sqrt(2x + 6) + 4 = x + 3What is the solution set of the equation above?

1. TRANSLATE the problem information

• Given equation: \(\sqrt{2\mathrm{x} + 6} + 4 = \mathrm{x} + 3\)
• Goal: Find all values of x that make this equation true

2. INFER the solution strategy

• The square root term makes this challenging to solve directly
• Strategy: Isolate the radical first, then square both sides to eliminate it
• Key insight: After squaring, we'll need to check our solutions since squaring can create "fake" solutions

3. SIMPLIFY by isolating the radical

• Subtract 4 from both sides: \(\sqrt{2\mathrm{x} + 6} = \mathrm{x} - 1\)
• This puts the radical by itself on the left side

4. SIMPLIFY by squaring both sides

• Square the left side: \((\sqrt{2\mathrm{x} + 6})^2 = 2\mathrm{x} + 6\)
• Square the right side: \((\mathrm{x} - 1)^2 = \mathrm{x}^2 - 2\mathrm{x} + 1\)
• Result: \(2\mathrm{x} + 6 = \mathrm{x}^2 - 2\mathrm{x} + 1\)

5. SIMPLIFY to standard quadratic form

• Move everything to one side: \(\mathrm{x}^2 - 2\mathrm{x} + 1 - 2\mathrm{x} - 6 = 0\)
• Combine like terms: \(\mathrm{x}^2 - 4\mathrm{x} - 5 = 0\)

6. SIMPLIFY by factoring the quadratic

• Look for two numbers that multiply to -5 and add to -4
• Those numbers are -5 and +1
• Factor: \((\mathrm{x} - 5)(\mathrm{x} + 1) = 0\)
• Solutions: \(\mathrm{x} = 5\) or \(\mathrm{x} = -1\)

7. APPLY CONSTRAINTS by checking both solutions

• For \(\mathrm{x} = 5\): \(\sqrt{2(5) + 6} + 4 = \sqrt{16} + 4 = 8\), and \(\mathrm{x} + 3 = 8\) ✓
• For \(\mathrm{x} = -1\): \(\sqrt{2(-1) + 6} + 4 = \sqrt{4} + 4 = 6\), and \(\mathrm{x} + 3 = 2\) ✗
• Only \(\mathrm{x} = 5\) works in the original equation

Answer: B. {5}

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor APPLY CONSTRAINTS skill: Students solve the quadratic correctly and get \(\mathrm{x} = 5\) and \(\mathrm{x} = -1\), but then skip the crucial verification step.

They assume that both solutions from the quadratic are automatically valid for the original radical equation, not realizing that squaring both sides can introduce extraneous solutions. Since \(\mathrm{x} = -1\) doesn't actually satisfy \(\sqrt{2\mathrm{x} + 6} + 4 = \mathrm{x} + 3\), including it makes their solution set incorrect.

This leads them to select Choice C ({-1, 5}) instead of the correct answer.

Second Most Common Error:

Weak SIMPLIFY execution: Students make algebraic errors when expanding \((\mathrm{x} - 1)^2\) or when collecting terms to form the quadratic equation.

For example, they might incorrectly expand \((\mathrm{x} - 1)^2\) as \(\mathrm{x}^2 - 1\) instead of \(\mathrm{x}^2 - 2\mathrm{x} + 1\), or make sign errors when rearranging terms. This leads to a wrong quadratic equation and completely different solutions.

This causes them to get stuck with incorrect values or leads to confusion and guessing.

The Bottom Line:

This problem tests whether students understand that algebraic operations (like squaring) can change the solution set of an equation, making verification essential. The mathematical work itself isn't too complex, but the conceptual understanding about extraneous solutions often trips students up.