How do you solve #2z^2-z+15=0# where #z in CC#?
1 Answer
Oct 14, 2017
# z = 1/4+-sqrt(119)/4i #
Explanation:
We have:
# 2z^2-z+15 = 0 #
We can complete the square:
# 2(z^2 - z/2+ 15/2) = 0 #
# :. z^2 - z/2+ 15/2 = 0 #
# :. ( z - 1/4)^2 - (1/4)^2+ 15/2 = 0 #
# :. (z - 1/4)^2 - 1/16+ 15/2 = 0 #
# :. (z - 1/4)^2 - 1/16+ 15/2 = 0 #
# :. (z - 1/4)^2 + 119/16 = 0 #
# :. (z - 1/4)^2 = - 119/16 #
# :. z - 1/4 = +-sqrt(119/16)i #
# :. z - 1/4 = +-sqrt(119)/4i #
# :. z = 1/4+-sqrt(119)/4i #
