Consider a degree four polynomial $$ f = a_4x^4 + a_3x^3 + a_2x^2 + a_1x+ a_0 \in \mathbb{R}[x] $$ with real coefficients. The discriminant $\Delta_f$ of $f$ is a homogeneous polynomials of degree six in the $a_i$.
If $\Delta_f > 0$ then either $f$ has four real roots or $f$ does not have any real root at all.
What conditions does one have to add to $\Delta_f > 0$ in order to characterize the degree four polynomials with no real roots?
Thanks a lot.
The answer is contained in this wonderful paper:
E. L. Rees. Graphical Discussion of the Roots of a Quartic Equation. The American Mathematical Monthly, Vol. 29, No. 2 (Feb., 1922), pp. 51-55
In which there is a full elaboration of the possibilities for the nature of the roots of a quartic polynomial.
The paper states that given a quartic in reduced (or sometimes called "depressed") form $x^4 + qx^2 + rx + s$, if the discriminant $\Delta$ is greater than zero, then the 4 roots are distinct, and either all real or all imaginary (which is what you claim in your question).
Additionally, if $q < 0, s > \frac{q^2}{4}$, then all the roots are imaginary; if $q < 0, s < \frac{q^2}{4}$, then all the roots are real; if $q \geq 0$ then all the roots are imaginary. So, you need to add the condition $q \geq 0 \lor (q < 0 \land s > \frac{q^2}{4})$ to make the quartic have no real roots (when $\Delta > 0$).
You can convert this condition back to deal with a "full" quartic, i.e., $f = \sum_{i=0}^4 a_i x^i$, by looking up the procedure of converting a quartic to its depressed/reduced form. There are algebraic expressions relating the coefficients of both forms. Hence you will get such a condition in terms of your $a_0, a_1, a_2, a_3, a_4$.
Edit: to be precise, "imaginary" above means non-real, i.e. a complex number $a+bi$ with $b \neq 0$.
