Given the equation 2 Square root (x minus 5) = 2, solve for x and identify if it is an extraneous solution.

2 root (x-5) = 2

Divide each side by 2 (in order to isolate the square root):

Root (x-5) = 1

Square both sides to remove the radical sign:

(x-5) = 1

Solve:

x=6

To be sure that it is not an extraneous solution, check the solution by replacing the value for the variable in the equation.

2 root (6-5) = 2

The square root of 1 is 1.

2 (1) = 2

This is correct.