Question:If \(12(\mathrm{y} - 4)^2 = 12(81)\), what is the negative solution for y?

1. SIMPLIFY the equation by removing common factors

• Given: \(12(\mathrm{y} - 4)^2 = 12(81)\)
• Notice both sides have factor of 12
• Divide both sides by 12: \((\mathrm{y} - 4)^2 = 81\)

2. CONSIDER ALL CASES when taking square roots

• When we have \((\mathrm{y} - 4)^2 = 81\), taking the square root gives us:
• \(\mathrm{y} - 4 = ±\sqrt{81}\)
• \(\mathrm{y} - 4 = ±9\)
• This means: \(\mathrm{y} - 4 = 9\) OR \(\mathrm{y} - 4 = -9\)

3. SIMPLIFY to solve both linear equations

• Case 1: \(\mathrm{y} - 4 = 9\) → \(\mathrm{y} = 13\)
• Case 2: \(\mathrm{y} - 4 = -9\) → \(\mathrm{y} = -5\)

4. APPLY CONSTRAINTS based on what the question asks

• The question specifically asks for "the negative solution"
• We have two solutions: \(\mathrm{y} = 13\) and \(\mathrm{y} = -5\)
• The negative solution is \(\mathrm{y} = -5\)

Answer: -5

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES reasoning: Students forget that taking the square root of both sides produces both positive and negative solutions. They might only consider \(\mathrm{y} - 4 = 9\), leading to \(\mathrm{y} = 13\) as their only solution.

This causes them to provide 13 instead of -5, or to get confused about why there should be a "negative solution" when they only found one answer.

Second Most Common Error:

Poor APPLY CONSTRAINTS execution: Students correctly find both solutions (\(\mathrm{y} = 13\) and \(\mathrm{y} = -5\)) but misread the question and provide the positive solution instead of the negative one.

This leads them to answer 13 when the question specifically asks for the negative solution.

The Bottom Line:

This problem tests whether students understand that quadratic-type equations typically have two solutions, and whether they can follow specific instructions about which solution to report. The algebra itself is straightforward once the ± aspect is recognized.