Question #9c18f

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Answer 1 · 2018-01-04T20:22:38

Please see below.

.

#x^4+3x^2-4=0#

Let's let #z=x^2#

#z^2+3z-4=0#

#(z+4)(z-1)=0#

#z+4=0#, #z=-4#

#z-1=0#, #z=1#

#x^2=-4#

This has no solutions in the real numbers domain. But it has solutions in the complex (imaginary) numbers domain:

#x=+-sqrt(-4)=+-sqrt((-1)(2)^2)=+-2sqrt(-1)=+-2i#

#x^2=1#

#x=+-1#