Let $x_{r+1} = g(x_r) = \alpha x_r + \beta x_r^2$, $r=0,1,\ldots$. Find coefficients $\alpha$ and $\beta$ such that the iterative sequence $(x_r)_{r \in \mathbb{N_0}}$ converges to a fixed real number $a > 0$ with quadratic convergence.
To have quadratic convergence: $g(a) = a, g'(a)=0,g''(a)\neq0$. The first two equations give: $$\alpha a + \beta a^2=a \Rightarrow \alpha + \beta a=1 \Rightarrow \beta = \frac{1-\alpha}{a}$$ and $$a + 2\beta a = 0 \Rightarrow \alpha = -2\beta a \Rightarrow \alpha = -2\frac{1-\alpha}{a}a \Rightarrow \alpha = 2.$$
So I found $\alpha$ and $\beta$ depends on $a$ and $\alpha$. Is it okay?
Newton's method converges quadratically to simple roots, the root $x=a$ of $f(x)=\frac1a-\frac1x$ is simple, and its Newton method is $$ x_{n+1}=x_n-\frac{\frac1a-\frac1{x_n}}{\frac1{x_n^2}}=2x_n-\frac{x_n^2}a $$