How do you solve #8x ^ { 2} + 96= 0#?

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Answer 1 · 2017-02-27T06:21:41

#sqrt 12# i

# 8x^2 + 96 = 0#

#8x^2 = -96#

#x^2 = -(96)/8#

#x = sqrt(-(96)/8) #

Since the squareroot of a negative number is imaginary

#x = sqrt (-12)# = #sqrt (-1*12)# = # sqrt(-1)# * #sqrt 12#

#sqrt(-1)# = i

#x = sqrt12 i#