Question:Consider the equation sqrt(m - 2x) = 19 - x, where m is a constant. If this equation has exactly...

1. TRANSLATE the problem requirements

• Given: \(\sqrt{m - 2x} = 19 - x\) needs exactly one real solution for \(x\)
• Find: minimum value of \(3m\)

2. INFER the constraint from square root properties

• Since the right side equals a square root: \(19 - x \geq 0\)
• This means we need \(x \leq 19\) for any valid solution

3. INFER a strategic substitution

• Let \(t = 19 - x\), so \(x = 19 - t\)
• Since \(t = 19 - x\) and \(x \leq 19\), we need \(t \geq 0\)
• Substituting: \(\sqrt{m - 2(19 - t)} = t\)
• SIMPLIFY: \(\sqrt{m - 38 + 2t} = t\)

4. SIMPLIFY by eliminating the square root

• Since both sides are non-negative, square both sides:
• \(m - 38 + 2t = t^2\)
• Rearrange: \(t^2 - 2t + (38 - m) = 0\)

5. INFER the connection to discriminant analysis

• This is a quadratic in \(t\) with discriminant \(\Delta = 4 - 4(38 - m) = 4m - 148\)
• For exactly one solution in \(x\), we need exactly one valid \(t \geq 0\)

6. CONSIDER ALL CASES based on discriminant value

Case 1: \(\Delta = 0\) (one repeated root)

• \(4m - 148 = 0 \Rightarrow m = 37\)
• Root: \(t = 2/2 = 1 \geq 0\) ✓
• This gives exactly one solution ✓

Case 2: \(\Delta \gt 0\) (two distinct real roots)

• \(m \gt 37\)
• Product of roots = \(38 - m\), Sum of roots = \(2\)
• If \(m \gt 38\): product \(\lt 0\), so one positive and one negative root ⟹ exactly one valid \(t\)
• If \(37 \lt m \lt 38\): product \(\gt 0\) and sum \(\gt 0\) ⟹ both roots positive ⟹ two solutions ✗

Case 3: \(\Delta \lt 0\) (no real roots)

• \(m \lt 37\): No solutions ✗

7. APPLY CONSTRAINTS to find valid m values

• Exactly one solution occurs when: \(m = 37\) or \(m \gt 38\)
• Minimum value: \(m = 37\)
• Therefore: \(3m = 3(37) = 111\)

Answer: 111

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor INFER reasoning: Students often try to solve the original equation directly by squaring both sides immediately, without recognizing that this approach makes the case analysis much more complicated because they don't establish the connection between having exactly one solution in \(x\) and having exactly one non-negative solution in \(t\).

Without the strategic substitution, they get bogged down trying to analyze when \(x^2 - 36x + (361 - m) = 0\) has exactly one solution satisfying both \(x \leq 19\) and \(x \leq m/2\), which involves much more complex constraint checking. This leads to confusion and incomplete analysis, causing students to guess or select an incorrect value.

Second Most Common Error:

Inadequate CONSIDER ALL CASES execution: Students may correctly set up the discriminant analysis but fail to properly analyze the sub-cases when \(\Delta \gt 0\). They might conclude that \(m \gt 37\) always works without recognizing that when \(37 \lt m \lt 38\), both quadratic roots are positive, giving two solutions instead of the required one.

This incomplete case analysis leads them to conclude that any \(m \gt 37\) works, giving a minimum of \(3m\) approaching 111 from above, causing confusion about the final answer.

The Bottom Line:

This problem challenges students to see that a clever substitution transforms a complex constraint problem into a more manageable discriminant analysis, requiring systematic case-by-case reasoning about when exactly one non-negative root exists.