How do you simplify #sqrt(68)#?
1 Answer
May 23, 2016
Explanation:
If
#sqrt(ab) = sqrt(a)sqrt(b)#
The prime factorisation of
#68=2xx2xx17#
So we have:
#sqrt(68) = sqrt(2^2*17) = sqrt(2^2)*sqrt(17) = 2sqrt(17)#
#sqrt(17) = [4;bar(8)] = 4+1/(8+1/(8+1/(8+1/(8+1/(8+1/(8+...))))))#
You can terminate this expansion at any point to give a rational approximation. For example:
#sqrt(17) ~~ [4;8,8] = 4+1/(8+1/8) = 268/65 ~~ 4.1231#