To solve question 6 from your exam:

Solve the equation:

$\sqrt{x - 1} + 7 = x$

Step-by-step solution:

1. Isolate the square root term: $\sqrt{x - 1} = x - 7$

2. Square both sides to eliminate the square root: $(\sqrt{x - 1})^2 = (x - 7)^2$ $x - 1 = (x - 7)(x - 7)$ $x - 1 = x^2 - 14x + 49$

3. Rearrange the equation: $0 = x^2 - 15x + 50$

4. Solve the quadratic equation using the quadratic formula: $x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(50)}}{2(1)}$ $x = \frac{15 \pm \sqrt{225 - 200}}{2}$ $x = \frac{15 \pm \sqrt{25}}{2}$ $x = \frac{15 \pm 5}{2}$

5. Calculate the solutions: $x_1 = \frac{15 + 5}{2} = 10$ $x_2 = \frac{15 - 5}{2} = 5$

6. Check for extraneous solutions by substituting both values into the original equation:

   • For $x = 10$: $\sqrt{10 - 1} + 7 = 10 \quad \text{True since} \quad \sqrt{9} + 7 = 10$
   • For $x = 5$: $\sqrt{5 - 1} + 7 = 5 \quad \text{False since} \quad \sqrt{4} + 7 = 9$

Final Answer:

Only $x = 10$ is a valid solution.

Would you like more details on this or any other part of the exam?

Here are 5 related questions for practice:

1. Solve $\sqrt{x + 2} + 5 = x$.
2. Solve $x^2 - 6x + 8 = 0$ by completing the square.
3. Find the domain of the function $f(x) = \sqrt{3x - 5}$.
4. Solve the inequality $\sqrt{x - 3} \leq 5$.
5. Simplify $\sqrt{x^4 - 16}$.

Tip: Always check for extraneous solutions when solving equations involving square roots, as squaring both sides can introduce invalid solutions.