If \((2\mathrm{y} + 6)^2 = 196\), what is the positive value of y + 3?

1. TRANSLATE the problem information

• Given information:
  • \((2\mathrm{y} + 6)^2 = 196\)
  • Need to find the positive value of \(\mathrm{y} + 3\)

2. SIMPLIFY by taking square roots

• Take the square root of both sides: \(\sqrt{(2\mathrm{y} + 6)^2} = \sqrt{196}\)
• This gives us: \(2\mathrm{y} + 6 = ±14\)
• Note: We must consider both the positive and negative square root

3. CONSIDER ALL CASES from the square root

• Case 1: \(2\mathrm{y} + 6 = 14\)
  • Subtract 6: \(2\mathrm{y} = 8\)
  • Divide by 2: \(\mathrm{y} = 4\)
  • Calculate \(\mathrm{y} + 3\): \(\mathrm{y} + 3 = 4 + 3 = 7\)
• Case 2: \(2\mathrm{y} + 6 = -14\)
  • Subtract 6: \(2\mathrm{y} = -20\)
  • Divide by 2: \(\mathrm{y} = -10\)
  • Calculate \(\mathrm{y} + 3\): \(\mathrm{y} + 3 = -10 + 3 = -7\)

4. APPLY CONSTRAINTS to select the final answer

• We found two possible values: \(\mathrm{y} + 3 = 7\) or \(\mathrm{y} + 3 = -7\)
• The problem asks for the positive value of \(\mathrm{y} + 3\)
• Therefore, our answer is 7

Answer: C) 7

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students often take only the positive square root, writing \(2\mathrm{y} + 6 = 14\) and solving to get \(\mathrm{y} = 4\), then \(\mathrm{y} + 3 = 7\). While this happens to give the correct final answer, it demonstrates incomplete understanding of square root properties. In other problems, this approach could lead to missing valid solutions.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students correctly find both \(\mathrm{y} + 3 = 7\) and \(\mathrm{y} + 3 = -7\), but fail to carefully read that the problem asks for the "positive value." They might select Choice A (-7) by picking the first value they calculated or by misreading the constraint.

The Bottom Line:

This problem tests whether students understand that squaring creates a quadratic relationship, even when the original equation looks linear. The key insight is recognizing that taking square roots always produces two cases that must both be analyzed before applying the final constraint.