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)

Related questions
See all questions in Simplification of Radical Expressions
Impact of this question
948 views around the world
You can reuse this answer
Creative Commons License