<s>[INST] Finding real matrices of order $2\times 2$ in matrix equation 
Finding all real matrices $X$ of order $2\times 2$ which satisfy the equation $X^2=\begin{pmatrix}
1 & 2 \\ 
 3 & 7
\end{pmatrix}$

My Try: Let $\displaystyle X=\begin{pmatrix} a & b\\
c & d \end{pmatrix}$. Then $\displaystyle X^2=\begin{pmatrix}a^2+bc & b(a+d)\\
c(a+d) & bc+d^2\end{pmatrix}=\begin{pmatrix}1 & 2\\ 3 & 7\end{pmatrix}$
So $a^2+bc=1\cdots (1)$ and $b(a+d)=2\cdots (2)$
And $c(a+d)=3\cdots (3)$ and $bc+d^2=7\cdots (4)$
But this is very complex method 
Could some help me to solve it some easy way .Thanks
 [/INST] You've already got
$$a^2+bc=1\tag1$$
$$b(a+d)=2\tag2$$
$$c(a+d)=3\tag3$$
$$bc+d^2=7\tag4$$
Now, $3\times (2)-2\times (3)$ gives
$$(3b-2c)(a+d)=0$$
Since $a+d\not=0$, we have
$$3b-2c=0,$$
i.e.
$$c=\frac 32b\tag5$$
Also, $(4)-(1)$ gives
$$(d-a)(d+a)=6$$
Multiplying the both sides by $b$ and using $(2)$ give
$$2(d-a)=6b\implies d=a+3b\tag6$$
So, from $(6)$, we have
$$(2)\implies 2ab=2-3b^2\tag7$$
Also, from $(5)$, we have
$$(1)\implies a^2+\frac 32b^2=1\implies 2a^2=2-3b^2\tag8$$
From $(7)(8)$, we get
$$2ab=2a^2\implies a(a-b)=0\implies a=0\quad\text{or}\quad a=b$$
Therefore, it is necessary that
$$(a,b,c,d)=\left(0,\pm\frac{\sqrt 6}{3},\pm\frac{\sqrt 6}{2},\pm\sqrt 6\right),\left(\pm\frac{\sqrt{10}}{5},\pm\frac{\sqrt{10}}{5},\pm\frac{3}{10}\sqrt{10},\pm\frac 45\sqrt{10}\right)$$
which is sufficient.
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