Well this is simple a calculator type problem...but if you are curious as the the algorithm used by simple calculators and such...

They use a Newtonian approximation until it surpasses the precision level of the calculator or computer program..

A newtonian approximation is an interative process that gets closer and closer to the actual answer to any mathematical problem...it is of the form:

x-(f(x)/(df/dx))

In a square root problem you wish to know:

x=√n where x is the root and n is the number

x^2=n

x^2-n=0

So f(x)=x^2-n and df/dx=2x so using the definition of the newton approximation you have:

x-((x^2-n)/(2x)) which simplifies further to:

(2x^2-x^2+n)/(2x)

(x^2+n)/(2x), where you can choose any starting value of x that you desire (though convergence to an exact (if possible) solution will be swifter the closer xi is to the actual value x)

In this case the number, n=95.54, so a decent starting value for x would be 10.

Using this initial x in (x^2+95.54)/(2x) will result in the following iterative sequence of x.

10, 9.777, 9.774457, 9.7744565, 9.7744565066299210578124802523397

The calculator result for my calc is: 9.7744565066299210578124802523381

So you see how accurate the newton method is in just a few iterations. :P

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3 years ago