How do you solve $x^{2} + 4 = 0$ $x^2 + 4 = 0$ by completing the square?

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Answer 1 · 2017-03-31T01:52:44

$x = \pm 2 i$ $x=+-2i$

$x^{2} + 4 = 0 \to x^{2} = - 4$ $x^2+4=0 -> x^2=-4$

Since $x^2 >= 0 forall x in RR$
$x$ in this example must be complex

$x=+-sqrt(-4)$

$x= +-2* sqrt(-1)$

$x = +-2i$

How do you solve x^2 + 4 = 0x2+4=0x^2 + 4 = 0 by completing the square?