The equation \((5\mathrm{x} + \mathrm{a})(4\mathrm{x} + \mathrm{b}) = 0\) has two real solutions, where a and b are positive constants....

1. TRANSLATE the problem information

• Given: \((5\mathrm{x} + \mathrm{a})(4\mathrm{x} + \mathrm{b}) = 0\) where a, b are positive constants
• Need to find: The constant k where the product of solutions equals \(\mathrm{k} \cdot \mathrm{ab}\)

2. INFER the solution approach

• Since we have a factored quadratic set equal to zero, we can use the zero-product property
• This means each factor must equal zero to find the individual solutions
• Then we'll find the product of these solutions

3. SIMPLIFY to find each solution

• From \(5\mathrm{x} + \mathrm{a} = 0\):
  \(5\mathrm{x} = -\mathrm{a}\), so \(\mathrm{x}_1 = -\mathrm{a}/5\)
• From \(4\mathrm{x} + \mathrm{b} = 0\):
  \(4\mathrm{x} = -\mathrm{b}\), so \(\mathrm{x}_2 = -\mathrm{b}/4\)

4. SIMPLIFY to find the product of solutions

• \(\mathrm{Product} = \mathrm{x}_1 \times \mathrm{x}_2 = (-\mathrm{a}/5) \times (-\mathrm{b}/4)\)
• Since negative times negative equals positive: \((-\mathrm{a}/5) \times (-\mathrm{b}/4) = \mathrm{ab}/20\)

5. INFER the value of k

• We know the product equals \(\mathrm{k} \cdot \mathrm{ab}\)
• So: \(\mathrm{ab}/20 = \mathrm{k} \cdot \mathrm{ab}\)
• Therefore: \(\mathrm{k} = 1/20\)

Answer: (B) 1/20

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may try to expand the factored form instead of using the zero-product property directly. They might expand \((5\mathrm{x} + \mathrm{a})(4\mathrm{x} + \mathrm{b})\) to get \(20\mathrm{x}^2 + (5\mathrm{b} + 4\mathrm{a})\mathrm{x} + \mathrm{ab} = 0\) and then struggle with applying the quadratic formula or making calculation errors in the expansion process. This indirect approach increases chances of arithmetic mistakes and may lead them to select Choice (A) (1/40) if they make errors in the expansion coefficients.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify that \(\mathrm{x}_1 = -\mathrm{a}/5\) and \(\mathrm{x}_2 = -\mathrm{b}/4\) but make sign errors when calculating the product. They might compute \((-\mathrm{a}/5) \times (-\mathrm{b}/4)\) as \(-\mathrm{ab}/20\) instead of \(+\mathrm{ab}/20\), forgetting that negative times negative equals positive. This leads them to get confused about the final coefficient and potentially guess among the remaining choices.

The Bottom Line:

This problem tests whether students can efficiently use the zero-product property instead of unnecessarily expanding the expression. The key insight is recognizing that factored form gives you direct access to the solutions, making the product calculation straightforward.