A rectangle has positive length L and positive width W. The length exceeds the width by 2 units. The perimeter...

1. TRANSLATE the problem information

• Given information:
  • Rectangle has positive length L and positive width W
  • Length exceeds width by 2 units
  • Perimeter equals \(\mathrm{L^2 - 2}\) units

• What this tells us:
  • \(\mathrm{L - W = 2}\) (length-width relationship)
  • \(\mathrm{2L + 2W = L^2 - 2}\) (perimeter equation)

2. INFER the solution approach

• We have two equations with two unknowns - this suggests using substitution
• From \(\mathrm{L - W = 2}\), we can express W in terms of L: \(\mathrm{W = L - 2}\)
• Substitute this into the perimeter equation to get one equation in L only

3. SIMPLIFY by substitution and algebraic manipulation

• Substitute \(\mathrm{W = L - 2}\) into \(\mathrm{2L + 2W = L^2 - 2}\):

\(\mathrm{2L + 2(L - 2) = L^2 - 2}\)

• Expand and collect terms:

\(\mathrm{2L + 2L - 4 = L^2 - 2}\)

\(\mathrm{4L - 4 = L^2 - 2}\)

\(\mathrm{0 = L^2 - 4L + 2}\)

4. SIMPLIFY using the quadratic formula

• For \(\mathrm{L^2 - 4L + 2 = 0}\), we have \(\mathrm{a = 1, b = -4, c = 2}\)

\(\mathrm{L = \frac{4 ± \sqrt{16 - 8}}{2}}\)

\(\mathrm{L = \frac{4 ± \sqrt{8}}{2}}\)

\(\mathrm{L = \frac{4 ± 2\sqrt{2}}{2}}\)

\(\mathrm{L = 2 ± \sqrt{2}}\)

5. APPLY CONSTRAINTS to select the valid solution

• If \(\mathrm{L = 2 + \sqrt{2}}\), then \(\mathrm{W = L - 2 = \sqrt{2}}\) (positive ✓)
• If \(\mathrm{L = 2 - \sqrt{2}}\), then \(\mathrm{W = L - 2 = -\sqrt{2}}\) (negative ✗)
• Since W must be positive, we choose \(\mathrm{L = 2 + \sqrt{2}}\) and \(\mathrm{W = \sqrt{2}}\)

Answer: D \(\mathrm{(2 + \sqrt{2}, \sqrt{2})}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students often struggle to set up the perimeter equation correctly, either forgetting the factor of 2 in the perimeter formula or misinterpreting "perimeter equals \(\mathrm{L^2 - 2}\)" as meaning something different.

This leads to incorrect equations like \(\mathrm{L + W = L^2 - 2}\) instead of \(\mathrm{2L + 2W = L^2 - 2}\), resulting in a different quadratic equation and wrong final answer. This causes confusion and may lead them to select an incorrect choice or guess.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students solve the quadratic correctly to get \(\mathrm{L = 2 ± \sqrt{2}}\), but fail to check which solution gives positive W.

They might choose \(\mathrm{L = 2 - \sqrt{2}}\) without realizing this gives \(\mathrm{W = -\sqrt{2} < 0}\), violating the condition that width must be positive. This may lead them to select Choice B \(\mathrm{(2 - \sqrt{2}, -\sqrt{2})}\).

The Bottom Line:

This problem combines algebraic manipulation with careful attention to real-world constraints. Success requires both setting up the correct system of equations AND checking that the mathematical solution makes physical sense for a rectangle.