Question:If \(19(\mathrm{n} + 4)^2 = 19(49)\), what is the negative value of n that is a solution to the equation?

1. INFER the best solution strategy

• Looking at \(19(n + 4)^2 = 19(49)\), I notice both sides have the common factor 19
• Strategy: Simplify by dividing both sides by 19, then use the square root method

2. SIMPLIFY by removing the common factor

• Divide both sides by 19:

\(19(n + 4)^2 \div 19 = 19(49) \div 19\)

\((n + 4)^2 = 49\)

3. INFER and apply the square root method

• To solve \((n + 4)^2 = 49\), I take the square root of both sides
• Key insight: \(\sqrt{49} = 7\), but when solving equations, square roots give us both positive and negative values

4. CONSIDER ALL CASES from the square root

• Taking square root: \(\sqrt{(n + 4)^2} = \pm\sqrt{49}\)
• This gives us: \(n + 4 = \pm 7\)
• Now I have two separate cases to solve:

Case 1: \(n + 4 = 7\)

• Subtract 4: \(n = 3\)

Case 2: \(n + 4 = -7\)

• Subtract 4: \(n = -11\)

5. TRANSLATE the question requirement and select the answer

• The question asks specifically for "the negative value of n"
• From my two solutions (\(n = 3\) and \(n = -11\)), the negative value is \(n = -11\)

Answer: -11

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students might not recognize that they should divide both sides by 19 first. Instead, they might try to expand \(19(49) = 19 \times 49 = 931\) and work with the equation \(19(n + 4)^2 = 931\), making the problem much more complicated than necessary.

This leads to confusion and abandoning systematic solution, resulting in guessing.

Second Most Common Error:

Inadequate CONSIDER ALL CASES execution: Students correctly simplify to \((n + 4)^2 = 49\) and find \(n + 4 = 7\), giving \(n = 3\). However, they forget that square roots produce both positive and negative solutions, so they miss the second case where \(n + 4 = -7\).

Since they only find \(n = 3\) (which is positive), they become confused when the question asks for the negative value and may guess or incorrectly conclude there's no negative solution.

The Bottom Line:

This problem tests whether students can efficiently simplify equations by removing common factors and whether they remember that square root operations in equation-solving contexts always produce two cases to consider.