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

### Step 1: Isolate the square root term
Subtract 2 from both sides:
\[
\sqrt{1 + x} = x - 5
\]

### Step 2: Square both sides to eliminate the square root
Square both sides of the equation:
\[
(\sqrt{1 + x})^2 = (x - 5)^2
\]
This simplifies to:
\[
1 + x = (x - 5)^2
\]

### Step 3: Expand the square on the right-hand side
\[
1 + x = (x - 5)(x - 5)
\]
Expanding the right-hand side:
\[
1 + x = x^2 - 10x + 25
\]

### Step 4: Rearrange into a quadratic equation
Move all terms to one side to set the equation equal to 0:
\[
0 = x^2 - 11x + 24
\]
This can be rewritten as:
\[
x^2 - 11x + 24 = 0
\]

### Step 5: Solve the quadratic equation
Factor the quadratic equation:
\[
(x - 3)(x - 8) = 0
\]
Thus, the solutions are:
\[
x = 3 \quad \text{or} \quad x = 8
\]

### Step 6: Check for extraneous solutions
Since we squared both sides in Step 2, we must check both solutions in the original equation.

1. **For \( x = 3 \):**
   \[
   2 + \sqrt{1 + 3} = 3 - 3
   \]
   \[
   2 + \sqrt{4} = 0
   \]
   \[
   2 + 2 = 0 \quad \text{(False)}
   \]
   Therefore, \( x = 3 \) is an extraneous solution.

2. **For \( x = 8 \):**
   \[
   2 + \sqrt{1 + 8} = 8 - 3
   \]
   \[
   2 + \sqrt{9} = 5
   \]
   \[
   2 + 3 = 5 \quad \text{(True)}
   \]
   Therefore, \( x = 8 \) is a valid solution.

### Final Answer:
The solution is \( x = 8 \).

Would you like further details or have any questions?

Here are five related questions:
1. How would you solve \( 1 + \sqrt{x} = x - 3 \)?
2. What is the solution to \( 3 + \sqrt{2 + x} = x - 1 \)?
3. Can you solve \( \sqrt{x - 2} = x - 4 \)?
4. How do you handle extraneous solutions in equations involving square roots?
5. What happens if the equation was \( 2 + \sqrt{x - 1} = x + 2 \)?

**Tip:** When solving equations with square roots, always check your solutions by substituting them back into the original equation to avoid extraneous results.

Learn how to solve the equation 2 + √(1 + x) = x - 3 with detailed steps and explanations. This problem involves equations with square roots and quadratic solving methods. Suitable for advanced high school students.

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