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Solve for solution and any extraneous solution

To solve the equation \(\sqrt{3x - 2} = x\), we need to eliminate the square root by squaring both sides. Here's the step-by-step process:

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

### Step 2: Rearrange the equation
Move all terms to one side to set the equation to zero:
\[
x^2 - 3x + 2 = 0
\]

### Step 3: Factor the quadratic equation
Factor the quadratic equation:
\[
(x - 1)(x - 2) = 0
\]
So, the solutions are:
\[
x = 1 \quad \text{or} \quad x = 2
\]

### Step 4: Check for extraneous solutions
We need to check if these solutions satisfy the original equation:

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

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

### Conclusion:
Both solutions \(x = 1\) and \(x = 2\) are valid. There are no extraneous solutions in this case.

Would you like more details or have any questions?

Here are some related questions to consider:
1. How do you solve quadratic equations by factoring?
2. What are extraneous solutions, and why do they occur?
3. How do you solve equations involving square roots?
4. What is the importance of checking solutions in the original equation?
5. Can you always factor quadratic equations easily, or are there other methods?

**Tip:** When solving equations involving square roots, always remember to check your solutions in the original equation to avoid including extraneous solutions.

Solve for solution and any extraneous solution in the equation \(\sqrt{3x - 2} = x\).

Discover how to solve the equation √(3x - 2) = x by squaring both sides and identifying any extraneous solutions. This problem involves key algebraic concepts such as square roots and quadratic equations. Ideal for students in Grades 9-11.

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