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.

Related questions
See all questions in Radical Equations
Impact of this question
1919 views around the world
You can reuse this answer
Creative Commons License