We are given the equation:

\[
\sqrt{3x + 2} = x
\]

To solve this equation using factoring, let's go through the following steps:

### Step 1: Square both sides of the equation
To eliminate the square root, square both sides:

\[
\left( \sqrt{3x + 2} \right)^2 = x^2
\]

This simplifies to:

\[
3x + 2 = x^2
\]

### Step 2: Rearrange the equation
Move all terms to one side of the equation to set it equal to zero:

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

### Step 3: Factor the quadratic equation
Now, we factor the quadratic equation:

\[
x^2 - 3x - 2 = (x - 2)(x + 1) = 0
\]

### Step 4: Solve for \( x \)
Set each factor equal to zero:

\[
x - 2 = 0 \quad \text{or} \quad x + 1 = 0
\]

Solving these gives:

\[
x = 2 \quad \text{or} \quad x = -1
\]

### Step 5: Check for extraneous solutions
Since we squared both sides in the process, we need to check if both solutions satisfy the original equation.

- For \( x = 2 \):
  \[
  \sqrt{3(2) + 2} = \sqrt{6 + 2} = \sqrt{8} \neq 2
  \]
  So, \( x = 2 \) is **not** a valid solution.

- For \( x = -1 \):
  \[
  \sqrt{3(-1) + 2} = \sqrt{-3 + 2} = \sqrt{-1}
  \]
  Since \( \sqrt{-1} \) is not a real number, \( x = -1 \) is also **not** valid.

### Conclusion:
There are no real solutions to this equation.

Would you like more details or have any questions?

Here are 5 questions to extend this problem:
1. What happens if we consider complex solutions for this equation?
2. How can we solve other types of square root equations?
3. How do we factor quadratic expressions in more complex cases?
4. What are extraneous solutions, and why do they occur when squaring equations?
5. How does factoring help in solving polynomial equations?

**Tip:** Always check for extraneous solutions when solving equations involving square roots.

Discover how to solve the square root equation sqrt(3x + 2) = x using factoring. The detailed step-by-step solution explains how to eliminate the square root, rearrange the equation into a quadratic form, and apply factoring techniques. Explore key algebraic concepts and learn why no real solutions exist for this equation. Suitable for students in Grades 9-11.

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