Let sqrt(3x + 12) = |x - 2|, where x is a real number. Which of the following is the...

1. APPLY CONSTRAINTS to find the domain

• The square root \(\sqrt{3x + 12}\) requires: \(3x + 12 \geq 0\)
• Solving: \(3x \geq -12\), so \(x \geq -4\)
• This domain constraint will be crucial for checking our final answers

2. INFER the solution strategy

• Notice both sides are non-negative: \(\sqrt{3x + 12} \geq 0\) and \(|x - 2| \geq 0\)
• Since both sides are non-negative, we can square both sides without losing or gaining solutions
• Squaring will eliminate both the radical and absolute value signs

3. SIMPLIFY by squaring both sides

• \((\sqrt{3x + 12})^2 = (|x - 2|)^2\)
• Left side: \(3x + 12\)
• Right side: \((x - 2)^2 = x^2 - 4x + 4\)
• Result: \(3x + 12 = x^2 - 4x + 4\)

4. SIMPLIFY to standard quadratic form

• Move all terms to one side: \(0 = x^2 - 4x + 4 - 3x - 12\)
• Combine like terms: \(0 = x^2 - 7x - 8\)

5. SIMPLIFY by factoring the quadratic

• Need two numbers that multiply to -8 and add to -7
• Those numbers are -8 and 1: \((-8)(1) = -8\), \((-8) + 1 = -7\)
• Factor: \((x - 8)(x + 1) = 0\)
• Solutions: \(x = 8\) or \(x = -1\)

6. APPLY CONSTRAINTS to verify solutions

• Check domain: Both \(-1 \geq -4\) ✓ and \(8 \geq -4\) ✓
• Verify in original equation:
  • \(x = 8\): \(\sqrt{24 + 12} = 6\) and \(|8 - 2| = 6\) ✓
  • \(x = -1\): \(\sqrt{9} = 3\) and \(|-3| = 3\) ✓

Answer: D. {-1, 8}

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when expanding \((x - 2)^2\) or when collecting like terms in the quadratic equation. For example, they might incorrectly expand \((x - 2)^2\) as \(x^2 - 4\) instead of \(x^2 - 4x + 4\), or make sign errors when moving terms to get the standard form. This leads to an incorrect quadratic equation and wrong factors, causing them to select an incorrect answer choice or abandon the problem entirely.

Second Most Common Error:

Inadequate APPLY CONSTRAINTS reasoning: Students find the correct quadratic solutions \(x = 8\) and \(x = -1\), but fail to verify these solutions in the original equation. They might assume that since the algebra worked out, both solutions must be valid. However, verification is essential with radical and absolute value equations because squaring can sometimes introduce extraneous solutions. Without this verification step, they might not realize they need to check both solutions work.

The Bottom Line:

This problem challenges students to manage multiple algebraic steps while maintaining careful attention to domain restrictions and solution verification. The combination of radicals and absolute values requires strategic thinking about when operations preserve equation equivalence.