How do you solve $x^{4} + x^{2} - 6 = 0$ $x^4+x^2-6=0$ ?

Question

8698 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-31T03:32:32

$x = \pm \sqrt{2}$ $x =+-sqrt(2)$

$(x^{2} + 3) (x^{2} - 2) = 0$ $(x^2 + 3)(x^2 - 2) = 0$

$x^{2} = - 3 and x^{2} = 2$ $x^2 = -3 and x^2 = 2$

$x = \pm \sqrt{- 3} and x = \pm \sqrt{2}$ $x = +-sqrt(-3) and x = +-sqrt(2)$

$x = \pm \sqrt{2}$ $x = +-sqrt(2)$

Hopefully this helps!

How do you solve x^4+x^2-6=0 x4+x2−6=0x^4+x^2-6=0 ?