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For $a>0$, $b>0$, consider the cubic equation

$$2ab x^3 + (a+b) x^2 - (a+b)x - 2 = 0,$$

which by Descartes' rule of sign, has exactly one positive real root $x_{+}>0$.

Can we write down $x_{+}$ as a "simple" function of parameters $a, b$? Obviously, simple is rather ill-defined, but I need something from where I can analyze the qualitative dependence of $x_{+}$ on parameters $a,b$. I tried Sage but the resulting expression was a mess.

If there are other indirect ways to infer the dependence of $x_{+}$ on $a,b$, than writing out the expression, I would be interested in learning that too.

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    You could depress the cubic and then use Cardano's formula. That would give you an explicit (but somewhat ugly) expression for $x_+$. – 2017-01-20

2 Answers 2

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We have $a>0,\,b>0$ and \begin{eqnarray} P(x)&=&2ab x^3 + (a+b) x^2 - (a+b)x - 2\\ P^\prime(x)&=&6abx^2+2(a+b)x-(a+b)\\ P^{\prime\prime}(x)&=&12abx+2(a+b) \end{eqnarray}

So we know that at the point $(0,-2)$ the graph is decreasing and concave up so the root will be greater than $\dfrac{\sqrt{(a+b)(a+b+6ab)}-(a+b)}{6ab}$ which is the $x$-coordinate of the minimum of the graph.

Given the limiting case of $b=0$ the cube term vanishes giving the parabola

\begin{equation} y=ax^2-ax+2 \end{equation}

which has positive zero $\dfrac{a+\sqrt{a(a+8)}}{2a}$. So we have bounds on the positive root

\begin{equation} \dfrac{\sqrt{(a+b)(a+b+6ab)}-(a+b)}{6ab}

Here is a corrected Desmos animation

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    This is great. And +1 for the animation. – 2017-01-20
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    @AbhishekHalder Thanks – 2017-01-20
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Not extremely simple, no. Well, maybe something like this might be useful:

Let's suppose $b$ is much larger than $a$. Then just looking at the terms with $b$ in the equation, we would have $x_+ = \frac{\sqrt{1+8a}-1}{4a}$. And in fact we can expand $x_+$ in a series in negative powers of $b$:

$$ x_+ = \frac{\sqrt{1+8a}-1}{4a} -{\frac {2\,\sqrt {1+8\,a}a+\sqrt {1+8\,a}+10\,a-1}{-2+2 \sqrt {1+8 \,a}-16\,a}} b^{-1} + \ldots $$