What is the method to show exactly one positive root of a Cubic equation? I have $a x^3 + b x^2 + c x + d =0$ and $a>0 , d<0$
$a, b, c, d$ is the function of all parameters.   I'm looking for an analytic solution of this cubic equation. How to prove that is cubic equation has exactly one positive root?

The question can be answered by drawing lots of diagrams of bendy, and not-so-bendy, cubic graphs and classifying them according to the signs ($0$, $+$, or $-$) of $P(0)$, $\alpha_i\;(i=1,2)$, and $P(\alpha_i)$, where the $\alpha_i$ are the real roots of the quadratic equation $P(x)=0$. Here the given condition $a>0$ is assumed throughout.
If $P(0)=0$, then $d=0$, and the analysis reduces to that of the simpler case of the quadratic equation $P'(x)=0$.
If $P(0)>0$, namely if $d>0$, then there is just one (repeated) positive root of $P(x)=0$ only in the special case when $P(\alpha_i)=0$ with $\alpha_i>0$ for some $i$.
If $P(0)<0$ (i.e. $d<0$), then there is just one root if either $P'(x)=0$ has no real roots, or if $\alpha_1$ and $\alpha_2$ are of different sign, or if they are both positive with $P(\alpha_1)$ and $P(\alpha_2)$ of the same sign.
