How do you simpilify ##sqrt(125x^2)##?

##sqrt(125x^2)=5sqrt(5)x##

A square root can also be seen as a power of ##1//2##, so we can rewrite our expression as

##(125x^2)^(1/2)##

In this case, the power is distributed across the factors, which is the same as saying that the square-root can be separated across the factors as follows

##125^(1/2) * (x^2)^(1/2) implies sqrt(125)*sqrt(x^2)##

In the second term, the powers simply multiply

##125^(1/2) * x^(2*1/2) = 125^(1/2) * x implies sqrt(125) * x##

Which gives us the rather un-surprising result that the square-root of ##x##-squared is simply ##x##. We can look at this as a process of looking for paired factors under the square root and pulling one of these to the outside, i.e. ##sqrt(x^2)=sqrt(ulx*ulx)=x##. Let's use this method on the last factor

##sqrt(125)*x = sqrt(5*ul5*ul5)*x=5sqrt(5)x##

Therefore

##sqrt(125x^2)=5sqrt(5)x##