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To solve the given radical equation \(\sqrt{x + 6} = x - 6\), we'll proceed by isolating \(x\) and checking for any extraneous solutions.

### Step 1: Square both sides of the equation
To eliminate the square root, we'll square both sides:
\[
(\sqrt{x + 6})^2 = (x - 6)^2
\]
This simplifies to:
\[
x + 6 = (x - 6)^2
\]

### Step 2: Expand the right side of the equation
Expanding the square on the right side:
\[
x + 6 = (x - 6)(x - 6) = x^2 - 12x + 36
\]

### Step 3: Rearrange the equation
Move all terms to one side to set the equation to zero:
\[
0 = x^2 - 12x + 36 - x - 6
\]
Simplify further:
\[
0 = x^2 - 13x + 30
\]

### Step 4: Solve the quadratic equation
Now, we'll solve the quadratic equation \(x^2 - 13x + 30 = 0\) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 1\), \(b = -13\), and \(c = 30\). Plugging in these values:
\[
x = \frac{13 \pm \sqrt{(-13)^2 - 4(1)(30)}}{2(1)}
\]
\[
x = \frac{13 \pm \sqrt{169 - 120}}{2}
\]
\[
x = \frac{13 \pm \sqrt{49}}{2}
\]
\[
x = \frac{13 \pm 7}{2}
\]
This gives two possible solutions:
\[
x_1 = \frac{13 + 7}{2} = \frac{20}{2} = 10
\]
\[
x_2 = \frac{13 - 7}{2} = \frac{6}{2} = 3
\]

### Step 5: Check for extraneous solutions
We must check each solution in the original equation:

- For \(x = 10\):
\[
\sqrt{10 + 6} = 10 - 6
\]
\[
\sqrt{16} = 4 \quad \text{which is true}
\]

- For \(x = 3\):
\[
\sqrt{3 + 6} = 3 - 6
\]
\[
\sqrt{9} = -3 \quad \text{which is false, since the square root of a number is non-negative}
\]

### Conclusion
The only valid solution is \(x = 10\).

Would you like more details or have any questions? Here are five related questions to consider:

1. How do you determine if a solution is extraneous in a radical equation?
2. What are other methods for solving radical equations?
3. Can a radical equation have more than one valid solution? If so, under what conditions?
4. How does the domain of the original equation affect possible solutions?
5. What is the importance of checking proposed solutions in equations involving radicals?

**Tip:** Always remember to check each solution in the original equation when solving radical equations, as squaring both sides can introduce extraneous solutions.

Solve the radical equation. Check all proposed solutions: \(\sqrt{x + 6} = x - 6\)

This problem involves solving the radical equation \(\sqrt{x + 6} = x - 6\) by squaring both sides, expanding, and solving the resulting quadratic equation. The solution also includes checking for extraneous solutions. Suitable for high school students in Grades 9-12.

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