The area of a rectangular garden is 216 square meters. The length of the garden is 6 meters less than...

1. TRANSLATE the problem information

• Given information:
  • Area of rectangular garden = 216 square meters
  • Length is "6 meters less than twice its width"

• What this tells us:
  • Area = length × width = 216
  • Length = 2(width) - 6

2. INFER the solution approach

• We have two relationships involving length and width, but we want to find just the width
• Strategy: Use substitution to eliminate length and create an equation with only width
• This will likely create a quadratic equation since we're multiplying width terms

3. Set up the equations using variables

Let \(\mathrm{w = width}\) and \(\mathrm{l = length}\) (both in meters)

• \(\mathrm{l \times w = 216}\)
• \(\mathrm{l = 2w - 6}\)

4. SIMPLIFY by substitution

Substitute \(\mathrm{l = 2w - 6}\) into the area equation:

\(\mathrm{(2w - 6) \times w = 216}\)

Distribute:

\(\mathrm{2w^2 - 6w = 216}\)

Rearrange to standard form:

\(\mathrm{2w^2 - 6w - 216 = 0}\)

Divide everything by 2 to make factoring easier:

\(\mathrm{w^2 - 3w - 108 = 0}\)

5. SIMPLIFY by factoring the quadratic

We need two numbers that multiply to -108 and add to -3

Think: \(\mathrm{-12 \times 9 = -108}\) and \(\mathrm{-12 + 9 = -3}\) ✓

So: \(\mathrm{(w - 12)(w + 9) = 0}\)

This gives us: \(\mathrm{w = 12}\) or \(\mathrm{w = -9}\)

6. APPLY CONSTRAINTS to select the valid answer

Since width represents a physical dimension, it must be positive:

\(\mathrm{w = 12}\) meters

Answer: B (12)

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Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students struggle to convert "6 meters less than twice its width" into the correct equation. They might write \(\mathrm{l = 2w + 6}\) (adding instead of subtracting) or get confused about the order of operations.

With the wrong relationship \(\mathrm{l = 2w + 6}\), they would get:

\(\mathrm{(2w + 6) \times w = 216}\)

\(\mathrm{2w^2 + 6w - 216 = 0}\)

\(\mathrm{w^2 + 3w - 108 = 0}\)

This factors to \(\mathrm{(w + 12)(w - 9) = 0}\), giving \(\mathrm{w = 9}\) or \(\mathrm{w = -12}\). Applying constraints gives \(\mathrm{w = 9}\).

This may lead them to select Choice A (9).

Second Most Common Error:

Poor INFER reasoning: Students set up the equations correctly but don't recognize they need to substitute. Instead, they might try to solve the system in a more complicated way or get stuck trying to factor with two variables present.

This leads to confusion and guessing among the answer choices.

The Bottom Line:

This problem requires students to smoothly transition from word relationships to algebraic equations, then work confidently with quadratic expressions. The key insight is recognizing that substitution creates a manageable single-variable quadratic equation.