How do I find the extraneous solution of #sqrt(x-1)=x-7#?

1 Answer
May 7, 2015

The extraneous solution is #x=5#

By squaring:
#x-1=x^2-14x+49 => x^2-15x+50=0#
#x_1=5, x_2=10 #

10 is an acceptable answer because it's grater than 1, so the sqrt is well defined, and it's greater than 10, so the result does make sense (we chose to consider the positive branch of the parabola for the definition of sqrt as a function)

5 is not an acceptable answer, because we should have #sqrt(4)=-2#, nonsense.

This happens because we chose a branch to define the function, if we chose the negative branch and we could define sqrt(a^2)=-a (because it's true that (-a)^2=a^2) 5 could be an acceptable solution.

This is even more evident when you work with complex numbers