How do you solve the quadratic $x^{2} + 4 = 0$ $x^2+4=0$ using any method?

Question

1848 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-26T18:43:42

$x = - 2 i$ $x=-2i$ or $x = 2 i$ $x=2i$

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

$\Leftrightarrow x^{2} - (- 4) = 0$ $hArrx^2-(-4)=0$ or

as $i^{2} = - 1$ $i^2=-1$ and hence

$x^{2} - {(2 i)}^{2} = 0$ $x^2-(2i)^2=0$ or

$(x + 2 i) (x - 2 i) = 0$ $(x+2i)(x-2i)=0$ or

$x + 2 i = 0$ $x+2i=0$ or $x - 2 i = 0$ $x-2i=0$

i.e. $x = - 2 i$ $x=-2i$ or $x = 2 i$ $x=2i$

How do you solve the quadratic x^2+4=0x2+4=0x^2+4=0 using any method?