What can we say about the largest solution of $x^{1+\alpha}-ax^{\alpha}-b=0$ compared with $x^2-ax-b=0$? 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!
 A: 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 $0<\alpha_L<\alpha_U$
$$
a+b<1 \Rightarrow x_1(\alpha_L)<x_1(\alpha_U)\\
a+b=1 \Rightarrow x_1(\alpha_L)=x_1(\alpha_U)\\
a+b>1 \Rightarrow x_1(\alpha_L)>x_1(\alpha_U)\\
$$
