Newton's method to solve for $f=ax^2-x+1=0$ [closed]

Let $\{x_k\}$ be sequence generated by Newton's Method for solving$ax^2-x+1=0$, where $0<a\leq\frac{1}{4}$. Suppose $x_0<\frac{1}{2a}$.

(a) show $x_k<x_{k+1}<\frac{1-\sqrt{1-4a}}{2a}$

(b) show $x_k$ converges to a root of the above equation and find this root.

$$f'(x)=2ax-1$$

So$$x_{k+1}=x_k-\frac{ax_k^2-x_k+1}{2ax_k-1}=\frac{ax_k^2-1}{2ax_k-1}$$

(a). Prove by induction

Base case: $x_1=\frac{ax_0^2-1}{2ax_0-1},2ax_0-1<0,ax_0^2-1>0$.

How should I continue the induction?

(b). Since $x_k$ is strictly increasing and bounded above, it converges by Monotone convergence theorem. Let $x_k\to x$, by taking limit on both sides of the sequence, $x=\frac{ax^2-1}{2ax-1}\implies ax^2-x+1=0$, but I got the same equation in the question. Are there other ways?

Thanks!