Question #9c18f

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Answer 1 · 2018-01-04T20:22:38

Please see below.

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

Let's let $z = x^{2}$ $z=x^2$

$z^{2} + 3 z - 4 = 0$ $z^2+3z-4=0$

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

$z + 4 = 0$ $z+4=0$ , $z = - 4$ $z=-4$

$z - 1 = 0$ $z-1=0$ , $z = 1$ $z=1$

$x^{2} = - 4$ $x^2=-4$

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

$x = \pm \sqrt{- 4} = \pm \sqrt{(- 1) {(2)}^{2}} = \pm 2 \sqrt{- 1} = \pm 2 i$ $x=+-sqrt(-4)=+-sqrt((-1)(2)^2)=+-2sqrt(-1)=+-2i$

$x^{2} = 1$ $x^2=1$

$x = \pm 1$ $x=+-1$