How do you find the square root of #6724# ?

1 Answer
May 3, 2017

Answer:

#sqrt(6724) = 82#

Explanation:

We can try factoring #6724# and identifying square factors:

#color(white)(000000)6724#
#color(white)(000000)"/"color(white)(00)"\"#
#color(white)(00000)2color(white)(000)3362#
#color(white)(000000000)"/"color(white)(00)"\"#
#color(white)(00000000)2color(white)(000)1681#
#color(white)(000000000000)"/"color(white)(00)"\"#
#color(white)(00000000000)41color(white)(00)41#

So:

#sqrt(6724) = sqrt(2^2*41^2) = 2*41=82#

That "worked", but I cheated slightly in knowing that #1681 = 41^2#

Let's try another approach...

Given #6724#, split off pairs of digits starting from the right to get:

#67|24#

Looking at the most significant pair of digits, note that:

#67 > 64 = 8^2#

Hence the square root of #6724# is a little larger than #80#

We can use the Babylonian method to find a better approximation:

Given a number #n# of which you want the square root and an approximation #a_i# to that root, a better approximation #a_(i+1)# is given by the formula:

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

So in our case, with #n=6724# and #a_0 = 80# we find:

#a_1 = (80^2+6724)/(2*80) = (6400+6724)/160 = 82.025#

Hmmm - that's suspiciously close to #82#. Does #82# work?

#82^2 = (80+2)^2 = 80^2+2*80*2+2^2 = 6400+320+4 = 6724#