How do you solve #(x + 2) ^ { 2} = - 27#?

1 Answer
Binayaka C.
Sep 16, 2017

Solution : #x=-2 + sqrt (27) i , x=-2 - sqrt (27) i #

Explanation:

#(x+2)^2=-27 or (x+2)= +- sqrt (-27) # or

# x+2= +- sqrt (27) i or x=-2 +- sqrt (27) i , [i^2=-1] #

# :. x=-2 + sqrt (27) i , x=-2 - sqrt (27) i #

Solution :#x=-2 + sqrt (27) i , x=-2 - sqrt (27) i # [Ans]

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