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To solve for \( x \) in the given equation:

\[
x = 1 + \sqrt{2x - 2}
\]

we can proceed as follows:

### Step 1: Isolate the square root term
Subtract 1 from both sides:

\[
x - 1 = \sqrt{2x - 2}
\]

### Step 2: Square both sides to eliminate the square root
\[
(x - 1)^2 = 2x - 2
\]

Expanding the left side:

\[
x^2 - 2x + 1 = 2x - 2
\]

### Step 3: Move all terms to one side
Rearrange to form a quadratic equation:

\[
x^2 - 4x + 3 = 0
\]

### Step 4: Factor the quadratic equation
\[
(x - 3)(x - 1) = 0
\]

### Step 5: Solve for \( x \)
This gives us two solutions:

\[
x = 3 \quad \text{and} \quad x = 1
\]

### Step 6: Check for extraneous solutions
Substitute each solution back into the original equation to confirm validity.

1. **For \( x = 3 \):**
   \[
   3 = 1 + \sqrt{2 \cdot 3 - 2}
   \]
   \[
   3 = 1 + \sqrt{6 - 2}
   \]
   \[
   3 = 1 + \sqrt{4}
   \]
   \[
   3 = 1 + 2
   \]
   This holds true, so \( x = 3 \) is a valid solution.

2. **For \( x = 1 \):**
   \[
   1 = 1 + \sqrt{2 \cdot 1 - 2}
   \]
   \[
   1 = 1 + \sqrt{2 - 2}
   \]
   \[
   1 = 1 + 0
   \]
   This also holds true, so \( x = 1 \) is also a valid solution.

### Final Answer
There are two solutions, and we should arrange them with \( x_2 > x_1 \):

\[
x_1 = 1, \quad x_2 = 3
\]

If you have questions or would like more details, please let me know!

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Here are some related questions that may interest you:

1. How do you determine if a quadratic equation has real or complex solutions?
2. What are extraneous solutions, and why do they occur in equations with square roots?
3. How can the quadratic formula be used to solve quadratic equations?
4. What is the significance of squaring both sides of an equation in solving it?
5. How can graphing the function help in verifying solutions?

**Tip:** Always check for extraneous solutions when solving equations involving square roots by substituting the solutions back into the original equation.

Solve for x. If there is only one solution, place 'N/A' in the second answer gap. If there are two solutions, ensure that x2 > x1.

Learn how to solve for x in the equation x = 1 + sqrt(2x - 2). This detailed solution includes isolating the square root, squaring both sides, and factoring to find two solutions. Suitable for students in Grades 8-10.

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