To solve the equation \( \sqrt{3x - 3} = x - 1 \) by factoring, let's go through the steps:

### Step 1: Square both sides of the equation.
Squaring both sides will eliminate the square root:
\[
(\sqrt{3x - 3})^2 = (x - 1)^2
\]
This simplifies to:
\[
3x - 3 = (x - 1)^2
\]

### Step 2: Expand the right-hand side.
Expand \( (x - 1)^2 \):
\[
3x - 3 = x^2 - 2x + 1
\]

### Step 3: Rearrange the equation.
Move all terms to one side of the equation:
\[
0 = x^2 - 2x + 1 - 3x + 3
\]
Simplifying:
\[
0 = x^2 - 5x + 4
\]

### Step 4: Factor the quadratic equation.
Factor \( x^2 - 5x + 4 \):
\[
0 = (x - 1)(x - 4)
\]

### Step 5: Solve for \( x \).
Set each factor equal to zero:
\[
x - 1 = 0 \quad \text{or} \quad x - 4 = 0
\]
This gives:
\[
x = 1 \quad \text{or} \quad x = 4
\]

### Step 6: Check for extraneous solutions.
Since we squared both sides of the equation, we need to check each solution in the original equation to make sure they are valid.

- For \( x = 1 \):
  \[
  \sqrt{3(1) - 3} = 1 - 1 \quad \Rightarrow \quad \sqrt{0} = 0
  \]
  This is true, so \( x = 1 \) is a valid solution.

- For \( x = 4 \):
  \[
  \sqrt{3(4) - 3} = 4 - 1 \quad \Rightarrow \quad \sqrt{12 - 3} = 3 \quad \Rightarrow \quad \sqrt{9} = 3
  \]
  This is also true, so \( x = 4 \) is a valid solution.

### Final Answer:
The solutions are:
\[
x = 1 \quad \text{and} \quad x = 4
\]

Would you like further details or explanations on any step?

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Here are 5 related questions to expand on this topic:
1. How do you factor a quadratic expression like \( x^2 - 5x + 4 \)?
2. Why is it necessary to check for extraneous solutions when solving square root equations?
3. What happens if an equation has complex roots instead of real roots?
4. How can you solve \( \sqrt{2x + 5} = x + 3 \) by a similar method?
5. How do you complete the square for a quadratic equation?

**Tip:** Always verify potential solutions when squaring both sides of an equation, as it may introduce extraneous solutions!

Learn how to solve the square root equation √(3x - 3) = x - 1 by factoring. This step-by-step solution explains key concepts like squaring both sides, expanding expressions, factoring quadratic equations, and checking for extraneous solutions. Suitable for Grades 8-10 students.

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