Question 30217
x^4 + 2x^2 + 4
let u = x^2
then
u^2 + 2u + 4 = 0
when using the quadratic formula, {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}, in order to get real roots, the discriminant, {{{b^2-4*a*c}}}, must be positive.
In our quadratic eqn above,
a = 1, b = 2, c = 4
bē-4ac = 4 - 4*1*4 = 4 - 16 = -12
i.e. the discriminant is negative, hence the eqn has no real solutions - only complex ones.
Since u is complex, then so also is x.

Fact: all polynomials of the nth degree have n solutions.
Fact: when complex solutions (to a polynomial) occur, they occur as conjugate pairs.
A cubic eqn is of the 3rd degree, hence has three solutions. It has either zero complex solutions or two complex solutions. Hence it has at least one real solution.

N.B. a+ib and a-ib are conjugate pairs. (change of sign for the imaginary component)