Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. When solving radical equations, we want to get the radical on one side by itself, and then whatever the index is, we're going to take it to that power. So in this first example, when we have sqrt X + 3 equaling 6, we're going to take that sqrt X + 3 and we're going to square it. If we square one side, we need to square the other. The square root and the square going to cancel. So X + 3 equal 36, X equal 33. We can always check these. Does sqrt 33 + 3 really equals 6? Sqrt 36 = 6. Yes it does check the next one we're going to square to get rid of the square root, and if we square one side, we're going to square the other square root and square cancel. So five Z - 2 is going to equal 9. Add the two across 5Z equal 11 Z equal 11 fifths. If we do our check, sqrt 5 * 11 fifths -2 equal 3, the five in the 5th part is going to cancel. So 11 - 2, which is 9 sqrt - 3 does equal 3. Here we have to get the square root by itself first, so we're going to start by adding seven to each side. If I have -4 + 7, that's going to give me 3 square one side to get rid of the square root square. The other X - 2 equal 9 X equal 11. When we check sqrt 11 - 2 - 7 equal -411 - 2 is 9. Sqrt 9 is three 3 - 7 is -4 so it checks this next one. We need to get the radical by itself on one side, so we're going to subtract 6 from each side to get rid of the square root. We're going to square, so we get y + 4 equaling 1Y equal -3. The check -3 + 4 + 6 = 7 sqrt 1 + 6 one plus 6 is indeed 7, the next one we need to get rid of a cube root, so we're going to cube each side. We cube one side, we cube the other. The cube root and the cube cancel, leaving us X - 2 three cube just 27. Add 2 and we get 29. So our check is the cube root of 29 -. 2 equal 3 cube root of 27 equal 3. Three does equal 3. This next one, it's a fourth index, so we're going to take it to the 4th power, and if I do one side to the fourth, we're going to do the other side to the 4th. X + 3 equal 81, subtract 3, so X = 78. Does the 4th 3 to 78 + 3 equal 3? We always check in the original. By the way, 78 + 3 is 81. The 4th root of 81 is 3. The one third power really means the cube root. And how do we get rid of a cube root? We cube it, and if we cube one side, we cube the other. So X + 5 equal 64, subtract a 559. So 59 + 5 to the one third power equal 459 + 5 is 64 to the one third is the cube root of 64, which is indeed 4. This next one 3 sqrt X + 12 equal 9. Start by subtracting the 12 from each side. Then we want to get the radical by itself, so I'm going to divide each side by three. Now how do we get rid of a square root? We square each side, square one side, square the other X equal 1. When we check it does 3 * 1 to the half plus 12 equal 9? Well, one to the half is just one. So 3 * 1 + 12 equal nine, 3 + 12 equal 9. No, this doesn't check. Now we actually, if we really are paying attention, we could have known right here it's going to be a no solution because a square root always equals positive or zero. So once we get the square root by itself, it has to equal a positive number or zero. If it doesn't, we can still do the computation, but the check will never occur. It won't ever be right. So this one's a no solution. Once I had that square root of X equaling -1 if I understood that the square root always had to equal a positive #0 I was done, and it was no solution. If we have more than one radical term, we need to get one of them on one side by itself, and we're going to square it. But if I square one side, we have to square the other. This one gets much more involved. When we square a square root, that portion cancels. But over here on the right hand side, we're going to actually have to foil because we have 2 + sqrt 2 X -5 times itself. So when we foil on the right side, 2 * 2 is 4 + 2 square roots of two X - 5 + 2 square roots of two X - 5 plus sqrt 2 X -5 ^2. So four X - 3 equaling 4 + 4 square roots two X - 5 + 2 X -5. I have to get everything to one side except for the radical. So if I combine my like terms, I'm going to get two X -, 1 plus 4 square roots of two X -, 5. Subtract the 2X from each side. So I'm going to get 4X minus 2X or 2X. Add the -1 to each side, so -2 equal 4 square roots of two X -, 5. Now we could actually at this point simplify. We could simplify by dividing everything through by A2. We get X - 1 equal to sqrt 2 X -5. If I now have the radical by itself, I can have a coefficient in front to help myself from having to have fractions. The left side is going to be X - 1 * X - 1. We're going to have to foil it again. The right side 2 ^2 is 4 sqrt 2 X -5 ^2 is going to be two X - 5. So we get X ^2 -, X -, X + 1 equaling 8 X -20. I'm going to take everything to one side. So X ^2 - 2, X plus one equal 8X minus twenty, X ^2 - 10, X plus 21 = 0 and factor it and X - 3 and X - 7. So we're going to get X - 3 = 0, X -7 = 0, X equals three, X = 7. We absolutely must check these, and we have to check it in the absolute original equation. So the square root of 4 * 3 - 3 does that equal 2 plus the square root 2 * 3 - 5? Well, 4 * 3 is 1212 - 3 is 9 sqrt 9 is 3. Two plus the 2 * 3 is 6 - 5 is 1, so 2 + sqrt 1. Two plus 1/3 does equal 3, so 3 works. What about the seven square root 4 * 7 - 3 equal 2 plus the square root 2 * 7 - 528 - 3 sqrt 25, which is 5 2 + sqrt 14 - 5 that sqrt 9 two plus three. This one also works. So our answer is three and seven here. We need to get one of the radicals one side by itself. It doesn't matter which one. I'm going to subtract the four to the other side. We're going to square root side. When we square the left side, the square and the square root cancel. When we square the right side, we have to foil it. So 2 * 2 is 4/2 square roots of 4 -, X plus another two square roots of 4 -, X plus the square root and the square are going to leave just 4 -, X. So if we combine our like terms, we get 10 -, X equaling 8 -, X + 4 square roots of 4 -, X. Have to get the radical all by itself. So I'm going to move that 8 - X to the other side and we're going to get 2 equaling 4 square roots of four minus XI. Might simplify here just to make life a little easier by dividing both sides by A2, so 1 = 2 square roots of 4 -, X. If I square each side now I'm going to get one equaling 4. The square root and the square are going to cancel, leaving me 4 -, X. If we distribute, we get 1 = 16 - 4 X -15 equal -4 X, so X equal 15 fourths. We have to check it back in the original. So does 4 + sqrt 10 -15 fourths equals 6 plus square root 4 - 15 fourths? Well, we need to get a common denominator of force, so we're going to have 40 fourths -15 fourths equals 6 plus square root, 16 fourths -15 fourths, 40 fourths -15 fourths is going to be 25 fourths, and 16 fourths -15 fourths is 1/4. So 4 + 5 halves equaling 6 + 1/2 five halves is really 2 1/2. So 6 1/2 = 6 1/2. This one does check. So our answers 15 fourths. Or you could think of that as four goes into 15 three times and 3/4 the next one. We need to get the radical one side by itself, doesn't matter which one. I'm going to leave it the square to six X + 7 because I think it's easier if we have positives. If we square one side, we have to square the other, remembering to foil. So six X + 7 equal 1 + 2 square roots of three X + 3 plus three X + 3. Combining our like terms, we get three X + 4 +2 square roots of three X + 3, taking all the X's to one side. So we're going to get 3X take the +4 to the other side plus 3, equaling 2 square roots, three X + 3. Now we're going to square one side, so we've got to square the other, and we do that to get rid of the square roots have to foil on the left side. So nine X ^2 + 9 X plus nine X + 9 equal 4 * 3 X +39 X squared plus 18X plus 9 is going to equal 12X plus 12-9 X squared. Getting everything to one side plus six X -, 3 = 0, we could divide everything through by a three. Now to keep our equation as simplified as possible. When we factor this, we get 3X and X and a + 1 and a - 1, so X is 1/3 or X equal -1 by setting each of those parentheses equal to 0. Then we have to check it. So sqrt 6 * 1/3 + 7 - sqrt 3 * 1/3 + 3. Does that equal 1? Well, 6 * 1/3 is 2 and 2 + 7 is 9 and sqrt 9 is three. 3 * 1/3 is one, 1 + 3 is 4 sqrt 4 is 2, and 3 - 2 is indeed one. How about -1 so square root 6 * -1 + 7 minus square root, 3 * -1 + 3 equal 1, so -6 + 7 is 1 - -3 + 3 is zero, 1 - 0 is 1. So both of those work OK on this last one. If G of X = sqrt X + sqrt X - 5 and G of X = 5, then those two things have to equal each other by the transitive property. I need to get one of those radicals by itself on one side, so I'm going to subtract the square root of X over. In order to get rid of a square root, I'm going to square it. But if I square one side, we have to square the other. The square root and the square cancel, leaving us X -, 5. The right side turns into 25 -, 10 square roots of X plus XI. Subtract the X from each side and I'm going to subtract the 25 from each side. So I'm going to get -30 equal -10 square root of XI. Might simplify at this point and divide each side by a -10 so 3 equals square root of ** equal 9. When I square each side to check it, does sqrt 9 + sqrt 9 - 5 really equal 5? Well sqrt 9 is three 9 - 5 is four 3 + 2 equal 5 so it does check. They will not always check just because the ones we have done right here as examples. Do I need to caution you that they don't always check? Thank you and have a wonderful day.