A rectangle has an area of 45 square inches. The length of the rectangle is 4 inches more than its...

1. TRANSLATE the problem information

• Given information:
  • Rectangle area = 45 square inches
  • Length = width + 4 inches
  • Need to find: width
• What this tells us: We have one equation with one unknown if we let width = \(\mathrm{w}\)

2. TRANSLATE to set up the equation

• Let \(\mathrm{w}\) = width in inches
• Then length = \(\mathrm{w + 4}\) inches
• Using Area = length × width: \(\mathrm{w(w + 4) = 45}\)

3. SIMPLIFY to standard quadratic form

• Expand the left side: \(\mathrm{w^2 + 4w = 45}\)
• Move everything to one side: \(\mathrm{w^2 + 4w - 45 = 0}\)

4. SIMPLIFY by factoring the quadratic

• Need two numbers that multiply to \(\mathrm{-45}\) and add to \(\mathrm{+4}\)
• Those numbers are \(\mathrm{+9}\) and \(\mathrm{-5}\): \(\mathrm{(9)(-5) = -45}\) and \(\mathrm{9 + (-5) = 4}\)
• Factor: \(\mathrm{(w + 9)(w - 5) = 0}\)

5. SIMPLIFY by solving each factor

• Set each factor equal to zero:
  • \(\mathrm{w + 9 = 0}\) → \(\mathrm{w = -9}\)
  • \(\mathrm{w - 5 = 0}\) → \(\mathrm{w = 5}\)

6. APPLY CONSTRAINTS to select the valid solution

• Since width must be positive in real-world context: \(\mathrm{w = 5}\) inches
• Verify: length = \(\mathrm{5 + 4 = 9}\), area = \(\mathrm{5 \times 9 = 45}\) ✓

Answer: C) 5

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students make algebraic errors when expanding \(\mathrm{w(w + 4) = 45}\), possibly writing \(\mathrm{w^2 + 4w = 45}\) incorrectly as \(\mathrm{w^2 + 4 = 45}\), or making sign errors when moving terms.

If they incorrectly get \(\mathrm{w^2 + 4 = 45}\), they'd solve \(\mathrm{w^2 = 41}\), giving \(\mathrm{w \approx 6.4}\), which doesn't match any answer choice. This leads to confusion and guessing.

Second Most Common Error:

Poor factoring execution in SIMPLIFY: Students struggle to factor \(\mathrm{w^2 + 4w - 45 = 0}\) correctly, either finding wrong factor pairs or making sign errors in the factoring process.

For example, they might incorrectly factor as \(\mathrm{(w + 5)(w - 9) = 0}\), leading to \(\mathrm{w = -5}\) or \(\mathrm{w = 9}\). Since width must be positive, they'd choose \(\mathrm{w = 9}\), leading them to select Choice D (9).

The Bottom Line:

This problem tests whether students can bridge word problems and quadratic algebra. The setup isn't too hard, but the multi-step algebraic manipulation creates several opportunities for calculation errors that can derail the solution.