Radicals and Absolute Values - Problem 2 3,313 views
Simplifying a higher degree square root with absolute values. The first problem we have is a 4th root and we are trying to simplify this down. So I’m going to do this all in one big step and walk you through what I’m thinking as I go through it.
So let’s go start with the 4th root of 32 so I know that 32 is a multiple of 2 so I just want to figure out how many 2’s go into it. So we have 2, 4, 8, 16 and then 32 makes 5. So we have 5 twos in 32, we need 4 in order to just take it out to the 4th root. So we can take out 1, 2 and we are left out with 1,2 left behind okay I’m going to make sure I throw my little 4 up there as well.
So moving down the road, 4th root of x to the 5th. X to the 5th is very similar to this which is with 2 the 5th, so we can take out a single x and we are left with a single x left behind. This one now we don’t know if x is positive or negative. So what we have to do is put around our absolute value making sure that this x is positive. We have an even route everything that comes out of it has to be positive.
4th root of y to the 10th. Y² to the 4th is y to the 8th which is close to y of 10ths so what we actually put out here is a y² and we are left with a y² on the inside. We don’t need an absolute value in this case because y² is always going to be positive, so no absolute value needed
Lastly is z to the 15th. Z³ to the 4th is z to the 12th with 3 left over, so we know that this is z³ leaving us with another z³. Z³ has to be positive, it's coming out of an even route in order to make sure it’s positive we have to throw around our absolute values.
So in this sense since we need absolute values around the x and the z because they are both odd powers coming out of an even route the y² is fine because it’s always going to be positive.
So whenever you are dealing with even routes no matter what degree it is 2, 4, 6, 8 so on and so forth, you're always going to have to think about what needs an absolute value when you take it up.