How do you find the square root of #6724# ?
1 Answer
Explanation:
We can try factoring
#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
Let's try another approach...
Given
#67|24#
Looking at the most significant pair of digits, note that:
#67 > 64 = 8^2#
Hence the square root of
We can use the Babylonian method to find a better approximation:
Given a number
#a_(i+1) = (a_i^2+n)/(2a_i)#
So in our case, with
#a_1 = (80^2+6724)/(2*80) = (6400+6724)/160 = 82.025#
Hmmm - that's suspiciously close to
#82^2 = (80+2)^2 = 80^2+2*80*2+2^2 = 6400+320+4 = 6724#