How do you simplify #sqrt(68)#?

Question

9930 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-23T21:49:13

#sqrt(68) = 2sqrt(17)#

If #a, b >= 0# then:

#sqrt(ab) = sqrt(a)sqrt(b)#

The prime factorisation of #68# is:

#68=2xx2xx17#

So we have:

#sqrt(68) = sqrt(2^2*17) = sqrt(2^2)*sqrt(17) = 2sqrt(17)#

#sqrt(17)# cannot be simplified any further, but since #17=4^2+1# is of the form #n^2+1# it has a very simple continued fraction expansion:

#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#