How do you simplify #sqrt(125j^12)#?

1 Answer
EZ as pi
Jan 26, 2017

#=5j^6 sqrt5#

Explanation:

It is helpful when finding roots, to write the number under the root as the product of its prime factors.

#sqrt(125j^12) = sqrt(5xx5xx5xx j^12#

#=sqrt(5xx5^2 j^12)#

Find the roots where possible:

#=5j^6 sqrt5#

(For indices- divide the index by the root)

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