How do you solve for all solutions, including complex solutions, to #x^4 + 4x^2 = 45#?

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Answer 1 · 2016-05-02T14:57:23

#x=+-sqrt(5)# or #x=+-3i#

OK, let's complete the square to start, adding #4# to both sides to get:

#x^4+4x^2+4 = 49#

That is:

#(x^2+2)^2 = 7^2#

Hence:

#x^2+2 = +-7#

Subtract #2# from both sides to get:

#x^2 = -2+-7#

That is:

#x^2 = 5# or #x^2 = -9#

If #x^2 = 5# then #x = +-sqrt(5)#

If #x^2 = -9# then #x = +-3i#