How do you simplify #(1/4)^(1/2)#?

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Answer 1 · 2017-01-29T10:35:49

#+-1/2#

#color(blue)("Demonstration of what this means using numbers")#

#5^(2/3) = root(3)(5^2)#

#5^(9/29)=root(29)(5^9)#
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#color(blue)("Answering the question")#

Given:# (1/4)^(1/2)#

This is the same as:#" "root(2)((1/4)^1)" " ->" "sqrt((1/4)^1)#

Note that anything raised to the power of 1 is itself so #(1/4)^1=1/4#
so now we have: #sqrt(1/4)#

This is the same as:#" "(sqrt(1))/(sqrt(4))=+-1/2#

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#color(blue)("Footnote")#

The answer has to be positive or negative.

#(-1/2)xx(-1/2)=+1/4#

#(+1/2)xx(+1/2)=+1/4#