How do you solve # sqrt(12-n)=n# and find any extraneous solutions?

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Answer 1 · 2016-08-28T20:46:02

#n=3#

Given:

#sqrt(12-n) = n#

Square both sides (noting that this may introduce spurious solutions):

#12-n = n^2#

This derived equation may (and actually does) have solutions which are not solutions of the original equation.

Subtract #(12-n)# from both sides to get:

#0 = n^2+n-12 = (n+4)(n-3)#

Hence #n=-4# or #n=3#

#n=-4# is not a solution of the original equation since:

#sqrt(12-(-4)) = sqrt(16) = 4 != -4#

#n=3# is a valid solution of the original equation:

#sqrt(12-3) = sqrt(9) = 3#