How do you simplify #(4n^(-1/2) xx 4n^(3/2))/(n^(3/2))#?

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Answer 1 · 2016-09-12T07:12:21

#(16 sqrtn)/n#

This question is much easier than it first appears, because there are like factors which can be cancelled.

#(4n^(-1/2) xx 4cancel(n^(3/2)))/cancel(n^(3/2))" "larr# attend to the negative index

=#16/n^(1/2)#

This can also be written as #16/sqrtn#

It is not good practice to have a radical in the denominator, so we apply a process called "rationalise the denominator"

#16/sqrtn xx sqrtn/sqrtn#

=#(16 sqrtn)/n#