How do you solve and check for extraneous solutions in #sqrt(6-x)-sqrt(x-6)=2#?

1 Answer
Adrian D.
Aug 3, 2015

There are no real valued solutions to the equation.

Explanation:

First note that the expressions in the square roots must be positive (restricting to real numbers). This gives the following constraints on the value of #x#:

#6-x>=0# => #6>=x#

and

#x-6>=0# => #x>=6#

#x=6# is the only solution to these inequalities. #x=6# does not satisfy the equation in the question, therefore there are no real valued solutions to the equation.

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