Solve: $x^{4} + 4 = 0$

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Solution

$x^{4} + 4 = 0$
$x^{4} + 4 + 4 x^{2} - 4 x^{2} = 0$
$x^{4} + 4 x^{2} + 4 - 4 x^{2} = 0$
$(x^{2})^{2} + 2.2 x^{2} + 2^{2} - (2 x)^{2} = 0$
$(x^{2} + 2)^{2} - (2 x)^{2} = 0$
$(x^{2} + 2 - 2 x) (x^{2} + 2 + 2 x) = 0$
$x^{2} - 2 x + 2 = 0, x^{2} + 2 x + 2 = 0$
$x = \frac{2 \pm \sqrt{4 - 8}}{2}, x = \frac{- 2 \pm \sqrt{4 - 8}}{2}$
$= \frac{2 \pm 2 i}{2}, \frac{- 2 \pm 2 i}{2}$
$∴ x = 1 + i, 1 - i, - 1 + i, - 1 - i$

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