Let k be a positive number such that 2k - 7 = 30/k. What is the value of k?5/25615/2

1. TRANSLATE the problem information

• Given equation: \(2\mathrm{k} - 7 = \frac{30}{\mathrm{k}}\)
• Constraint: k is positive
• Find: the value of k

2. INFER the solution strategy

• The fraction 30/k creates complexity that needs to be eliminated first
• Since \(\mathrm{k} ≠ 0\) (it's positive), we can multiply both sides by k
• This will transform the equation into a more manageable quadratic form

3. SIMPLIFY to eliminate the fraction

Multiply both sides by k:

\(\mathrm{k}(2\mathrm{k} - 7) = \mathrm{k}\left(\frac{30}{\mathrm{k}}\right)\)

Distribute on left side and simplify right side:

\(2\mathrm{k}^2 - 7\mathrm{k} = 30\)

4. SIMPLIFY to standard quadratic form

Move all terms to one side:

\(2\mathrm{k}^2 - 7\mathrm{k} - 30 = 0\)

5. SIMPLIFY by factoring the quadratic

To factor, find two numbers that multiply to \((2)(-30) = -60\) and add to -7:

• Those numbers are -12 and 5

Rewrite the middle term using these numbers:

\(2\mathrm{k}^2 - 12\mathrm{k} + 5\mathrm{k} - 30 = 0\)

Factor by grouping:

\(2\mathrm{k}(\mathrm{k} - 6) + 5(\mathrm{k} - 6) = 0\)

\((2\mathrm{k} + 5)(\mathrm{k} - 6) = 0\)

6. SIMPLIFY to find potential solutions

Set each factor equal to zero:

• \(2\mathrm{k} + 5 = 0\) → \(\mathrm{k} = -\frac{5}{2}\)
• \(\mathrm{k} - 6 = 0\) → \(\mathrm{k} = 6\)

7. APPLY CONSTRAINTS to select final answer

• Since the problem states k is positive:
  • \(\mathrm{k} = -\frac{5}{2}\) is negative (rejected)
  • \(\mathrm{k} = 6\) is positive (accepted)

Answer: C. 6

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students attempt to solve the rational equation without first eliminating the fraction, trying to work directly with \(2\mathrm{k} - 7 = \frac{30}{\mathrm{k}}\). They might try cross-multiplication incorrectly or get confused by the mixed linear and rational terms, leading to algebraic mistakes that produce incorrect values.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students correctly solve the quadratic and find both \(\mathrm{k} = -\frac{5}{2}\) and \(\mathrm{k} = 6\), but fail to eliminate the negative solution. Since choice A is 5/2 (the positive version of -5/2), they might select Choice A (5/2) thinking this is correct.

The Bottom Line:

This problem tests whether students can strategically handle rational equations by converting them to familiar quadratic form, then remember to apply real-world constraints to select appropriate solutions.