1. **State the problem:** Solve the equation $x - 11 = \sqrt{41 - 4x}$ for $x$. 2. **Formula and rules:** To solve equations involving square roots, isolate the square root term and then square both sides to eliminate the radical. Remember to check for extraneous solutions after squaring. 3. **Isolate and square:** $$x - 11 = \sqrt{41 - 4x}$$ Square both sides: $$ (x - 11)^2 = (\sqrt{41 - 4x})^2 $$ $$ (x - 11)^2 = 41 - 4x $$ 4. **Expand and simplify:** $$ (x - 11)^2 = x^2 - 22x + 121 $$ So, $$ x^2 - 22x + 121 = 41 - 4x $$ 5. **Bring all terms to one side:** $$ x^2 - 22x + 121 - 41 + 4x = 0 $$ $$ x^2 - 18x + 80 = 0 $$ 6. **Factor the quadratic:** $$ x^2 - 18x + 80 = (x - 10)(x - 8) = 0 $$ 7. **Solve for $x$:** $$ x - 10 = 0 \Rightarrow x = 10 $$ $$ x - 8 = 0 \Rightarrow x = 8 $$ 8. **Check for extraneous solutions:** - For $x=10$: $$ 10 - 11 = -1 $$ $$ \sqrt{41 - 4(10)} = \sqrt{41 - 40} = \sqrt{1} = 1 $$ Left side $\neq$ right side, so $x=10$ is extraneous. - For $x=8$: $$ 8 - 11 = -3 $$ $$ \sqrt{41 - 4(8)} = \sqrt{41 - 32} = \sqrt{9} = 3 $$ Left side $\neq$ right side, so $x=8$ is extraneous. 9. **Re-examine the original equation:** The left side can be negative but the right side is always non-negative. So, the equation $x - 11 = \sqrt{41 - 4x}$ implies $x - 11 \geq 0$. 10. **Apply domain restriction:** $$ x - 11 \geq 0 \Rightarrow x \geq 11 $$ Check $x=10$ and $x=8$ do not satisfy this. 11. **Conclusion:** No real solutions satisfy the equation. **Final answer:** No real solution.