A rectangle has a length that is 7 units more than its width. If the area of the rectangle is...

1. TRANSLATE the problem information

• Given information:
  • Length = \(\mathrm{width + 7}\) units
  • Area = \(\mathrm{144}\) square units
  • Need to find: width

• Let \(\mathrm{w = width}\), then \(\mathrm{length = w + 7}\)

2. TRANSLATE the area condition into an equation

• Using \(\mathrm{Area = length \times width}\):
  • \(\mathrm{144 = (w + 7) \times w}\)
  • \(\mathrm{144 = w^2 + 7w}\)

3. SIMPLIFY to standard quadratic form

• Rearrange: \(\mathrm{w^2 + 7w - 144 = 0}\)
• Now we need to solve this quadratic equation

4. SIMPLIFY by factoring the quadratic

• Look for two numbers that multiply to -144 and add to 7
• Try 16 and -9: \(\mathrm{(16)(-9) = -144}\) ✓ and \(\mathrm{16 + (-9) = 7}\) ✓
• Factor: \(\mathrm{(w + 16)(w - 9) = 0}\)
• Solutions: \(\mathrm{w = -16}\) or \(\mathrm{w = 9}\)

5. APPLY CONSTRAINTS to select the valid answer

• Since width must be positive in a geometric context: \(\mathrm{w = 9}\)
• Verify: If \(\mathrm{width = 9}\), then \(\mathrm{length = 16}\), and \(\mathrm{area = 9 \times 16 = 144}\) ✓

Answer: B) 9

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to set up the relationship between length and width correctly. They might write \(\mathrm{length = 7}\) instead of \(\mathrm{length = width + 7}\), leading to the equation \(\mathrm{7w = 144}\), which gives \(\mathrm{w \approx 20.6}\). Since this doesn't match any answer choice exactly, this leads to confusion and guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{w^2 + 7w - 144 = 0}\) but struggle with factoring. They might attempt to use the quadratic formula but make calculation errors, particularly with the discriminant \(\mathrm{\sqrt{49 + 576} = \sqrt{625} = 25}\). Arithmetic mistakes here could lead them to incorrect values that might seem to match Choice A (7) or Choice C (16).

The Bottom Line:

This problem combines algebraic manipulation with geometric reasoning. The key challenge is translating the word relationship into the correct mathematical expression, then executing the quadratic solution accurately while remembering that geometric measurements must be positive.