EDIT.
However, the square root of a real-valued general symmetric positive definite $2 \times 2$ matrix can be calculated exactly:
$$
\begin{bmatrix} a & b \\ b & c \end{bmatrix}^2 = \begin{bmatrix} A & B \\ B & C \end{bmatrix} \quad \Longrightarrow \\
\begin{bmatrix} a & b \\ b & c \end{bmatrix}\begin{bmatrix} a & b \\ b & c \end{bmatrix} =
\begin{bmatrix} a^2+b^2 & b(a+c) \\ b(a+c) & b^2+c^2 \end{bmatrix} = \begin{bmatrix} A & B \\ B & C \end{bmatrix}
$$
It follows that:
$$
\begin{cases} a^2+b^2 = A \\ b(a+c) = B \\ b^2+c^2 = C \end{cases} \quad \Longrightarrow \quad
\begin{cases} a^2 = A - b^2 \\ b^2(a^2+2ac+c^2) = B^2 \\ c^2 = C - b^2 \end{cases}
$$
Using the determinants gives an additional equation:
$$
(ac-b^2)^2 = AC-B^2 \quad \Longrightarrow \quad ac = \sqrt{AC-B^2}+b^2
$$
Making smart combinations already gives the end-result:
$$
b^2(a^2+2ac+c^2) = B^2 \quad \Longrightarrow \\ b^2([A - b^2]+2[\sqrt{AC-B^2}+b^2]+[C - b^2]) = B^2
\quad \Longrightarrow \\ b^2(A+2\sqrt{AC-B^2}+C) = B^2 \quad \Longrightarrow \\ b = \frac{B}{\sqrt{A+2\sqrt{AC-B^2}+C}}
$$
Together with
$$
a = \sqrt{A-b^2} \quad ; \quad c = \sqrt{C-b^2}
$$
So here comes a much simpler solution. First form the transpose times the original matrix:
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^T\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} =
\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} =
\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} A & B \\ B & C \end{bmatrix} =
\begin{bmatrix} a & b \\ b & c \end{bmatrix}^2
$$
With the above formulas:
$$
b = \frac{B}{\sqrt{A+2\sqrt{AC-B^2}+C}} = \frac{1}{\sqrt{2 + 2\sqrt{1} + 1}} = 1/\sqrt{5} \\
a = \sqrt{A-b^2} = \sqrt{2 - 1/5} = 3/\sqrt{5} \\
c = \sqrt{C-b^2} = \sqrt{1 - 1/5} = 2/\sqrt{5}
$$
So we have, indeed:
$$
\sqrt{\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}^T\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}} =
\begin{bmatrix} 3/\sqrt{5} & 1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
At last, the orthogonal matrix is found with:
$$
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
\begin{bmatrix} 3/\sqrt{5} & 1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}^{-1} =
\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ -1/\sqrt{5} & 3/\sqrt{5} \end{bmatrix} / 1 =
\begin{bmatrix} 2/\sqrt{5} & -1/\sqrt{5} \\ 1/\sqrt{5} & 2/\sqrt{5} \end{bmatrix}
$$
Which leads to the same conclusion:
$$
\begin{bmatrix}
1 & 0 \\
1 & 1
\end{bmatrix} =
\begin{bmatrix}
2/\sqrt{5} & -1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
\begin{bmatrix}
3/\sqrt{5} & 1/\sqrt{5} \\
1/\sqrt{5} & 2/\sqrt{5}
\end{bmatrix}
$$