The length of a rectangle is 5 centimeters greater than 2 times its width w centimeters. The area of the...

1. TRANSLATE the problem information

• Given information:
  • Width = \(\mathrm{w}\) centimeters
  • Length = '5 centimeters greater than 2 times the width' = \(\mathrm{2w + 5}\)
  • Area = 12 square centimeters

• What we need to find: The value of \(\mathrm{w}\)

2. TRANSLATE the area relationship into an equation

• Using the rectangle area formula: \(\mathrm{Area = length \times width}\)
• Substituting our expressions: \(\mathrm{12 = (2w + 5) \times w}\)
• This gives us: \(\mathrm{12 = 2w^2 + 5w}\)

3. SIMPLIFY to standard quadratic form

• Rearrange the equation: \(\mathrm{2w^2 + 5w - 12 = 0}\)
• This is now in the standard form \(\mathrm{ax^2 + bx + c = 0}\) where:
  • \(\mathrm{a = 2, b = 5, c = -12}\)

4. SIMPLIFY using the quadratic formula

• Apply the quadratic formula: \(\mathrm{w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\)
• Substitute values: \(\mathrm{w = \frac{-5 \pm \sqrt{5^2 - 4(2)(-12)}}{2(2)}}\)
• Calculate the discriminant: \(\mathrm{w = \frac{-5 \pm \sqrt{25 + 96}}{4}}\)
• Simplify: \(\mathrm{w = \frac{-5 \pm \sqrt{121}}{4} = \frac{-5 \pm 11}{4}}\)

5. CONSIDER ALL CASES to find both solutions

• First solution: \(\mathrm{w = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2}}\)
• Second solution: \(\mathrm{w = \frac{-5 - 11}{4} = \frac{-16}{4} = -4}\)

6. APPLY CONSTRAINTS to select the valid answer

• Since width represents a physical dimension, it must be positive
• Therefore: \(\mathrm{w = \frac{3}{2}}\) centimeters

Answer: B (3/2)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to convert '5 centimeters greater than 2 times its width' into the correct algebraic expression. They might write the length as \(\mathrm{5 + 2w}\) instead of \(\mathrm{2w + 5}\), or worse, interpret it as something like \(\mathrm{5w + 2}\).

When they set up the area equation incorrectly, they get the wrong quadratic equation entirely, leading to solutions that don't match any of the given answer choices. This causes confusion and often results in guessing.

Second Most Common Error:

Inadequate APPLY CONSTRAINTS reasoning: Students correctly solve the quadratic equation and find both \(\mathrm{w = \frac{3}{2}}\) and \(\mathrm{w = -4}\), but they don't recognize that width must be positive in a real-world context. They might select an answer choice that corresponds to using the negative solution in some way, or they get confused about which solution to use.

The Bottom Line:

This problem requires students to seamlessly move between word problem translation and quadratic equation solving - two skills that are often taught separately. The key insight is recognizing that even though you get two mathematical solutions, only one makes sense in the physical context of measuring a rectangle.