How do you solve #sqrt(x^2-28)-1=x#?
Start by isolating the radical on one side of the equation.
Any possible solution to this equation must satisfy the conditions
#x^2 - 28>=0# #->#for real numbers, you can only take the square root from positive numbers. #x-1>=0 => x>=1# #->#the square root of a real number can only be a positive number.
In order for the expression under the radical to be positive, you need
So, square both sides of the equation to get rid of the radical
This is equivalent to
As you can see, this is an extraneous solution because it doesn't meet the condition