. These terms over here on the right are called conjugates, so we're going to multiply by its conjugate. If I have 7 plus root 6, the conjugate is the exact same thing, but we're changing just the single sign in the middle. And if I do the bottom, I've got to do the top, so the conjugate is 7 plus root 6 is 7 minus root 6. Now we're going to not multiply the numerator out until the very end. In case something can cancel, we are going to multiply the denominators out seven times, seven 7 ^2 - 7 root 6 + 7, root 6 -, sqrt 6 ^2. The whole reason for the conjugate is to make all the radicals in the denominator cancel. So we're going to have 3 * 7 minus root 6 still in the numerator 7 ^2, 49, the -7 root 6, and the +7 root 6 cancel and sqrt 6 ^2 is going to leave us 6. So we're going to get 3 * 7 minus root 6 all over 43. The three and the 43 cannot reduce. So we can leave our answer like that. Or you could have it as 21 -, 3 root 6 / 43. Both are fine answers. What's the conjugate for this one? It's going to be sqrt a + sqrt b sqrt a + sqrt b. We're not going to distribute out in the numerator, we're going to distribute the denominator. So sqrt a * sqrt a is sqrt a ^2 plus square root of AB, minus square root of AB, minus square root of B squared. So we're going to get sqrt b * sqrt a + sqrt b. The a sqrt a ^2 is going to cancel a, the square root of AB and the negative square root of AB are going to cancel, and minus sqrt b ^2 is going to give me minus B. That's our final answer. Or you could have it as square root of AB plus B over A -, b. Now be careful, make sure you don't think that you can cancel the BS there. It's not a monomial, so we can't cancel. To have it be a monomial, we look back at what we had a moment ago before we distributed that square root of B out, and the parenthesis wouldn't be exactly the same as anything on the bottom. Here. For root 3 plus the three square roots of two, the conjugate everything stays exactly the same except the sign in the middle. Just the sign in the middle changes. When we foil, we're going to get 28 root 6 + 21 root 4 + 16 root 9 + 12 root 6. On the top and the bottom, we're going to get 16 square roots of nine, plus 12 root 6 -, 12 root 6 -, 9 square roots of four. So 28 root 6 and 12 root 6 is going to give us 40 sqrt 6 + sqrt 4 is 2, and sqrt 9 is just three. In the bottom we're going to get 16 * 3. The outer and inner terms better cancel if we did everything right -9 * 2. So 40 sqrt 6 + 42 + 48 all over 48 - 18. So keeping going 42 and 48's going to give us 90 + 40 square roots of 6 / 48 -. 18 is going to be 30. Now we could factor out a ten 9 + 4 square roots of six all over 30. 10 / 30 is going to reduce to 1/3, so 9 + 4 root 6 / 3. Or if you wanted, you could have it as 3 + 4 square roots of 6 / 3. Because three went into 9 three times. We want to rationalize the numerator this time, so we're going to do the conjugate of the numerator instead of the conjugate of the denominator. So sqrt 6 + 3 times sqrt 6 + 3. We foil the top. We get sqrt 6 ^2 + 3, root 6 - 3, root 6 - 3 ^2 / sqrt 18 + 3, sqrt 3 + 7, sqrt 6 + 21. Well, sqrt 6 ^2 is 6. The outer and inner terms better cancel. 3 ^2 is 9. Sqrt 1818 is going to be 2 * 9 or 2 * 3 * 3. So we're going to have three square roots of two. There were two threes, so I can take it out. Sqrt 2, we don't know. Plus three square roots of 3 + 7 square roots of 6 plus 20 one 6 -, 9 is -3 / 3 root 2 + 3 root 3 + 7, root 6 + 21. This time we have different indices. There's a understood 2 and an understood 4. The easiest way I can tell you to do this is to think about getting common indices. If I have a two and a four, the common indicee would be 4. Well, what did I have to do to this two to make it turn into a 4I multiplied by two 2 * 2? So I'm going to multiply the understood exponent of A1 here, also by two, so we get a squared. This second term already had a fourth, so we're going to have a ^3. Once the indices are the same, we add the exponents. 4 goes into five one time with one leftover. Now a different way to do it is to think about putting it into fractional form. 8 1/2 * 8 of the 3/4. Once those exponents or once those bases are the same, we add the exponents. What do we have to do to add 1/2 and 3/4? We have to get common denominators. 2 fourths plus 3/4. That gives us 5 fourths. So 4 went into five one time with one leftover four and three. The common indices going to be 12. What did we have to do to four to get 12? We multiplied by three. So if I multiply there by three, I'm going to multiply the exponent by three and get A to the 9th. Here we had a three. We had to multiply by 4 to get 12. If I do the index, I'm going to do the exponent, so A to the 8th. Once the indices are the same, we add the exponents. 12 goes into 17 one time with five leftover, so a the 12th root of A to the 5th. Here's an understood 2 and an understood three or and a three. So 2 and 3 is going to have a common indices of six. What did I do to this two to get six I * 3? So I've got to do each and every exponent by three. So we're going to have 2 ^3 X to the 9th, Y to the 9th. Then this three we multiplied by two. I'm going to think of four as 2 ^2. So I'm going to multiply that square by A2, the understood one there and the two there. So we're going to get 2 to the 4th, X ^2 Y to the fourth. Once the radican's the same, we're going to get 2 to the 7th, X to the 11th, Y to the 13th. 6 goes into 7/1 full time with one leftover. Six goes into 11/1 full time with five leftover. Six goes into 13 twice with one leftover. So that's our answer here. We're going to divide, but the same concept, common indices going to be 10. So what did I do to this five I * 2? So I'm going to multiply everything, all the exponents by two. I'm going to get X to the 6th, Y to the 8th down here. The understood two I've got to multiply by 5, so I'm going to multiply each exponent by 5, SX to the fifth, Y to the 5th, 10th root. Once the bases are the same, we subtract the exponents when we're dividing. So we get XY cubed. The 10th root of XY cubed here 3 and 4 is going to give us a common indices of 12. What did we have to do to that Three? We had to multiply by 4. If we do the denominator or if we do the index, we have to do the exponent. What we do to the four in the bottom, we multiply it by three. If we do the index, we have to do the exponent. There was an understood one on that exponent. Once we get the bases the same, then we subtract. So the 12th root of 2 + 5 X to the 5th. This last one, multiply it out first, then get your common indices. It'll make the work much easier. So here's an understood 3 and a 2. So we're going to have a 6th as our indices for those first two. And here's a three and A5 SA 15 index for the last 2-3 got multiplied by two, so we're going to have X to the 4th y ^2 that understood 2 got multiplied by a three, so we're going to have X ^3, y ^3. This three got multiplied by 5 S X10 Y to the fifth. The five got multiplied by a three, so X ^3 y to the 9th once those common index. So we get X to the 7th, Y to the fifth minus the 15th root of X to the 13th, Y to the 14th. 6 went into 7/1 full time with one leftover. 6 doesn't go into 5 evenly because 5 is smaller than 615 doesn't go into 13, and 15 doesn't go into 14. So our final answer is X the 6th root of XY to the fifth minus 15th root of X to the 13th, Y to the 14th. Thank you and have a wonderful.