Question:\(\mathrm{x - 29 = (x - a)^2 (x - 29)}\)Which of the following are solutions to the given equation, where...

1. CONSIDER ALL CASES based on the equation structure

• Given equation: \(\mathrm{x - 29 = (x - a)^2 (x - 29)}\)
• The factor \(\mathrm{(x - 29)}\) appears on both sides, creating two scenarios:
  • Case 1: \(\mathrm{x - 29 = 0}\) (which means \(\mathrm{x = 29}\))
  • Case 2: \(\mathrm{x - 29 \neq 0}\) (we can divide by this factor)

2. SIMPLIFY Case 1: When x = 29

• Substitute \(\mathrm{x = 29}\) into the original equation:
  \(\mathrm{29 - 29 = (29 - a)^2 (29 - 29)}\)
  \(\mathrm{0 = (29 - a)^2 \cdot 0}\)
  \(\mathrm{0 = 0}\) ✓
• This is always true, so \(\mathrm{x = 29}\) is definitely a solution

3. SIMPLIFY Case 2: When x ≠ 29

• Since \(\mathrm{x - 29 \neq 0}\), divide both sides by \(\mathrm{(x - 29)}\):
  \(\mathrm{\frac{x - 29}{x - 29} = \frac{(x - a)^2 (x - 29)}{x - 29}}\)
  \(\mathrm{1 = (x - a)^2}\)
• Take the square root of both sides: \(\mathrm{x - a = \pm 1}\)
• Therefore: \(\mathrm{x = a + 1}\) or \(\mathrm{x = a - 1}\)

4. INFER the complete solution set

• Combining both cases: \(\mathrm{x \in \{29, a + 1, a - 1\}}\)

5. APPLY CONSTRAINTS to check each given option

• Option I (\(\mathrm{x = a}\)): Substituting into original equation:
  \(\mathrm{a - 29 = (a - a)^2 (a - 29) = 0 \cdot (a - 29) = 0}\)
  This requires \(\mathrm{a - 29 = 0}\), so \(\mathrm{a = 29}\)
  But we're given \(\mathrm{a \gt 30}\), so this is impossible. Option I is NOT a solution.
• Option II (\(\mathrm{x = a + 1}\)): This matches one solution from Case 2 ✓
• Option III (\(\mathrm{x = 29}\)): This matches the solution from Case 1 ✓

Answer: C (II and III only)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES reasoning: Students see the factor \(\mathrm{(x - 29)}\) on both sides and immediately divide without considering what happens when \(\mathrm{x = 29}\).

They go directly to: \(\mathrm{1 = (x - a)^2}\), getting solutions \(\mathrm{x = a \pm 1}\), and miss that \(\mathrm{x = 29}\) is also a solution. When checking the options, they incorrectly conclude that only option II works, leading them to select an answer that doesn't include option III.

This may lead them to select Choice A (I and II only) or causes confusion when their solution set doesn't match the given options.

Second Most Common Error:

Inadequate APPLY CONSTRAINTS execution: Students correctly find the complete solution set \(\mathrm{\{29, a + 1, a - 1\}}\) but fail to check whether \(\mathrm{x = a}\) actually works given the constraint \(\mathrm{a \gt 30}\).

They see that \(\mathrm{x = a}\) would require \(\mathrm{a = 29}\) but don't recognize this contradicts \(\mathrm{a \gt 30}\), so they incorrectly include option I in their answer. This may lead them to select Choice B (I and III only) or Choice D (I, II, and III).

The Bottom Line:

This problem tests whether students can systematically handle equations where division by zero is a concern while also applying given constraints to verify which theoretical solutions are actually valid.