Question 54534
√(6x+1)= x-1 Solve. 
I got 6x+1=x^2-2x+1
Subtracted 6x and 1 and got 0=x^2-8x+0
x(x-8)
x=0
x=8
Now you have to check these answers to see if one
or both are extraneous.
Checking x=0 in the original problem you get:
sqrt(1)=-1. So x=0 is extraneous.
Checking x=8 in the original problem you get:
sqrt(49)=8-1=7
That is a good result.  x=8 is your solution.
You might ask why you got an extraneous answer.
It happened when you squared both sides of the
equation.  When you did that you introduced a 
root that was not part of the original equation.
Cheers,
Stan H.