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)
You can work this out without MacLaurin, starting from

$$p^2(x)\approx(1+x)q^2(x).$$

Then with

$$p(x)=(1+ax)^2,q(x)=(1+bx)^2,$$
we have
$$1+2ax+a^2x^2\approx1+(2b+1)x+(b^2+2b)x^2+b^2x^3,$$
and we identify
$$2a=2b+1,\\a^2=b^2+2b$$

which gives the solution

$$\sqrt{1+x}\approx\frac{1+\frac34x}{1+\frac14x}.$$