hi,

another numerical method is newton rapson method which is successive aproximation iterative method.


lets xn+1 be the n+1'th' guess of root the new aproximation is given by


xn+1 = xn - f(xn)/f'(xn) ----(1)


in your case let a be the real nmuber you are finding the root for.let x be the root of a, which gives


then
x^2 = a

and
x^2-a = 0

then
f(x) = x^2 - a and f'(x) = 2x


puting the value of f(x) and f'(x) gives following


xn+1 = xn + (xn^2-a)/2xn

start with xn = a/2



abhishek