How do you solve the quadratic with complex numbers given #x^4+13x^2+36=0#?

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Answer 1 · 2016-09-28T12:09:59

#x=+-2i# or #+-3i#

#x^4+13x^2+36=0#

Let #z=x^2 -> z^2+13z+36=0#

#(z+4)(z+9)=0#

Hence: #z=-4# or #-9#

#:. x^2= -4# or #-9#

#x = sqrt(-4)# or #sqrt(-9)#

#x=+-2i# or #+-3i#