Simplifying a more complicated square root. So for this particular problem what we’re looking at is a square root which has both a constant term and a bunch of variables in it. And what we could do is split it up so we’re dealing with just the 75 square root of the x, square root of the y, square root of the z. We could split that square root up into 4 different ones. But what I’m going to do is just to tackle them all at once. And what you can do is just by tackling them at once is look at each of those components but without rewriting it as a bunch of different radicals.
First let’s start at 75. What perfect square goes into 75? Hopefully you can see that there’s a 25 in there, so we know that’s 25 times 3. Square root of 25 we can take out and we have a 5 on the outside and how I tend to write this is I’ll just write my square root over here and then sort of fill in the gaps as I go. This 75 can break up into the 5 on the outside. We’re still left with the 3 on the inside.
Next going down the line, x³. What perfect square goes into x³? Well we know that x² has a square root of x and we know that x³ is just one more x. So what that really tells us is we can pull out an x and we are left with a single x inside.
Going down to the y term; square root of y to the 8th. What times what gives us y to the 8th? Y to the 4th. And that’s actually a perfect square because y to the 4th times y to the 4th is y to the 8th so we don’t have any left over terms inside.
Lastly moving down to the z. What perfect square goes into z to the 5th? Z to the 4th which is z²,squared so we can pull out a z² and we are left with a single z on the inside.
We were able to simplify this all in one step. One way you can tell that you’ve actually simplified it completely is if that the root, so this is the square root so there’s a little invisible 2 right out front, if that is larger than every single power in here. So here we have x to the 1st, z to the 1st and the square root. So the square root is actually larger than everything here. If we had z to the third inside, I know I could take out at least one more z because these powers are backwards. This power has to be bigger than the power on the inside.