How do you simplify #sqrt65#?

1 Answer
Mar 14, 2016

Answer:

#65 = 5*13# has no square factors, so #sqrt(65)# is the simplest form.

Explanation:

If a radicand (the part under the root sign) of a square root has a square factor, then it can be simplified:

#sqrt(a^2b) = abs(a) sqrt(b)#

or if you know that #a >= 0#, more simply:

#sqrt(a^2b) = a sqrt(b)#

For example, #sqrt(24) = sqrt(2^2*6) = 2sqrt(6)#

In our example, we find #65 = 5 * 13# has no square factors, so cannot be simplified in this way.

If you like, you can reexpress it:

#sqrt(65) = sqrt(5)sqrt(13)#

but that is not (as far as I know) considered 'simpler'.

#color(white)()#
Note that #sqrt(65)# is an irrational number. That is, it cannot be expressed as a fraction #p/q# for integers #p# and #q#. As a result, its decimal expansion does not terminate or recur.

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Bonus

#65 = 64 + 1 = 8^2 + 1#

is of the form #n^2 + 1# with #n = 8#.

As a result, the square root can be expressed as a very simple continued fraction ...

#sqrt(65) = [8;bar(16)] = 8+1/(16+1/(16+1/(16+1/(16+...))))#

You can use this to give you good approximations for #sqrt(65)#, by truncating the continued fraction after a few terms.

For example,

#[8; 16] = 8+1/16 = 129/16 = 8.0625#

#[8; 16, 16] = 8+1/(16+1/16) = 8+16/257 = 2072/257 ~~ 8.0622568#

Actually #sqrt(65)# is closer to #8.06225774829854965236#