Question:\(15\mathrm{x}^2 + (\mathrm{d}+45)\mathrm{x} + 3\mathrm{d} = 0\)In the given equation, d is a positive constant. Given that one solution is...

1. TRANSLATE the problem information

• Given equation: \(15x^2 + (d+45)x + 3d = 0\)
• d is a positive constant
• One solution is an integer
• Other solution has form kd (where k is what we're finding)

2. INFER a strategic approach

• Since one root is an integer, we can test small integer values
• The coefficient structure (involving d and 45) suggests trying \(x = -3\)
• We can use Vieta's formulas to connect the roots

3. SIMPLIFY by testing the suspected integer root

• Test \(x = -3\):
  \(15(-3)^2 + (d+45)(-3) + 3d\)
  \(= 135 - 3d - 135 + 3d = 0\) ✓
• This works for any value of d, confirming \(r_1 = -3\)

4. INFER the relationship between roots using Vieta's formulas

• Sum of roots: \(r_1 + r_2 = -\frac{d+45}{15}\)
• Since \(r_1 = -3\) and \(r_2 = kd\):
  \(-3 + kd = -\frac{d+45}{15} = -\frac{d}{15} - 3\)

5. SIMPLIFY to solve for k

• From \(-3 + kd = -\frac{d}{15} - 3\):
  \(kd = -\frac{d}{15}\)
• APPLY CONSTRAINTS: Since \(d \gt 0\), we can divide both sides by d:
  \(k = -\frac{1}{15}\)

6. INFER verification using product of roots

• Product should equal \(\frac{d}{5}\): \((-3)(kd) = (-3)(-\frac{d}{15}) = \frac{3d}{15} = \frac{d}{5}\) ✓

Answer: (B) \(-\frac{1}{15}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that the integer root constraint provides a strategic starting point. Instead, they might attempt to use the quadratic formula directly with the parameter d, leading to complex expressions that are difficult to work with. This often results in confusion and guessing rather than systematic solution.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students may correctly identify \(x = -3\) as the integer root but make algebraic errors when manipulating the equation \(kd = -\frac{d}{15}\). A common mistake is failing to properly divide by d or incorrectly handling the negative signs, potentially leading them to select Choice (C) \(\frac{1}{15}\) by dropping the negative sign.

The Bottom Line:

This problem rewards strategic thinking over brute force calculation. The key insight is recognizing that the integer root constraint, combined with the specific coefficient structure, points toward testing \(x = -3\) rather than attempting complex parameter manipulation from the start.