well there is brute force: for example, sqrt(10)

3^2 = 9, 4^2 = 16; therefore the solution to sqrt(10) lies between 3 and 4:

3.5^2 = 12.25; therefore the solution to sqrt(10) lies between 3 and 3.5

3.1^2 = 9.61; therefore the solution to sqrt(10) lies between 3.1 and 3.5

3.2^2 = 10.24; therefore the solution to sqrt(10) lies between 3.1 and 3.2

3.15^2 = 9.9225; therefore the solution to sqrt(10) lies between 3.15 and 3.2

3.16^2 = 9.9856; therefore the solution to sqrt(10) lies between 3.16 and 3.2

3.17^2 = 10.0489; therefore the solution to sqrt(10) lies between 3.16 and 3.17

and so on, until your fingers bleed. Or you make a mistake calculating one of those squares (which is very likely without a calculator) and fuck everything up :

there is another algorithm which I don't know very well but is a better than this, which was taught to students back in the day but has fallen out of favor since you can buy a calculator that can do this for you for about $1:
http://www.homeschoolmath.net/teaching/square-root-algorithm.php

/e actually looking at that algorithm, I don't think it is much better, lol

This post was edited by Azrad on Nov 1 2013 11:38am

thanks alot!
very helpful ^.^

gonna send you a donation

There are 2 ways to compute decimals of a square root that I'm aware of, without using exponantial and logarithm functions, and in a direct way (also dichotomy method works fine of course).

The first is using continuous fractions :
every real number X can be expressed via a sequence of integers a1, a2, a3, ... such as : X = a1 + 1/ ( a2 + 1/( a3 + ... ))

If your real number is an algebraic number (like any square root), this sequence is periodic.

Let's try with square root of 5 ( sqrt5 ) :
4 < 5 < 9, hence a1 = 2.

sqrt5 = 2 + 1/x2
x2 = 1 / ( sqrt5 - 2 ) = ( sqrt5 + 2) / 1 = sqrt5 + 2 = 4 + 1/x3
sqrt5 = 2 + 1/x3

hence x2 = x3 = x4 = ... = sqrt5 + 2
and finally a1 = 2, a2 = a3 = a4 = ... = 4

That is to say :

sqrt5 = 2 + 1/ (4+1/(4+1/(4+1/(...)))) = [2,4,4,4,4,4,...]

Now if you want some decimals of sqrt5, just choose a rank for your approximation.

Example :
sqrt5 ~ 2 + 1/(4+1/(4+1/4))
sqrt5 ~ 2 + 1/(4+1/(17/4))
sqrt5 ~ 2 + 1/(4+4/17)
sqrt5 ~ 2 + 1/(72/17)
sqrt5 ~ 2 + 17/72
sqrt5 ~ 161/72

It it possible to prove that, in any case, |sqrt5 - 161/72| < 1/72²
(so your first 3 digits must be correct).

You can go as high as you want to compute every decimals of sqrt5, you are just left with a simple division.
It can be effective if you need a large number of digits, but it's a lot of work if you're just interested in the first 2 or 3 digits...

A second method consists in using the approximation : (1+x)² ~ 1+2x, as x approaches zero.
You can compute as follow :

variable A = 5 (the number you want to know the square root)
variable B = 0 (a kind of first remainder)
variable C = 2 (since we know that 2 is the integer part of sqrt5)

A := A - BC - C²
B := B + 2C
C := [A/B]
where [...] stands for the decimal developement of A/B, stopped at the first non-zero decimal.

step 1 :
A = 5 - 0*2 - 2² = 1
B = 0 + 2*2 = 4
C = [1/4] = 0.2

step 2 :
A = 1 - 4*0.2 - 0.2² = 0.16
B = 4 + 2*0.2 = 4.4
C = [0.16/4.4] = 0.03

step 3 :
A = 0.16 - 4.4*0.03 - 0.03² = 0.0271
B = 4.4 + 2*0.03 = 4.46
C = [0.0271 / 4.46] = 0.006

etc

and finally sqrt5 is the sum of all C's :
2 + 0.2 + 0.03 + 0.006 + ... ~ 2.236

This second method can be used to compute the decimals of a square root with only a pen and a paper.
It used to be taught in smaller classes and is very quick with some practice.

Good luck !

http://en.wikipedia.org/wiki/Methods_of_computing_square_roots

personally i'd do what azrad did.

aye, already solved with Azrad's method,

well but thank you very much! .. epecially thanks to feanur for his effort

This post was edited by Breee on Nov 2 2013 04:01am

so i know ur already done with ur problem but honestly guessing and checking isnt really going to impress ur teacher. try this out:
http://www.wikihow.com/Find-a-Square-Root-Without-a-Calculator
it honestly blew my mind for how close u can get and its really simple to calculate

I think this is exactly the method from the link of azrad & i used it ^^

but ty ^^