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.

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