Solve (sqr)(x-5) +7= 11 and identify whether it is extraneous

The solution to the equation √(x-5) + 7 = 11 is x = 21 and it is not extraneous. 

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Thank you for posting your question with clear formatting for the square root.

   

Subtract 7 from both sides

   

Square both sides of the equation.
This would be the step that can introduce extraneous solutions.
It will not always introduce them.

   

A solution to an equation only works if the solution results in a true statement when it put it in the original equation. 

To check whether this proposed solution of x = 21 is extraneous, substitute it back into the original equation and see if we get a true statement.

   

This resulted in a true statement as 11 is equal to 11.

The solution to the equation is x = 21 and it is not extraneous because the solution works in the original equation.

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Step-by-step explanation:

5 6/12 - 2 4/12=

(5-2) (6-4)/12=

3 2/12

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