Question

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

Answer 1

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.

   $\sqrt{x-5} + 7 = 11$

Subtract 7 from both sides

   $\sqrt{x-5} = 4$

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

   $\begin{aligned} \left( \sqrt{x-5} \right)^2 &= (4)^2 \\ x - 5 &= 16 \\ x &= 16 + 5 && (\text{\footnotesize Add 5 to both sides}) \\ x &= 21 \end{aligned}$

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.

   $\begin{aligned} \sqrt{x-5} + 7 &= 11 \\ \sqrt{21-5} + 7 &\stackrel{?}{=} 11 \\ \sqrt{16} + 7 &\stackrel{?}{=} 11 \\ 4 + 7 &\stackrel{?}{=} 11\\ 11 &= 11\quad \checkmark \end{aligned}$

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.