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?
 A: $(x-1)(x-2)(x-3)=x^3-6x^2+11x-6$ has $a>0,d<0$ but it has three positive roots.
Added: to get exactly two positive roots, use $(x-1)^2(x-2)=x^3-4x^2+5x-2.$
A: Let us denote the polynomial by $P(x)=ax^3+bx^2+cx+d$. Then as $a>0$, so $\lim_{x\rightarrow\infty}P(x)=+\infty$. Also $P(0)=d<0$. As any polynomial is a continuous function, its graph must intersect the $x-$axis at  some finite point greater than zero (this is called the intermediate value property of continuous functions which says that, if $f$ is continuous over an interval $[a,b]$, then $f$ takes all values between $f(a)$ and $f(b)$). So the equation has at least one positive solution. But there is no way to deduce from what is given, that it has exactly one solution.
A: The standard method to define
that the cubic polynomial 
\begin{align} 
f(x)&=ax^3+bx^2+cx+d
\end{align}  
has only one real root
is to check that
its discriminant
$\Delta<0$:
\begin{align} 
\Delta &= 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2
,
\end{align}  
Obviously, just the two conditions $a>0$, $d<0$
is not enough to define even if there is only one real root.
A: 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.
