\( 1.1 .5 \quad \sqrt { x - 1 } + 3 = x - 4 \)

1. Write the equation: \[ \sqrt{x-1}+3=x-4 \] 2. Isolate the square root by subtracting 3 from both sides: \[ \sqrt{x-1}=x-7 \] 3. Determine the domain of the equation. Since a square root is defined only for nonnegative arguments, we require: \[ x-1\geq0 \quad\Rightarrow\quad x\geq1. \] Also, because \(\sqrt{x-1}\geq0\), the right-hand side must be nonnegative: \[ x-7\geq0 \quad\Rightarrow\quad x\geq7. \] Thus, we only consider solutions with \(x\geq7\). 4. Square both sides to eliminate the square root: \[ (\sqrt{x-1})^2=(x-7)^2 \quad\Rightarrow\quad x-1=(x-7)^2. \] 5. Expand the right-hand side: \[ x-1=x^2-14x+49. \] 6. Rearrange to form a quadratic equation: \[ 0=x^2-14x+49-x+1 \quad\Rightarrow\quad x^2-15x+50=0. \] 7. Solve the quadratic equation using the quadratic formula: \[ x=\frac{15\pm\sqrt{15^2-4\cdot1\cdot50}}{2\cdot1}=\frac{15\pm\sqrt{225-200}}{2}=\frac{15\pm\sqrt{25}}{2}. \] \[ x=\frac{15\pm5}{2}. \] 8. This gives two potential solutions: \[ x=\frac{15+5}{2}=10\quad \text{or} \quad x=\frac{15-5}{2}=5. \] 9. Check the solutions against the domain \(x\geq7\): - \(x=5\) does not satisfy \(x\geq7\). - \(x=10\) satisfies \(x\geq7\). 10. Verify \(x=10\) in the original equation: \[ \sqrt{10-1}+3=\sqrt{9}+3=3+3=6, \] \[ 10-4=6. \] Both sides equal 6, confirming the solution. The final solution is: \[ \boxed{10} \]