How do you solve #z^2 +12 = -z#?

1 Answer
May 22, 2018

Answer:

#-1/2±sqrt(47)/2i#

Explanation:

Use the quadratic formula

#z^2+12=-z <=> z^2+z+12=0#

#z=(-1±sqrt(1^2-4(12)(1)))/2=-1/2±sqrt(47)/2i#

Alternatively, you could complete the square,

#z^2+z+12=0#
#z^2+z+1/4 + 47/4 =0#
#(z+1/2)^2=-47/4#
#z+1/2=+-sqrt(47)/2i#
#z=-1/2+-sqrt(47)/2i#

Which yields the same result (the quadratic formula is derived from completing the square)