We are given the equation:

$\sqrt{x - 8} = \sqrt{x} - 2$

To solve this, follow these steps:

Step 1: Isolate one of the square roots

We already have one square root isolated on the left side:

$\sqrt{x - 8} = \sqrt{x} - 2$

Step 2: Square both sides of the equation

To eliminate the square root on the left side, square both sides of the equation:

$(\sqrt{x - 8})^2 = (\sqrt{x} - 2)^2$

This simplifies to:

$x - 8 = (\sqrt{x} - 2)^2$

Step 3: Expand the right-hand side

Now expand the square on the right-hand side:

$x - 8 = (\sqrt{x})^2 - 2 \cdot 2 \cdot \sqrt{x} + (-2)^2$

$x - 8 = x - 4\sqrt{x} + 4$

Step 4: Simplify the equation

Now, simplify the equation:

$x - 8 = x - 4\sqrt{x} + 4$

Subtract $x$ from both sides:

$-8 = -4\sqrt{x} + 4$

Step 5: Solve for $\sqrt{x}$

Now isolate the square root term:

$-8 - 4 = -4\sqrt{x}$

$-12 = -4\sqrt{x}$

Divide both sides by $-4$:

$3 = \sqrt{x}$

Step 6: Solve for $x$

Square both sides of the equation to get rid of the square root:

$3^2 = x$

$x = 9$

Step 7: Verify the solution

To verify, substitute $x = 9$ back into the original equation:

$\sqrt{9 - 8} = \sqrt{9} - 2$

$\sqrt{1} = 3 - 2$

$1 = 1$

Since both sides are equal, $x = 9$ is the correct solution.

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Would you like a more detailed explanation or any clarification?

Here are some related questions:

1. How do you solve equations involving square roots with multiple terms?
2. How can you check if your solution is extraneous in square root equations?
3. What happens if you encounter a negative result when squaring both sides?
4. Can you solve this equation using substitution methods?
5. How would you approach solving equations with higher-order roots (e.g., cube roots)?

Tip: Always check for extraneous solutions when squaring both sides of an equation. Squaring both sides can sometimes introduce solutions that do not satisfy the original equation.