To solve the equation $\sqrt{x - 12} + 2 = \sqrt{x}$, we can follow these steps:

Step 1: Isolate one of the square roots

First, subtract $2$ from both sides to isolate $\sqrt{x - 12}$:

$\sqrt{x - 12} = \sqrt{x} - 2$

Step 2: Square both sides

Next, square both sides to eliminate the square roots:

$(\sqrt{x - 12})^2 = (\sqrt{x} - 2)^2$

This simplifies to:

$x - 12 = (\sqrt{x})^2 - 2 \times \sqrt{x} \times 2 + 2^2$

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

Step 3: Simplify the equation

Now, subtract $x$ from both sides:

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

Subtract $4$ from both sides:

$-16 = -4\sqrt{x}$

Step 4: Solve for $\sqrt{x}$

Divide both sides by $-4$:

$\sqrt{x} = 4$

Step 5: Solve for $x$

Now, square both sides to solve for $x$:

$x = 4^2 = 16$

Step 6: Check for extraneous solutions

We substitute $x = 16$ into the original equation to verify:

$\sqrt{16 - 12} + 2 = \sqrt{16}$

$\sqrt{4} + 2 = 4$

$2 + 2 = 4$

Since both sides are equal, $x = 16$ is a valid solution.

Final Answer: $x = 16$

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Do you want more details, or have any questions about the steps?

Here are 5 related questions:

1. What if the equation had a different constant instead of 12?
2. How would the solution change if the equation was $\sqrt{x + 12} + 2 = \sqrt{x}$?
3. Can you solve $\sqrt{x - 5} + 1 = \sqrt{x + 1}$?
4. What happens if there’s no solution for similar equations involving square roots?
5. How can you check for extraneous solutions in more complex equations?

Tip: Always check for extraneous solutions when squaring both sides in equations with square roots.