\(\sqrt{\mathrm{x - 10}} = (\mathrm{x - k})\sqrt{\mathrm{x - 10}}\)In the equation above, k is a constant such that k gt...

Brief Solution

Concepts tested: Algebraic equation solving, case analysis, constraint application

Primary process skills: Simplify, Apply Constraints

Essential Steps:

• Rearrange \(\mathrm{x - 29 = (x - a)(x - 29)}\) and consider cases where \(\mathrm{x = 29}\) and \(\mathrm{x ≠ 29}\)
• For \(\mathrm{x ≠ 29}\): divide by \(\mathrm{(x - 29)}\) to get \(\mathrm{x = a + 1}\)
• For \(\mathrm{x = 29}\): equation becomes \(\mathrm{0 = 0}\) (always true)
• Test \(\mathrm{x = a}\): requires \(\mathrm{a = 29}\), but constraint \(\mathrm{a \gt 30}\) makes this impossible

Final Answer: C) II and III only

Top 3 Faltering Points

1. Constraint Violation - Phase: Selecting Answer → Choice B (I and III only)
   • Process skill failure: Apply Constraints
   • Students incorrectly conclude \(\mathrm{x = a}\) is valid without checking that it requires \(\mathrm{a = 29}\), violating the given constraint \(\mathrm{a \gt 30}\).
2. Incomplete Case Analysis - Phase: Devising Approach → Choice A (I and II only)
   • Process skill failure: Consider All Cases
   • Students miss that \(\mathrm{x = 29}\) makes the equation \(\mathrm{0 = 0}\), focusing only on the algebraic solution \(\mathrm{x = a + 1}\).
3. Division Error - Phase: Executing Approach → Various
   • Computational error: Algebraic manipulation
   • Students incorrectly divide by \(\mathrm{(x - 29)}\) without considering the case where \(\mathrm{x = 29}\), leading to missing this valid solution.

Detailed Solution

This problem tests your ability to solve an equation involving a parameter while respecting given constraints. Let's work through it systematically.

Starting equation: \(\mathrm{x - 29 = (x - a)(x - 29)}\)

Process Skill: SIMPLIFY - Let's rearrange this equation to better understand its structure. Notice that both sides contain the factor \(\mathrm{(x - 29)}\), which suggests we should consider what happens when this factor equals zero.

Case Analysis Approach:

Since we have \(\mathrm{(x - 29)}\) appearing on both sides, we need to consider two distinct cases:

Case 1: x = 29

Substituting \(\mathrm{x = 29}\):

\(\mathrm{29 - 29 = (29 - a)(29 - 29)}\)

\(\mathrm{0 = (29 - a) × 0}\)

\(\mathrm{0 = 0}\)

This equation is always true regardless of the value of a.
Process Skill: INFER - This tells us that \(\mathrm{x = 29}\) is always a solution to our equation.

Case 2: x ≠ 29

Since \(\mathrm{x ≠ 29}\), we can divide both sides by \(\mathrm{(x - 29)}\):

\(\mathrm{\frac{(x - 29)}{(x - 29)} = \frac{(x - a)(x - 29)}{(x - 29)}}\)

\(\mathrm{1 = x - a}\)

\(\mathrm{x = a + 1}\)

Process Skill: APPLY CONSTRAINTS - We need to verify that this solution is valid. Since \(\mathrm{a \gt 30}\), we have \(\mathrm{x = a + 1 \gt 31}\), which means \(\mathrm{x ≠ 29}\), so our division was legitimate.

Testing the Given Options:

Option I: x = a

Substituting \(\mathrm{x = a}\) into the original equation:

\(\mathrm{a - 29 = (a - a)(a - 29)}\)

\(\mathrm{a - 29 = 0 × (a - 29)}\)

\(\mathrm{a - 29 = 0}\)

This would require \(\mathrm{a = 29}\).
Process Skill: APPLY CONSTRAINTS - However, we're told that \(\mathrm{a \gt 30}\), so a cannot equal 29. Therefore, \(\mathrm{x = a}\) is NOT a solution.

Option II: x = a + 1

We already found this is a solution from our Case 2 analysis above. Let's verify by substitution:

\(\mathrm{(a + 1) - 29 = ((a + 1) - a)((a + 1) - 29)}\)

\(\mathrm{a - 28 = (1)(a - 28)}\)

\(\mathrm{a - 28 = a - 28}\) ✓

Option III: x = 29

We already verified this in Case 1 above. ✓

Process Skill: CONSIDER ALL CASES - We've systematically examined both the algebraic solution \(\mathrm{(x = a + 1)}\) and the special case solution \(\mathrm{(x = 29)}\), ensuring we haven't missed any valid solutions.

Conclusion: The solutions are II \(\mathrm{(a + 1)}\) and III \(\mathrm{(29)}\), making the answer C) II and III only.

Detailed Faltering Points Analysis

Errors while devising the approach:

• Incomplete Case Analysis: Students may focus solely on algebraic manipulation and miss that \(\mathrm{x = 29}\) creates a special case where \(\mathrm{0 = 0}\), automatically satisfying the equation. This leads to missing option III.
• Overlooking Factor Structure: Not recognizing that \(\mathrm{(x - 29)}\) appears on both sides suggests considering when this factor equals zero, leading to incomplete solution sets.

Errors while executing the approach:

• Invalid Division: Dividing by \(\mathrm{(x - 29)}\) without first checking \(\mathrm{x = 29}\) case, which is mathematically incorrect and causes students to miss a valid solution.
• Algebraic Manipulation Errors: Making arithmetic mistakes when expanding \(\mathrm{(x - a)(x - 29)}\) or when solving the resulting equation.
• Substitution Errors: Incorrectly substituting test values back into the original equation, leading to wrong validation of potential solutions.

Errors while selecting the answer:

• Constraint Violation: The most critical error is concluding that \(\mathrm{x = a}\) is valid without checking that this requires \(\mathrm{a = 29}\), which violates the constraint \(\mathrm{a \gt 30}\). This leads students to incorrectly choose options containing I.
• Partial Solution Recognition: Students may find one correct solution (like \(\mathrm{x = a + 1}\)) and select answer choices containing only that solution, missing other valid solutions like \(\mathrm{x = 29}\).