How do you simplify #sqrt(5/4)#?

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Answer 1 · 2016-02-13T17:52:48

#sqrt(5)/2#

Write as :#" " (sqrt(5))/(sqrt(4)) " "=" "(sqrt(5))/(sqrt(2^2)) " "=" " 1/2xxsqrt(5)#

Or #" "sqrt(5)/2#

Answer 2 · 2016-02-13T18:32:44

#sqrt5/2#

#sqrt5/(4)#

since #sqrt(a/b)# = #sqrta/sqrtb#,

we can evaluate first the numerator, the upper part of a fraction

#sqrt5#, as it is not a "perfect square", perfect squares are like
#2 * 2=4#
#3*3=9 #
#4*4=16#
and so on...

since #sqrt5# is not a perfect square, it will still remain not simplified.

then we proceed to the denominator, the lower part of a fraction.
#sqrt4#

since #4# is a perfect square, #2*2=4#

we can get #sqrt4 = 2#,

plugging all the answers, we get

#= sqrt5/2#

:)