Why is square root by long division found so? We were taught the long division method of finding square root in junior classes. The  logic behind the method used to be unclear, it remains so even now! However, we learnt and practiced the algorithm well. To me it appears that the identity $$(a+b)^2=a^2+2ab+b^2$$ lies behind the method. However this seems to be only the minimal idea. So extensions and generalizations are welcome here. See the picture below as I can't type this stuff. Some one else may edit it properly.

 A: For $3$ or $4$ digit numbers you were using $$(10a+b)^2=100a^2+(20a+b)b$$
For $5$ or $6$ digit numbers you would have used the messier $$(100a+10b+c)^2=10000a^2+(2000a+ 100b)b+(200a+ 20 b+c)c$$
So you would separate the number to be square-rooted into $100$s, in your example $27,04$
and it looked like
       ?  ?
     ------ 
     |27,04
 

The largest square less than or equal to $27$ is $25=5^2$ so you write
       5  ?
     ------ 
   5 |27,04
     |25
     ---
     | 2,04

You then double the $5$ you have written at the top to give $10$ and then add an extra digit $X$ so $10X \times X$ is a large as possible but does not exceed $204$: $102 \times 2= 204$ works exactly
       5  2
     ------ 
   5 |27,04
     |25
     ---
 102 | 2,04
     | 2,04
     ------
          0

So $\sqrt{2704} =52$
If the result had not been exact, you could have continued the same way, bringing down two more digits (possibly $00$).
My school thought this a waste of time and instead taught us to use logarithm tables.
A: This is based on my comments to the question.
Consider that you have evaluate the square root of $a$. Let $x$ be the square root and $h$ be the least significant digit of the square root and $b$ be the number formed from $x$ by removing this least significant digit $h$. Then we have $x=10b+h$. (for example consider $a=144,x=12,b=1,h=2$).
The process of long division allows you to find $h$ given $a, b$. Thus we can find the next digit of square root if previous digits are known.
Our job is to find $h$ such that $(10b+h)^2=a$ or $$(2\cdot 10b+h)h=a-(10b)^2$$ the subtraction in right hand side is done by placing digits accordingly (for $a=144,b=1$ you place the digit $1$ of $b$ below $1$ of $a$ and get $44$ so that this is essentially doing $a-(10b)^2$).
Now analyze the left hand side. $2\cdot 10b$ involves doubling the number $b$ and multiplying further by $10$ to make space for next digit $h$. We have to choose next digit $h$ so that the new number $(20b+h)$ when multiplied by $h$ equals the number obtained after subtraction on right hand side. This is exactly what we do in the long division process (try to see for the example $a=144,b=1$ and you can easily guess that $21\cdot 1=21,22\cdot 2=44$).

Consider the same in terms of polynomials and roots. Let us solve $x^2-a=0$ digit by digit. We guess the root $b$ (which is the approximate square root of first two digits of $a$) and find an equation whose roots are $t=x-b$. This equation is clearly $$(t+b) ^2-a=0$$ Next multiply the roots by $10$ so that $z=10t$ gives $$(z+10b)^2-100a=0$$ or $$z(20b+z)=100a - 100b^2$$
Let us see that $100a$ is essentially about getting two more digits from $a$ and then subtracting $100b^2$ from it means that we subtract $b^2$ from first two digits of $a$ and bring in next two digits from $a$. The guess for $z$ is based on $z(20b+z)$ which is doubling $b$, creating next digit $z$ and multiplying it by $z$.
We can continue the process by subtracting $z$ and multiplying by $10$ and get another equation. The pattern remains same and can be continued till any desired number of digits.

The mysterious part in the long division process is doubling the divisor at each step and creating space for quotient as next digit. I have tried to explain that part in both the approaches above.
A: Let us do it with a trinomial
$$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$$ see the fig.

A: I’ll just add the following intuition to the other answers.
Suppose you are extracting the square root of $x$. First you guess, say, $x_0$. If $x_0^2\neq x$, then you must correct your guess by adding some new term $x_1$ to your initial guess $x_0$. Your guess for $\sqrt{x}$ is thus updated to $x_0+x_1$, whose square is $x_0^2+\color{red}{2x_0x_1+x_1^2}$.
The key point of the algorithm is that incrementing your guess by $x_1$ increments the square of your guess by the amount in red, which can be written $\color{red}{x_1(2x_0+x_1)}$. This fact helps you determine by how much to increment your guess.
Now apply this idea to $\sqrt{2704}$. First we guess $50$, since $50^2=2500$ gets us pretty close. The gap between the squares is $204$. If we increment our guess of $50$ by some amount $a$, we’ll increment the square of our guess by $a(2\cdot50+a)$. We’d like this to be as close to $204$ as possible. Notice $a=1$ increments by $101$, but $a=2$ increments by $204$: perfect! So we increment our initial guess of $50$ by $2$, and since that fills the gap perfectly we’re done! We’ve discovered $2704$ is a perfect square, and we iteratively arrived at $52$ as the square root.
It all boils down to noting how the square of your estimate changes when you change your estimate.
A: Good explanation from geimetey. My original submission has been done by my friend Zaffar. That's because I am not familiar with formalities if stack exchange.
Few points I clarify in this context:
(1) The division method can be extended for any n-root. Sat for cube root :
Start with a number bB+s where B is the base; a and b are integers. (bB +a) ^3 =(bB)^3 + 3a(bB)^2 + 3(bB)a^2 + a^3.
Let us group three at a time from right. So, two groups. Left group has only (bB)^3 and right group has 3a(bB)^2 + 3(bB)a^2 + a^3. One has to find suitable b such that right group matches. Fir actual numbers, one has to search largest close number for a and b at every step.
Generalising the geometry approach, this can thenbe compared with funding suitable 3D visualisation where in stead of area, it willill be  volume.
(2) For any n-root, the method is similar. For visualisation, one can conceive n dimensional hypercubes.
