Question

Simplify radical expression r/169x(P4)yz(P7)
r=the radical signt, p=means power to right of number

Answer 1

square root of 169x (to the power of 4) yz (to the power of 7)

Answer 2

Only z is the 7th power or all of yz is to the 7th?

Answer 3

only z

Answer 4

I was thinkg it was 13x(to the power of 2)z(to the power of 3) square root of yz but was not sure

Answer 5

Well, anything inside of a root can be factored and split into multiple roots. For example, if I have
\[\sqrt{96}\]
I can factor that and make each factor a square root.

\[\sqrt{96}=\sqrt{2}*\sqrt{2}*\sqrt{2}*\sqrt{2}*\sqrt{2}*\sqrt{3}\]

The point behind showing this is if we can factor our square root in a way that a bunch of roots cancel, we can get everything simplified. Now, the letters are the easiest to do. I can get a square root to go away if I have any even power. Because sqrt(x^6) would be x^3, sqrt(y^2) would be y, etc.

For x's, we already have an even power of 4. Becayse of that, I can take it out of the sqrt by just cutting the power in half and making it x^2. As for z^7, we cant reduce all of that, but we can reduce some of it.
\[x^{2}\sqrt{169y}*\sqrt{z^{6}}*\sqrt{z}\]Because I can split z^7 into z^6 times z^1, I can make the z^6 portion int just z^3. The extra z has to stay inside of the sqrt, though, cant take it out. So now we have:
\[x^{2}z^{3}\sqrt{169yz}\]The only thing that can be reduced now is the 169. There is no real easy way to break it down into factors, so I dont mind saying its simply 13. So that is everything we can possibly simplify, all but 1 y and 1 z, leaving us with:

\[13x^{2}z^{3}\sqrt{yz}\]