How do you simplify #root4 (40)#?

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Answer 1 · 2017-09-18T16:01:25

#2.514866859#

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#root4(40)=2.514866859#

Answer 2 · 2017-09-18T19:04:44

About the most you can do is:

#root(4)(40) = sqrt(2sqrt(10))#

The prime factorisation of #40# is:

#40 = 2*2*2*5#

Since there are no #4#th powers, it is not possible to "move a factor" outside the radical.

It is possible to move one factor "half way" out, since we do have a square factor.

So we find:

#root(4)(40) = sqrt(sqrt(2^2*10)) = sqrt(sqrt(2^2)*sqrt(10)) = sqrt(2sqrt(10))#