20 Jun ##sqrt(125x^2)=5sqrt(5)x## A square root can also be seen as
##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##
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