How do you find all the real and complex roots of #z² + 4z + 8 = 0#?

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Answer 1 · 2016-02-06T02:43:39

#z=-2+-2i#

Complete the square:

#z^2+4z=-8#

Add #4# to both sides since #x^2+4z+4=(z+2)^2#.

#z^2+4z+4=-8+4#

#(z+2)^2=-4#

Take the square root of both sides.

#z+2=+-sqrt(-4)#

#z+2=+-2i#

#z=-2+-2i#

This can also be solved through the quadratic formula:

#z=(-4+-sqrt(16-32))/2=(-4+-4i)/2=-2+-2i#