How do you write #5x^(5/2)# as a radical form?

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Answer 1 · 2016-04-23T15:36:44

#5sqrt(x^5)# or #5x^2sqrt(x)#

If #x >= 0# then:

#5x^(5/2) = 5x^(5*1/2) = 5(x^5)^(1/2) = 5sqrt(x^5)#

If you prefer:

#5x^(5/2) = 5x^(2+1/2) = 5x^2x^(1/2) = 5x^2sqrt(x)#

#color(white)()#
Footnote

Fractional exponents like this tend to work well with non-negative numbers, but can misbehave horribly for negative or Complex numbers.

For example:

#(-1)^(3/2) = -i != i = ((-1)^3)^(1/2)#

or even:

#((-1)^(3/2))^(2/3) = (-i)^(2/3) = 1/2-sqrt(3)/2i != -1 = (-1)^(3/2 * 2/3)#