Question #9c18f

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Answer 1 · 2018-01-04T20:22:38

Please see below.

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

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:

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

$x^2=1$

$x=+-1$