How do you find square root of 800?

Answer 1

If you have a non-scientific calculator you can use the Newton-Raphson method.

Let f(x) = x2 - 800, and f'(x) = 2x

[f'(x) is the derivative of f(x) but you do not need to know that to use the N-R method.]

Make a guess at the square root of 800, and call is x0.

Then calculate

xn+1 = xn - f(xn)/f'(xn) for n = 1, 2, 3, ...

Provided you made a reasonable choice for the starting point, the iteration will very quickly converge to the true answer. Even if it is not so good:

Suppose you start with x0 = 20 (a pretty poor choice since 202 is 400, which is nowhere near 800).

Even so, x3 = 2.843 is less than 1 in 24 thousand from the true value and x4 has an error of less than 1 in 30 billion.