How do you find the square root of 16562?

The ancient greeks used to compute square roots by sucessive approximations.

Given a number #N# for which they need to compute the square root and given an initial approximation #q_0# they proceed as follows:

#(q_0+ delta q_0)^2=N# or
#q_0+2 q_0 delta q + (delta q_0)^2 = N#

So they were searching for an approximation variation #delta q_0# with the purpose of correct the initial guess #q_0#. They supposed also that #delta q_0# being small, much smaller would be #(delta q_0)^2# so they used the approximation

#q_0+2 q_0 delta q_0 approx N#

solving for #delta q_0# they got

#delta q_0 = ((N/q_0) - q_0)/2#

once corrected #q_0# they got #q_1 = q_0 + delta q_0#
and then follow with

#delta q_1 = ((N/q_1) - q_1)/2# etc.

Let us apply that process for calculation of square root of

#N =16562#

our initial guess will be #q_0 = 400#

#delta q_0 =((16562/400)-400)/2 =-179.298#

so #q_1 = 400-179.298# and calculating #delta q_1#

#delta q_1=-72.83015# so #q_2 = 400-179.298-72.83015#

calculating #delta q_2#

#delta q_2 = -17.93516# so #q_3 = 400-179.298-72.83015-17.93516#

In the third iteration we get

#delta_3 = -1.23779# obtaining a result of

#sqrt(16562) approx 128.699# which is a satisfactory result.

To find square root of #16562#, we should first factorize it.

From divisibility rules, it is apparent that it is divisible by #2# and dividing by #2#, we get #8281#.

#8281# is clearly not divisible by #3# and #5#, but is divisible by #7#. Dividing by #7#, we get #1183#, which is again divisible by #7#and dividing it by #7# we get #169#, which is #13xx13#.

Hence, #16562=2xx7xx7xx13xx13# and hence

#sqrt16562=sqrt(2xxbar(7xx7)xxbar(13xx13))#

= #7xx13xxsqrt2=91sqrt2=91xx1.4142=128.6922#

If you are not sure of the numbers build a factor tree to find the squared prime numbers.

Thus #sqrt(16562) =sqrt(2xx7^2xx13^2)#

#=7xx13xxsqrt(2) = 91sqrt(2)~~128.69# to 2 decimal places