While I was trying to solve the equation:
$\begin{bmatrix} x_{1} & x_{2}\\ x_{3} & x_{4} \end{bmatrix}^2=\begin{bmatrix} 7 & 10\\ 15 & 22 \end{bmatrix}$
which turns in the four equations:
$x_{1}^2+x_{2}x_{3}=7$
$x_{1}x_{2}+x_{2}x_{4}=10$
$x_{1}x_{3}+x_{3}x_{4}=15$
$x_{2}x_{3}+x_{4}^2=7$
From Wolfram Alpha I found the following four solutions:
First Solution: $x_{1}=-1,x_{2}=-2,x_{3}=-3,x_{4}=-4$
Second Solution: $x_{1}=1,x_{2}=2,x_{3}=3,x_{4}=4$
Third Solution: $x_{1}=-3 \cdot \sqrt{\frac{3}{11}},x_{2}=\frac{-10}{\sqrt{33}},x_{3}=-5 \cdot\sqrt{\frac{3}{11}},x_{4}=-8 \cdot\sqrt{\frac{3}{11}}$
Fourth Solution: $x_{1}=3 \cdot \sqrt{\frac{3}{11}},x_{2}=\frac{10}{\sqrt{33}},x_{3}=5 \cdot\sqrt{\frac{3}{11}},x_{4}=8 \cdot\sqrt{\frac{3}{11}}$
So my question how to find these solutions without wolfram alpha ?
What you're after is a square root of the matrix$$A=\begin{bmatrix}7&10\\15&22\end{bmatrix}.$$This matrix has two eigenvalues: $\frac12\left(29\pm5\sqrt{33}\right)$. It turns out that $\left(-3+5\sqrt{33},6\right)$ is an eigenvector corresponding to the eigenvalue $\frac12\left(29+5\sqrt{33}\right)$ and that $\left(-3-\sqrt{33},6\right)$ is an eigenvector corresponding to the eigenvalue $\frac12\left(29-5\sqrt{33}\right)$. So, if$$P=\begin{bmatrix}-3+\sqrt{33}&-3-\sqrt{33}\\6&6\end{bmatrix},$$then$$P^{-1}.A.P=\begin{bmatrix}\frac12\left(29+5\sqrt{33}\right)&0\\0&\frac12\left(29-5\sqrt{33}\right)\end{bmatrix}.$$Therefore, the square roots of $A$ are the matrices of the form $P.S.P^{-1}$, where $s$ is one of the matrices:
Compute these four matrices and you shall get the solutions that WolphramAlpha gave you. It will help you to do this to use the fact that$$\sqrt{\frac12\left(29\pm5\sqrt{33}\right)}=\frac12\left(\pm5+\sqrt{33}\right).$$