Is there a way to figure out a square root with out a calculator?

Question

I have figure out a way to take the sqaure of any number with this formual : first =10*A2-10 - lets say it 2 ==== 10*2 - 10 = 10 this is N. ok then ((19*n+(n*n+n))/100)+1 ----- 19*10+(10*10+10) = 190+110 = 300/100 = 3 +1 = 4 --- it works all the time my question is is there a way to do it backwords ??
please help.

Answer 1

First of all, your "way" to find the square of a number requires that you actually square a number. Note the n*n term in the middle. So you are really better off simply squaring the number in the first place. If you know a little algebra, its not hard to show how it works. If your original number was m, let n = 10m-10.
((19*n+(n*n+n))/100)+1
= (n*n+20n)/100+1
= (n*n+20n+100)/100
= (n+10)(n+10)/100
(remember that n = 10m-10, so n+10 = 10m)
= (10m)(10m)/100
= m*m

There are several ways to find a square root with only pencil and paper. See my reference for one of them.

Answer 2

A number of ways. You could use logarithms:

Let's say you have a number x.

ln (x ^ 1/2) = (1/2) ln x

so if you have a table of logarithms handy, you could find the natural logarithm of x, divide it by two, and then find out which number the result is the natural logarithm of. You'll end up with something between two of the values on the table, giving you a halfway decent approximation which you can then improve on using linear interpolation.

Namely: If you have a function f (the square root function, in this case) whose value you know with reasonable accuracy for two points a and b, setting

f( r a + (1 - r) b) = r f(a) + (1 -r) f(b) for 0 < r < 1

approximating the graph of the curve from x=a with x=b by a secant line from (a,f(a)) to (b,f(b)). The logarithmic function is smooth, monotonic with a derivative that goes to zero as x -> infinity, and the x values in the log tables come pretty close together, so you should get a very good approximation doing this.

You could also use a Taylor expansion of the square root function around a point at which the value is known (something of the form (n^2,n)), and get greater accuracy at the expense of doing more work. Be careful, though, about the radii of convergence. If, say, you are trying to compute the first few digits of the square root of 29, and you're doing a Taylor expansion around (4,2), the results are going to break down swiftly. Make sure that the x value your expansion is based on is closer to the x value whose square root you're trying to compute than it is to zero, preferably much closer.

I recall that there are other ways as well, but those are the first two that come to mind.

Answer 3

The "square root" with the 5 on top is not a square root at all but it is the 5th root "a is a square root of b" means a×a = b "a is a fifth root of b" means a×a×a×a×a = b To find the fifth root of 32, you can write 32 in the form offator product: 32 = 2×16 = 2×(2×8) = 2×2×(2×4) = 2×2×2×2×2 (=2^5) therefore 5th root of 32 is 2

Answer 4

There have been several methods devised to calculate square roots. The trouble has always been the irrational numbers.
Try your method with 1.4, 1.41, 1.414... each is an approximation of the square root of 2. Which one gives you the correct answer? Depends on how many digits of accuracy you need.

Check out Newton's method, and Halley's method, on Wiki.