The quartic equation ax^4+bx^2+c=0 can be taken as a quadratic equation in terms of x^2, of which the roots can be determined by the root formula

x^2= \dfrac{-b\pm\sqrt{b^2-4ac} }{2a}

In order for x^2 to have real roots, the discriminant must be great than 0, then

In order for x to have 4 distinct real roots, x^2 must be a positive real number, then we get

\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}>0

Then we get

b< 0 \;\text{and}\; ac >0

So choice D is the correct.

PS: it's also possible for the equation has all real roots when b^2-4ac=0 and b<0 (a>0), then the equation has 2 pair of equal real roots (not distinct 4 real roots).