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
 A: Here is what I tried. Let $A=\left(\begin{array}{cc}1&2\\3&7\\\end{array}\right)$. Then $\chi_A(x)=x^2-8x+1$ which has two roots $\lambda_1=4+\sqrt{15}$ and $\lambda_2=4-\sqrt{15}$. So there exists $P$ invertible such that $$P^{-1}AP=  \left(\begin{array}{cc}\lambda_1&0\\0&\lambda_2\\\end{array}\right)=:D.$$
Now for $X$ a given matrix, $X^2=A\iff Y^2=D$ where $Y=P^{-1}XP$. Hence we just have to find $Y$ satisfying $Y^2=D$, and the only possible solutions to this are the four matrix of the form $$\left(\begin{array}{cc}\pm\sqrt{\lambda_1}&0\\0&\pm\sqrt{\lambda_2}\\\end{array}\right).$$
To see this, you can see the link given in the comments. One way to prove this would be as follows. Let $\mu_1$ and $\mu_2$ be two complex eigenvalues of $Y$ (maybe equal). Then $$Y\sim\left(\begin{array}{cc}\mu_1&\alpha\\ 0&\mu_2\\ \end{array}\right)$$
for some $\alpha$ so  $$Y^2\sim\left(\begin{array}{cc}\mu_1^2&\alpha^\prime\\ 0&\mu_2^2\\ \end{array}\right).$$
From here we deduce that $\mu_1\not=\mu_2$, and for example $\mu_i^2=\lambda_i$. Futhermore, $\mu_1$ can't be a complex number (not real I mean) because we would have $\mu_2 = \bar{\mu_1}$ and $(\mu_1\mu_2)^2=det(D)=1=\vert \mu_1\vert^2$ which is not the case ($\vert \lambda_i\vert\not=1$). This shows that $Y$ has two real eigenvalues, so is diagonalizable. Because $D$ and $Y$ commute they can be simultaneously diagonalizable, which means that $Y$ is in fact diagonal, and so of the form $\left(\begin{array}{cc}\pm\sqrt{\lambda_1}&0\\0&\pm\sqrt{\lambda_2}\\\end{array}\right).$
Finally the solutions of the problem are the matrix of the form 
$$P\left(\begin{array}{cc}\pm\sqrt{\lambda_1}&0\\0&\pm\sqrt{\lambda_2}\\\end{array}\right)P^{-1}.$$
You just have to find $P$ (you have $P= (X_1~X_2)$ where $PX_i=\lambda_i X_i,~i=1,2$, so there are just two linear systems to solve).
I hope I haven't done any mistake, sorry for my bad English I'm not so good at it!
A: 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.
A: The characteristic equation of $X$ is, by definition, $\det (X - t I_2) = 0$. The unknown is $t$, and $I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$. Since $X$ is square of dimension $2$, this equation will be of degree $2$; more specifically, it will be
$$t^2 - (\operatorname{tr} X) t + \det X = 0 \ .$$
By the Cayley-Hamilton theorem, $X$ too will verify the above equation, therefore
$$X^2 - (\operatorname{tr} X) X + (\det X) I_2 = 0 \ .$$
In this equation (in $X$, this time), we already know $X^2$. On the other hand, we have
$$(\det X)^2 = \det (X^2) = \det \begin{pmatrix} 1 & 2 \\  3 & 7 \end{pmatrix} = 1 \ ,$$
whence we deduce that $\det X = \pm 1$.
This means that our equation in $X$ becomes
$$\begin{pmatrix} 1 & 2 \\  3 & 7 \end{pmatrix} \pm \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = (\operatorname{tr} X) X \ .$$
This has two possible cases:


*

*$\det X = -1$, therefore $(\operatorname{tr} X) X = \begin{pmatrix} 0 & 2 \\  3 & 6 \end{pmatrix}$. Taking the trace, we get $(\operatorname{tr} X) ^2 = 6$, whence $\operatorname{tr} X = \pm \sqrt 6$, so $X = \pm \frac 1 {\sqrt 6} \begin{pmatrix} 0 & 2 \\  3 & 6 \end{pmatrix}$;

*$\det X = 1$, therefore $(\operatorname{tr} X) X = \begin{pmatrix} 2 & 2 \\  3 & 8 \end{pmatrix}$. Taking the trace, we get $(\operatorname{tr} X) ^2 = 10$, whence $\operatorname{tr} X = \pm \sqrt {10}$, so $X = \pm \frac 1 {\sqrt {10}} \begin{pmatrix} 2 & 2 \\  3 & 8 \end{pmatrix}$.
It is now easy to check that all the above 4 matrices are indeed solutions of the given equation. They are also the only ones, by the above analysis.
