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.

Solve the equation: \(\sqrt{x - 12} + 2 = \sqrt{x}\).

Learn how to solve the radical equation \(\sqrt{x - 12} + 2 = \sqrt{x}\) with a detailed step-by-step solution. This problem covers key algebra concepts such as isolating radicals, squaring both sides, and checking for extraneous solutions. Ideal for high school students in Grades 9-12.

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