What can we say about the largest solution of $x^{1+\alpha}-ax^{\alpha}-b=0$ compared with $x^2-ax-b=0$?

Question

I wish to know if there is any comparison between the positive roots (if they exist) of lets say, \begin{equation} x^{1+\alpha}-ax^{\alpha}-b=0 \end{equation} where, $\alpha\geq0$ and $b\geq 0$ now, lets say that the positive root of this equation is, i.e., $x_1$. Now quadratic is given by, \begin{equation} x^{2}-ax-b=0 \end{equation} the positive root of this is lets say, $x_1^{'}$. I want to find the condition on lets say $\alpha,a\text{ and },b$ such that the positive root of first equation is smaller than the positive root of quadratic i.e., $x_1 \leq x_1^{'}$. Is there such a comparison? Thanks for your time and consideration!

Answer 1

Let's start by introducing the root $x_1(\alpha)$ as a function of $\alpha$. Rewriting and implicit differentiation yields:

$$ x_1(\alpha)^{\alpha+1}-ax_1(\alpha)^{\alpha}-b=0\\ \Leftrightarrow x_1(\alpha)-a-bx_1(\alpha)^{-\alpha}=0\\ \Rightarrow x_1'(\alpha)=-\frac{bx_1(\alpha)\ln(x_1(\alpha))}{\alpha b+x_1(\alpha)^{\alpha+1}} $$ Since we have $\alpha\geq0$ and $b\geq0$ the denominator stays positive. Studying the sign of $-x\ln(x)$ we see $$ \cases{x_1'(\alpha)<0 \mbox{ for } x_1(\alpha)>1\\x_1'(\alpha)=0\mbox{ for } x_1(\alpha)=1\\x_1'(\alpha)>0\mbox{ for } x_1(\alpha)<1} $$ So depending on the solution of the quadratic equation we have for $\alpha_L<2<\alpha_U$ $$ x_1(2)<1 \Leftrightarrow a+b<1 \Rightarrow x_1(\alpha_L)<x_1(2)<x_1(\alpha_U)\\ x_1(2)=1 \Leftrightarrow a+b=1 \Rightarrow x_1(\alpha_L)=x_1(2)=x_1(\alpha_U)\\ x_1(2)>1 \Leftrightarrow a+b>1 \Rightarrow x_1(\alpha_L)>x_1(2)>x_1(\alpha_U)\\ $$ The result is not restricted to the quadratic reference point, in fact it holds that for $0<\alpha_L<\alpha_U$ $$ a+b<1 \Rightarrow x_1(\alpha_L)<x_1(\alpha_U)\\ a+b=1 \Rightarrow x_1(\alpha_L)=x_1(\alpha_U)\\ a+b>1 \Rightarrow x_1(\alpha_L)>x_1(\alpha_U)\\ $$