What is #x# in #x^2 = 17#?

1 Answer
George C.
Dec 27, 2015

#x = +-sqrt(17) ~~ +-4.123#

Explanation:

The equation #x^2 = 17# has two solutions, which are the square roots of #17#. The principal square root of #17# is the positive one, which is what we mean when we write #sqrt(17)#.

#sqrt(17)# is an irrational number. That is, it cannot be represented as #p/q# for some integers #p# and #q#.

As a result, its decimal expansion does not eventually terminate or recur.

It does have a simple expansion as a so called "continued fraction":

#sqrt(17) = [4;bar(8)] = 4+1/(8+1/(8+1/(8+1/(8+...))))#

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