How do you simplify #sqrt(257)#?

1 Answer
Jul 29, 2015

Answer:

#257# is prime, so #sqrt(257)# cannot be simplified.

Explanation:

If the radicand has a square factor, then you can move the square root of that factor outside of the square root.

For example, #sqrt(12) = sqrt(2^2*3) = sqrt(2^2)sqrt(3) = 2sqrt(3)#

If #n# has no square factor (e.g. #n# is prime), then #sqrt(n)# cannot be simplified in this way.

In your case you can note that #257# is #256+1# and #sqrt(256) = 16#. So #16# makes a very good first approximation to the square root for use with a method like Newton Raphson.

So you can put #a_0 = 16# and iterate using:

#a_(i+1) = a_i + (n-a_i^2)/(2a_i)#

or suchlike to quickly find a very good approximation.

In our case #a_1 = 16+1/32 = 16.03125# while #sqrt(257) ~= 16.03122#