QUESTION STEM:Consider the quadratic equation 5x^2 + kx + 180 = 0, where k is a real parameter.The equation has...

1. TRANSLATE the problem requirement

• Given: Quadratic equation \(5x^2 + kx + 180 = 0\)
• Need: Greatest integer k where equation has no real solutions
• TRANSLATE "no real solutions" to mathematical condition: discriminant \(\lt 0\)

2. INFER the discriminant approach

• For quadratic \(ax^2 + bx + c = 0\), discriminant \(D = b^2 - 4ac\)
• Here: \(a = 5, b = k, c = 180\)
• Strategy: Set up \(D \lt 0\) and solve for k

3. SIMPLIFY the discriminant inequality

• Calculate discriminant: \(D = k^2 - 4(5)(180) = k^2 - 3600\)
• Set up inequality for no real solutions: \(k^2 - 3600 \lt 0\)
• SIMPLIFY: \(k^2 \lt 3600\)
• Take square root of both sides: \(|k| \lt 60\)
• This gives us: \(-60 \lt k \lt 60\)

4. APPLY CONSTRAINTS to find the answer

• We need the greatest INTEGER in the open interval (-60, 60)
• Since \(k \lt 60\) (not \(k \leq 60\)), the value \(k = 60\) is excluded
• Greatest integer less than 60 is 59

5. Verify the answer

• Check \(k = 59\):
  \(D = 59^2 - 3600\)
  \(= 3481 - 3600\)
  \(= -119 \lt 0\) ✓
• Check \(k = 60\): \(D = 60^2 - 3600 = 0\) (gives one real solution, not zero)

Answer: 59

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak APPLY CONSTRAINTS reasoning: Students correctly solve to get \(-60 \lt k \lt 60\), but then incorrectly conclude that \(k = 60\) is acceptable since "60 is less than or equal to 60."

They miss that the interval is open (\(k \lt 60\), not \(k \leq 60\)), meaning \(k = 60\) is excluded. At \(k = 60\), the discriminant equals zero, giving exactly one real solution rather than no real solutions. This leads them to answer 60 instead of the correct 59.

Second Most Common Error:

Poor TRANSLATE execution: Students misinterpret "no real solutions" and set discriminant \(\geq 0\) instead of discriminant \(\lt 0\).

This leads them to solve \(k^2 - 3600 \geq 0\), getting \(k \leq -60\) or \(k \geq 60\), and then selecting 60 as the "greatest integer." This fundamental misunderstanding of the relationship between discriminant and solution types causes them to work with the wrong inequality from the start.

The Bottom Line:

This problem tests whether students truly understand what "no real solutions" means mathematically and can correctly apply interval constraints. The key insight is recognizing that discriminant = 0 gives one real solution (not zero), so we need strictly negative discriminant, creating an open interval that excludes the boundary values.