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Find a solution and extraneous solution if any

To solve the equation \(\sqrt{x - 2} = \sqrt{3x - 1}\), we will follow these steps:

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

### Step 2: Rearrange the equation to isolate \(x\).
\[
x - 3x = -1 + 2
\]
\[
-2x = 1
\]
\[
x = -\frac{1}{2}
\]

### Step 3: Check for extraneous solutions.
Substitute \(x = -\frac{1}{2}\) back into the original equation:
- Left side: \(\sqrt{-\frac{1}{2} - 2} = \sqrt{-\frac{1}{2} - \frac{4}{2}} = \sqrt{-\frac{5}{2}}\) (This is not a real number, so it's invalid.)
- Right side: \(\sqrt{3(-\frac{1}{2}) - 1} = \sqrt{-\frac{3}{2} - 1} = \sqrt{-\frac{5}{2}}\) (This is also not a real number.)

Since \(\sqrt{x - 2}\) and \(\sqrt{3x - 1}\) must both be real numbers, \(x = -\frac{1}{2}\) is not a valid solution. 

### Conclusion:
There is no real solution to this equation, and the solution \(x = -\frac{1}{2}\) is extraneous.

Would you like to go over any specific steps or have any questions?

Here are five related questions to further your understanding:
1. What is an extraneous solution, and why does it occur?
2. How would you solve an equation involving higher-degree radicals?
3. Can you think of a scenario where squaring both sides of an equation might introduce an extraneous solution?
4. What would happen if the equation had a cubic root instead of a square root?
5. How can you verify your solution when solving radical equations?

**Tip:** Always check your solutions when solving radical equations by substituting them back into the original equation to ensure they are valid.

Solve the equation: √(x - 2) = √(3x - 1). Find the solution and check for any extraneous solutions.

Learn how to solve the radical equation √(x - 2) = √(3x - 1) and identify any extraneous solutions. This problem involves algebraic manipulation, squaring both sides of the equation, and verifying potential solutions. Suitable for students in Grades 10-12.

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