Question #9c18f

1 Answer
Jan 4, 2018

Please see below.

Explanation:

.

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