$\frac{1}{k^{2}}r^{4}-r^{2}+\frac{\varrho S}{f}(-\omega)=0$
Solve the characteristic equation for r.
Using Mathematica I get:
In the end, one should be able to show that, if we name the real roots of the characterisitc equation as "q" and the imaginary roots as "p", then:
$q=\sqrt{p^2+k^2}$
Can anyone help me ?
The following assumes that $\varrho, S, f, \omega \gt 0$ (which seems a sensible assumption in the context).
Let $s = r^2$ and $t = \frac{\varrho S}{f}(-\omega) \lt 0$, then the equation reduces to the quadratic:
$$\frac{1}{k^{2}}s^{2}-s+t=0$$
Since $t \lt 0$ the quadratic will always have $2$ real roots of opposite signs, $s_1 \lt 0$ and $s_2 \gt 0$. It follows by Vieta's formulas that: $$s_1 + s_2 = k^2 \tag{1}$$
The roots in $r$ will then be $r_{1,2} = \pm i \sqrt{-s1}\;$ and $r_{3,4} = \pm \sqrt{s_2}\;$. Let $p = \sqrt{-s_1}\;$ be the imaginary part of $r_1$, and $q=\sqrt{s_2}$ the positive real root $r_3$. Substituting $s_1 = -p^2$ and $s_2 = q^2$ into $(1)$:
$$q^2 - p^2 = k^2 \quad \iff \quad q = \sqrt{p^2 + k^2}$$