Question: [![This is the problem](https://i.stack.imgur.com/UaVxr.png)](https://i.stack.imgur.com/UaVxr.png)

I'm not sure how they found the B matrix for this problem. I have no idea what they did since I keep getting 2,0,0 for the first column and that's already wrong.

Answer: 
Using the fact that $\sqrt{1+x}=1+\frac x{1+\sqrt{1+x}}$, we get
$$
\begin{align}
\sqrt{1+x}
&=1+\cfrac x{2+\cfrac x{2+\cfrac x{2+\cfrac x{2+\cdots}}}}
\end{align}
$$
Which gives the approximants
$$
\begin{align}
\color{#C00000}{1+\frac12x}&=1+\frac12x\\
\color{#00A000}{\frac{1+\frac34x}{1+\frac14x}}&=1+\frac12x-\frac18x^2+\frac1{32}x^3-\frac1{128}x^4+O\left(x^5\right)\\
\color{#5555FF}{\frac{1+x+\frac18x^2}{1+\frac12x}}&=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac1{32}x^4+O\left(x^5\right)\\
\color{#C0A000}{\frac{1+\frac54x+\frac5{16}x^2}{1+\frac34x+\frac1{16}x^2}}&=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+O\left(x^5\right)\\
\sqrt{1+x}&=1+\frac12x-\frac18x^2+\frac1{16}x^3-\frac5{128}x^4+O\left(x^5\right)
\end{align}
$$
[![enter image description here](https://i.stack.imgur.com/EC0A0.png)](https://i.stack.imgur.com/EC0A0.png)
Since apparently no condition is made that the order of approximation should be maximal for the degree, you do not need to employ Pade approximants. So you can just simply set
$$
1+x=\frac{1+y}{1-y}\iff y=\frac{x}{2+x}
$$
and use Taylor polynomials of equal degree for $\sqrt{1\pm y}$.

---

In first order, you would get $\sqrt{1\pm y}\approx 1\pm\frac12y$ and
$$
\frac{2+y}{2-y}=\frac{4+2x+x}{4+2x-x}=\frac{4+3x}{4+x}.
$$

---

Second order has $\sqrt{1\pm y}\approx 1\pm\frac12y-\frac18y^2$ which leads to
$$
\frac{8+4y-y^2}{8-4y-y^2}=\frac{32+32x+8x^2+(8x+4x^2)-x^2}{32+32x+8x^2-(8x+4x^2)-x^2}=\frac{32+40x+11x^2}{32+24x+3x^2}
$$