What is the simplest radical form for #sqrt(145)#?

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Answer 1 · 2017-03-28T20:56:21

#sqrt145#

There is no simple form for this.

Let's try using the factors of #145#

#sqrt145=sqrt145*sqrt1#

#sqrt145=sqrt29*sqrt5#

This cannot be broken into any simpler forms so there is no simple from for #sqrt145#

Answer 2 · 2017-03-28T21:03:28

#sqrt(145)#

The prime factorisation of #145# is:

#145 = 5*29#

Since this has no square factors, there is no simpler radical form than #sqrt(145)#.

Note however that #145 = 12^2+1# is of the form #n^2+1#

As a result, its square root has a very simple form as a continued fraction:

#sqrt(145) = [12;bar(24)] = 12+1/(24+1/(24+1/(24+1/(24+1/(24+...)))))#