How do you solve #sqrt(x-6)+2=6#?

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Answer 1 · 2016-07-25T08:06:57

This equation has one solution #x=22#

We start from the equation #sqrt(x-6)+2=6#

Before any calculations we have to find the domain of the equation. It contains roots so the expression under sqare root sign must be positive:

#x-6>=0#

#x>=6#

The domain is: #D=<6;+oo)#

Now we can solve the equation.
First we move #2# to the right:

#sqrt(x-6)=6-2#

#sqrt(x-6)=4#

If the square root of an expression is #4# then the expression must have the value of #16#, so we get:

#x-6=16#

#x=22#

The calculated value lies in the domain so we can write the answer:

The equation has one solution: #x=22#