Why is $a_n(x) \neq 0$ for $a_n(x) = c_1 x a_{n-1}(x) + c_2 x a_{n-2}(x)$ if the discriminant of the characteristic polynomial $\Delta_{\lambda} > 0$?

Question

Lets define the sequence $a_0 = 1$, $a_1 = c_1x$ and $a_n = c_1 x a_{n-1} + c_2 x a_{n-2}$ with $c_{1,2} \in \mathbb{N}$ and $x \in \mathbb{R}$.

Then the characteristic polynomial is:

$\lambda^2 - c_1 x \lambda - c_2 x = 0$

and the discriminant $\Delta_{\lambda}$ of this characteristic polynomial is:

$\Delta_{\lambda} = x \left( c_1^2 x + 4 c_2 \right)$

Question: Why is there no solution for $x$ for $c_1 x a_{n-1}(x) + c_2 x a_{n-2}(x) = 0$ if $\Delta_{\lambda} > 0$ ?

My steps so far:

The roots of the characteristic polynomial are:

$\lambda_{1,2} = \frac{c_1 x \pm \sqrt{\Delta_{\lambda}}}{2}$

$\lambda_1 = \frac{c_1 x + \sqrt{\Delta_{\lambda}}}{2}$; $\lambda_2 = \lambda_1 - \sqrt{\Delta_{\lambda}}$

If $a_n = A_1 \lambda_1^n + A_2 \lambda_2^n$ is a solution, then the following needs to be true:

$a_0 = A_1 \lambda_1^0 + A_2 \lambda_2^0 = A_1 + A_2 = 1$

$a_1 = A_1 \lambda_1^1 + A_2 \lambda_2^1 = A_1 \lambda_1 + A_2 \lambda_2 = c_1 x$

Solving for the coefficients $A_1$ and $A_2$, I get:

$A_1 = \frac{1}{2} + \frac{c_1 x}{2 \sqrt{\Delta_{\lambda}}}$

$A_2 = \frac{1}{2} - \frac{c_1 x}{2 \sqrt{\Delta_{\lambda}}}$

So the general solution for $a_n$ is:

$a_n = \left( 1 + \frac{c_1 x}{\sqrt{\Delta_{\lambda}}} \right) \frac{\left(c_1 x + \sqrt{\Delta_{\lambda}} \right)^n}{2^{n+1}} + \left( 1 - \frac{c_1 x}{\sqrt{\Delta_{\lambda}}} \right) \frac{\left(c_1 x - \sqrt{\Delta_{\lambda}} \right)^n}{2^{n+1}}$

Still, the question is: Why is there no solution to the following equality if $\Delta_{\lambda} > 0$ ?

$\left(\frac{c_1 x}{\sqrt{\Delta_{\lambda}}} + 1 \right) \frac{\left(c_1 x + \sqrt{\Delta_{\lambda}} \right)^n}{2^{n+1}} = \left( \frac{c_1 x}{\sqrt{\Delta_{\lambda}}} - 1 \right) \frac{\left(c_1 x - \sqrt{\Delta_{\lambda}} \right)^n}{2^{n+1}}$

$\left(\frac{c_1 x}{\sqrt{\Delta_{\lambda}}} + 1 \right) \left(c_1 x + \sqrt{\Delta_{\lambda}} \right)^n = \left( \frac{c_1 x}{\sqrt{\Delta_{\lambda}}} - 1 \right) \left(c_1 x - \sqrt{\Delta_{\lambda}} \right)^n$

So far, I see that if $\Delta_{\lambda} > 0$:

$\left(\frac{c_1 x}{\sqrt{\Delta_{\lambda}}} + 1 \right) > \left(\frac{c_1 x}{\sqrt{\Delta_{\lambda}}} - 1 \right)$

Answer 1

The real roots of the polynomial $a(x)$ will be found in the negative real axis because all the coefficients are non-negative. So lets investigate for negative $x$ where the following inequalities hold:

$\left| -|x| \frac{c_1}{\sqrt{\Delta_{\lambda}}} + 1 \right| < \left| -|x| \frac{c_1}{\sqrt{\Delta_{\lambda}}} - 1 \right|$

$\left| -|x| c_1 + \sqrt{\Delta_{\lambda}} \right| < \left| -|x| c_1 - \sqrt{\Delta_{\lambda}} \right|$

and so:

$\left| \left( -|x| \frac{c_1}{\sqrt{\Delta_{\lambda}}} + 1 \right) \left( -|x| c_1 + \sqrt{\Delta_{\lambda}} \right)^n \right| < \left| \left( -|x| \frac{c_1}{\sqrt{\Delta_{\lambda}}} - 1 \right) \left( -|x| c_1 - \sqrt{\Delta_{\lambda}} \right)^n \right|$

which means that:

$\left( -|x| \frac{c_1}{\sqrt{\Delta_{\lambda}}} + 1 \right) \left( -|x| c_1 + \sqrt{\Delta_{\lambda}} \right)^n \neq \left( -|x| \frac{c_1}{\sqrt{\Delta_{\lambda}}} - 1 \right) \left( -|x| c_1 - \sqrt{\Delta_{\lambda}} \right)^n$

so that it is proofed that $a_n(x) \neq 0$ for all $x < - \frac{4 c_2}{c_1^2}$.