Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College solving radical equations. We need to get the radical on one side by itself. If it's a square root, we're going to understand that there's no number in the index, and hence the index is an understood too. And to get rid of the index, we'd have to take it to that power. So if it's a sqrt 2, we're going to square it. If it's a fourth root, we're going to take it to the 4th power. So number one example, we already have the radical on one side by itself. So we're going to square each side to get rid of the square root. So the square root and the square cancel five Z -, 2 equal 9. We're going to add 2. We get 11. We're going to divide by 5 S 11 fifths is our answer. Now we actually need to check these. So sqrt 5 * 11 fifths -2 equal 3. The five and the fifths are going to cancel. We get 11 - 2 and sqrt 11 - 2 is 9 and sqrt 9 is 3. So that one does check this next example. I need to get the radical by itself first, so we're going to add 70 each side. Then because it's X - 2, I have to square each side in order to get rid of the square root. So we get X -, 2 equal 9 or X equal 11. I'll let you check that one. For the next one, we're going to take it to the 4th power because the index was four. And if I do it to one side, we have to do it to the other. So X + 3 equal 81 or X = 78. Once again, make sure you check it. The 4th root of 81 or 78 + 3 equal 3. The 4th root of 81 is indeed 3. This next one when the exponent is written as. When the exponent is written as a fraction, we know that that is going to actually be a cube root, because the number in the denominator is the index and the number in the numerator is the power. In this case, X + 5 to the 1st and anything to the first power is really just itself. So to get rid of this, we're going to cube each side. Cube one side, cube the other. So we get X + 5 equaling 64 or X equal 59. Once again, go ahead and check that, giving you a few to try the next example. We have two radical terms, and when we have two radical terms, we have to get one by itself on one side first. When we look here, this one is already by itself, so we're going to square to get rid of the square root. But if I square one side, we have to square the other, and the other side is a binomial. There's 2 terms. So over here on the right side, we're going to actually have to foil it out. So on the left side, it turns into pretty easy square root, square cancel. But on the right side, we're going to foil it. So we're going to get 4 + 2 square roots of two X -, 5 plus another two square roots of two X -, 5 + sqrt 2 X -5 ^2. So the square root and square down there on the end are going to cancel. So we're going to get four X -, 3 equaling this 4 + 4 square roots of two X -, 5 by combining like terms, +2 X -5 S continuing to combine like terms. I'm going to leave the radical down at the end, and I'm going to bring the 2X and the -5 and the +4 all up in the front. So we get something that looks like this just by combining like terms. Now I'm going to subtract the 2X from each side and I'm going to add 1 to each side. And I'm going to see that I have an equation now where every term there are three terms are all divisible by two. And I like to keep my equation small because I do a lot of the math in my head. So I'm going to divide everything through by two X -, 1 equal. Now this whole thing on the right hand side, 4 sqrt 2 X -5 is all really one term. So when I divide that whole term by two, I'm going to get 2 sqrt 2 X -5. Now to get rid of the square root, I've got to square it. And if I'm going to square one side, I've got to square the other. So the left side we're going to have to foil again X ^2 -, X -, X + 1. The right side is going to turn into 4, the square root and the square going to cancel. So two X -, 5. Continuing to solve this one, we get X ^2 - 2, X plus one equaling eight, X - 20. Taking everything to one side, we're going to get X ^2 - 10, X plus 21 = 0. Factoring, we get X - 3 and X - 7 equaling 0, so X equal 3 and seven. Now we absolutely have to check. So 4 * 3 - 3 equal 2 + 2 * 3 - 512 - 3 is 9 ^2 and 9 is 3. 2 + 6 - 5 is sqrt 1, so that one does indeed check. If we do 7, we get 4 * 7 - 3 equal 2 + sqrt 2 * 7 - 5 S 2 four times seven. 2828 - 3 is 25, sqrt 25 is 5. 14 - 9 is 3. No 14 - 5 is 9, sqrt 9 is 3. So this one does check also, they won't always. So let's look at the next example. And I'm actually going to go ahead and delete all this because I'm going to need a little more room. Feel free to write in the margins of your problems. So on this next one, the very first step is I've got to get one of the radicals by itself on one side. So I'm going to subtract the four. So I'm going to get sqrt 10 -, X equaling 2 + sqrt 4 -, X I'm going to square each side, remembering I have to foil on the right side. So I get 10 - X equaling 4 + 2 square roots 4 -, X + 2 square roots 4 -, X plus 4 - X. So 10 minus X 4 + 4 is 8 -, X + 2 no +4 combining like terms sqrt 4 -, X. This case I'm going to take everything to one side except for the radical. So I'm going to subtract 8 and I'm going to add X and I get 2 = 4 square roots of four minus XI. Like small numbers, both sides are divisible by two, so I'm going to divide both sides by two. Now we have to square again. So when I square the left side, I square the right side. I get one equaling 4 square root and square cancels so 4 - X so 1 = 16 -, 4 X so -15 equal -4 ** equals 15 fourths. So when I check this one, we're going to get 4 + sqrt 10 - 15 fourths equaling 6 + sqrt 4 - 15 fourths. 10 we could think of as 40 fourths -15 fourths equaling 6 + 4 is 16 fourths -15 fourths, so 40 -. 15 is 25 fourths and we can square root each of those separately. So we get 5 halves and two goes into 5 twice with one leftover. So 5 halves is 2 1/2 or 6 1/2 on the left side, and on the right side we're going to get 6 + sqrt 1/4 or 6 + 1/2 which is 6 1/2. And that does check a couple for you to try. And then our last example is going to be G of X equal 5. So we're going to put five in for our G of X. And we're going to get sqrt X + sqrt X -, 5. Just like before, I need to get one of the square roots by itself. So I'm going to subtract that square root of X. It doesn't matter which one. I'm going to square each side. When I square the left side, I got to remember to foil. So I'm going to get 25 -, 5 square roots of X -5 square roots of X and sqrt X * sqrt X is just going to give me an X right side. The X minus fives are going to cancel. So we get 25 -, 10 square roots of X + X equal X -, 5. In this case, we're going to get the radical by itself again. So we're going to subtract 25 from each side, and we're going to subtract X from each side. I like small numbers, so I'm going to divide each side by a -10. In this case, when I square each side, I get sqrt X ^2 equaling 3 ^2 or X equal 9. When I check it, 5 = sqrt 9 + sqrt 9 - 5 sqrt 9 is three, 9 - 5 is 4, sqrt 4 is 2. This one does indeed check. Thank you and have a wonderful day.