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17 April, 03:02

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

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  1. 17 April, 04:16
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    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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