Write as an imaginary number. Answer is j/12?

#sqrt(-4/16#

1 Answer
Dec 15, 2016

#sqrt(-4/16)=color(magenta)(i/2)#

Explanation:

#sqrt(-4/16)#
#color(white)("XXX")=sqrt(-1) * sqrt(4/16)#

#color(white)("XXX")=sqrt(-1) * sqrt(1/4)#

#color(white)("XXX")= sqrt(-1) * sqrt(1)/sqrt(4)#

#color(white)("XXX")= i * 1/2 or 1/2 i or i/2#

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I replaced your #j# with an #i# since from what I have observed here, #i# is the more common symbol used here for #sqrt(-1)# (although I have seen #j# used elsewhere).

I think the #1# in your suggested answer #j/12# was just a typo.