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I have a univariate polynomial: $$f(r) = r^4 - \frac{1}{a^2} r^2 - \frac{2 \mu}{a^2} r+ \frac{\mu^2 k^2}{a^2}$$ which for $f(r) = 0$ has the following roots: $$r_1 = -\frac{1}{2} \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}-\frac{1}{2} \sqrt{\frac{4}{3 a^2}-\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}-\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}-\frac{4 u}{a^2 \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}}}$$ The second root being, $$r_2=\frac{1}{2} \sqrt{\frac{4}{3 a^2}-\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}-\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}-\frac{4 u}{a^2 \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}}}-\frac{1}{2} \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}$$

Third root,

$$r_3=\frac{1}{2} \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}-\frac{1}{2} \sqrt{\frac{4}{3 a^2}-\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}-\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}+\frac{4 u}{a^2 \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}}}$$ and the fourth root is, $$r_4=\frac{1}{2} \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}+\frac{1}{2} \sqrt{\frac{4}{3 a^2}-\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}-\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}+\frac{4 u}{a^2 \sqrt{\frac{2}{3 a^2}+\frac{\sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}{3 \sqrt[3]{2} a^2}+\frac{\sqrt[3]{2} \left(12 a^4 u^2+12 a^2 k^2+1\right)}{3 a^2 \sqrt[3]{72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2+\sqrt{\left(72 a^2 \left(a^2 u^2+k^2\right)+108 a^2 u^2-2\right)^2-4 \left(12 a^4 u^2+12 a^2 k^2+1\right)^3}-2}}}}}$$

Which are obtained by the usual manner, now my question is regarding the process through which we can check the nature of the roots i.e. whether the given roots is positive, negative or Real, imaginary etc.?

I know that just by looking at the polynomial I can say that there are at most two positive roots available using the Descartes rule of sign but to determine which of the following are the positive roots is a question that still remains.

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It's easier to write the polynomial as $a^2r^4-r^2-2\mu r+\mu^2k^2$. One can use Wikipedia's article on quartic functions. If I'm not mistaken, the discriminant of your polynomial is: $$\Delta=16a^2\mu^2\left(16a^4\mu^4k^6-8a^2\mu^2k^4-36a^2\mu^2k^2-27a^2\mu^2+k^2+1\right).$$ You'll also need to consider the following polynomials (as defined in Wikipedia): $$P=-8a^2,\quad R=-16a^4\mu ,\quad \Delta_0=1+12a^2\mu^2 k^2,\quad D=-16a^4(4a^2+1).$$ For example, if $\mu=0\implies \Delta=0$. Since $P<0$, $D<0$ and $\Delta_0=1\neq0$, "there are a real double root and two real simple roots". You can do other cases following Wikipedia's rules.

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  • $\begingroup$ I see, appreciate the article, helps me get a new insight on the nature of the roots of the equation. But, is there a way, I can study the given 4 roots by making some manipulations until I obtain a 'recognizable' expression? As in an expression which can be gauged to be positive or negative in in itself (by taking some fixed values of the constants here) $\endgroup$
    – mnuizhre
    Feb 8 at 14:16
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    $\begingroup$ I don't know. In the Wikipedia's article there's a method called the Ferrari's solution for solving depressed quartic equations as yours. But it seems really difficult to apply to your problem due to the unkown $a,\mu$ and $k$. $\endgroup$
    – moqui
    Feb 8 at 21:40

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