On polynomial equation of fourth order depending on two parameters and bound on a maximal root

Question

I would like to apologize in advance for a too technical question. Let us consider the following fourth order polynomial equation in $x$: \begin{eqnarray} F(x) \equiv (2 a p +2)x^4+ (6a(1-a)p^2+(6-12a)p-6)x^3 \nonumber \\ + p(2(a-2)(a-1)a p^2 + 3(5a^2-9a+2)p +12a-18)x^2 \nonumber \\ - p^2 ((a-2)(4a^2 - 7a +1)p +9a^2-25a+18)x \nonumber \\ + (a-2)(a-1)(2a-3)p^3 =0, \quad (1) \end{eqnarray} where real parameters $a$ and $p$ obey inequalities: \begin{equation} 0 < a < 2, \quad p > 0. \quad (2) \end{equation}

It is necessary to prove that for any $a$ and $p$ obeying (2), the equation (1) has one and only one solution $x = x_{*} = x_{*}(a,p)$ which satisfies the relation \begin{equation} x_{*} > 1; \qquad (3) \end{equation} moreover, for all $a$ and $p$ (see (2)) this solution obeys the following bound \begin{equation} x_{*} > x_0 \equiv \frac{1}{2} [(a - 1)p + 3/2 + \sqrt{d}], \quad (4) \end{equation}
where $d = (1 - a)^2 p^2 + (3 - a)p + 9/4$.

Comment. The existence of the root satisfying (3) can be readily proved. Indeed, due to (2) we have \begin{equation} F(1) = 2(p+1)^2((a - 2)p-2) < 0. \qquad (5) \end{equation}
Since $(2ap+2) > 0$ we get
\begin{equation} F(x) \to + \infty , \qquad (6) \end{equation} as $x \to + \infty$. This implies the existence of $x_{+} > 1$ such that \begin{equation} F(x_{+}) > 1. \qquad (7) \end{equation} Applying the intermediate value theorem to the continuous function $F(x)$ on closed interval $[1, x_{+}]$ with boundary values given by (5) and (7) we find that there exists $x_{*}$ belonging to interval $(1, x_{+} )$ such that
\begin{equation} F(x_{*} ) = 0. \qquad (8) \end{equation} So, the existence of the solution to equation (1) satisfying $x_{*} > 1$ is proved. Analogous consideration gives us an existence of another root $x_{**}$ ( $F(x_{**} ) = 0$ ) which satisfies $x_{**} < 1$. P.S. This task appears in solving certain physical problem and here all inequalities have certain physical meanings.

Answer 1

The existence of the root $x_{*} > 1$ was proved above.

Now we prove the uniquenes of the root $x_{*} > 1$. Let us suppose that there exists another root $x_{2,*} > 1$. Without loss of generality we put $x_{2,*} > x_{*}$. An elementary graphical analysis leads us to three possibilities for our quartic polynomial $F(x)$: $$ i) \quad F'(x_{*}) = 0, \quad F'(x_{2,*}) > 0, $$ $$ii) \quad F'(x_{*}) > 0, \quad F'(x_{2,*}) = 0,$$ $$iii) \quad F'(x_{*}) > 0, \quad F'(x_{2,*}) < 0, \quad F'(x_{3,*}) > 0, \quad $$
where $x_{3,*}$ is the third root: $F(x_{3,*}) = 0$,
obeying (without loss of generality) $x_{3,*} > x_{2,*} $.

In any case by Rolle's theorem we get that there exist two stationary points of the function $F(x)$ : $ 1 < x_{1} < x_{2}$, which obey the cubic equation $$F'(x_{i} ) = 0,$$
$i = 1,2$. But according to the Lemma proved in my answer to my other question
$$https://mathoverflow.net/questions/474831$$ $$/a-real-root-of-a-cubic-equation-for-a-stationary-point$$ this is imposible.

Thus, we come to a contradiction which proves the uniqueness of the root $x_{*} > 1$ of our quartic eqution.

Now let us prove that $x_{*} > x_0$, where $x_0$ is defined in equation (4) of the question.

Let us consider the auxiliary function (this hint just follows from the physical entity of the question) $$v(x) = \frac{H^a x^2 \left[ap(x-1) + p+x\right]}{2 \Delta_0},$$ where $H = 1 + p/x$, and quadratic polynomial $$\Delta_0 = x^2 + ((1 - a)p - 3/2)x + (2a-3) p/2$$ has a biggest real root $x_0 > 1$. The function $v(x)$ is well defined on interval $(x_0, + \infty)$ and tends to $+ \infty$ in two limits: when $x \to x_0$ or $x \to + \infty$. Thus, it has global minimum at some point $x_{0,*}$. We get that $v'(x_{0,*}) = 0$.

But due to identity (which could be readily verified) $$ \frac{dv}{dx} = \frac{H^{a-1}}{4 \Delta_0^2} F,$$ we get that $F(x_{0,*}) = 0$.

Thus, there exists a root $x_{0,*}$ of our quartic equation which obeys $1< x_0 < x_{0,*}$. But due to uniqueness of a root obeying $x > 1$ we get $x_{0,*} = x_{*}$. Thus, $1< x_0 < x_{*}$. Hence our proposition is totally proved.