What is #sqrt(49n^3)#?

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Answer 1 · 2017-05-25T00:26:07

See a solution process below:

First, use this rule for radicals to simplify the constant:

#sqrt(a * b) = sqrt(a) * sqrt(b)#

#sqrt(49n^3) => sqrt(49 * n^3) = sqrt(49) * sqrt(n^3) =>#

#7sqrt(n^3)#

Use this same rule to simplify the #n# term:

#7sqrt(n^3) => 7sqrt(n^2 * n) => 7(sqrt(n^2) * sqrt(n)) =>#

#7nsqrt(n)#

Answer 2 · 2017-05-25T00:35:20

#7nsqrt(n)#

#sqrt(49n^3)# can be rewritten as follows

#sqrt(49n^3)->sqrt(ul(7*7)*ul(n*n)*n)#

See the underlined parts in the square root? This means we can take out the two #color(red)(7's)# and the two #color(red)(n's)# to get

#7nsqrt(n)#, which is in its most simplified form