Square root of Matrix $A=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}$ Find Square root of Matrix $A=\begin{bmatrix} 1 &2 \\   3&4  \end{bmatrix}$ without using concept of Eigen Values and Eigen Vectors
I assumed its square root as $$B=\begin{bmatrix} a &b \\   c&d  \end{bmatrix}$$
hence
$$B^2=A$$ $\implies$
$$\begin{bmatrix} a^2+bc &b(a+d) \\   c(a+d)&d^2+bc  \end{bmatrix}=\begin{bmatrix} 1 &2 \\   3&4  \end{bmatrix}$$
So
$$a^2+bc=1 \tag{1}$$
$$b(a+d)=2 \tag{2}$$
$$c(a+d)=3 \tag{3}$$
$$d^2+bc=4 \tag{4}$$  From $(2)$ and $(3)$ we get
$$\frac{b}{c}=\frac{2}{3}$$
Let $b=2k $ and $c=3k$ Then
$$a^2=1-6k^2$$ and $$d^2=4-6k^2$$  So
$$B=\begin{bmatrix} \sqrt{1-6k^2} &2k\\   3k&\sqrt{4-6k^2}  \end{bmatrix}$$ 
So
$$B^2=\begin{bmatrix} 1 &2k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \\   3k \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)&4  \end{bmatrix}=\begin{bmatrix} 1 &2 \\   3&4  \end{bmatrix}$$
Now we have to solve for $k$ using equation
$$ \left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right)=\frac{1}{k} \tag{5}$$ Also
$$\left(\sqrt{1-6k^2}+\sqrt{4-6k^2}\right) \times \left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3$$ So from $(5)$
$$\left(\sqrt{4-6k^2}-\sqrt{1-6k^2}\right)=3k \tag{6}$$
Subtracting $(6)$ from $(5)$ we get
$$2\sqrt{1-6k^2}=\frac{1}{k}-3k$$  Squaring Both sides we get
$$33k^4-10k^2+1=0$$ we get
$$k^2=\frac{5 \pm i \sqrt{8}}{33}$$ From this we get $k$. is there any other simpler way to find ?

Eliminate $bc$ from the first equation (the uppercase letters are the RHS),
$$a^2+\frac{BC}{(a+d)^2}=A,$$ and from the first and the last, $$d^2-a^2=D-A.$$
Let $p:=(d+a)/2,m:=(d-a)/2$, giving
$$(p-m)^2+\frac{BC}{4p^2}=A,\\mp=4(D-A)$$
and multiplying by $4p^2$,
$$4(p^2-mp)^2+BC=4(p^2-4(D-A))^2+BC=4Ap^2.$$
This biquadratic equation will give the values of $p$, then $m, a$ and $d$, followed by $b$ and $c$.
