https://file.aiquickdraw.com/mathbot/file/1729096654340gsjavgn0.pdf

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.

Solve the equation √𝑥 − 1 + 7 = 𝑥

Step-by-step guide to solving the square root equation √x - 1 + 7 = x. Learn how to isolate the square root, solve the quadratic equation, and check for extraneous solutions. Suitable for high school students in Grades 9-12.

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