A rectangle has a width of x inches and a length of \(\mathrm{(x + 3)}\) inches. If the diagonal of...

1. TRANSLATE the problem information

• Given information:
  • Rectangle width: \(\mathrm{x}\) inches
  • Rectangle length: \(\mathrm{(x + 3)}\) inches
  • Rectangle diagonal: \(\mathrm{15}\) inches
• Find: the value of \(\mathrm{x}\)

2. INFER the geometric relationship

• A rectangle's diagonal creates a right triangle with its width and length
• This means we can use the Pythagorean theorem: \(\mathrm{width^2 + length^2 = diagonal^2}\)

3. TRANSLATE this relationship into an equation

• Using the Pythagorean theorem:
  \(\mathrm{x^2 + (x + 3)^2 = 15^2}\)

4. SIMPLIFY the equation through algebraic expansion

• Expand \(\mathrm{(x + 3)^2}\): \(\mathrm{x^2 + (x^2 + 6x + 9) = 225}\)
• Combine like terms: \(\mathrm{2x^2 + 6x + 9 = 225}\)
• Move all terms to one side: \(\mathrm{2x^2 + 6x - 216 = 0}\)
• Divide by 2: \(\mathrm{x^2 + 3x - 108 = 0}\)

5. SIMPLIFY by factoring the quadratic

• Look for two numbers that multiply to \(\mathrm{-108}\) and add to \(\mathrm{3}\)
• Those numbers are \(\mathrm{12}\) and \(\mathrm{-9}\): \(\mathrm{(x + 12)(x - 9) = 0}\)
• Therefore: \(\mathrm{x = -12}\) or \(\mathrm{x = 9}\)

6. APPLY CONSTRAINTS to select the valid solution

• Since width cannot be negative in a real rectangle: \(\mathrm{x = 9}\)

Answer: 9

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that this is a Pythagorean theorem problem. They might think about perimeter instead of the diagonal relationship, or they might not see that the diagonal creates a right triangle with the rectangle's sides. This leads to confusion about which formula to use, causing them to get stuck and guess randomly.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when expanding \(\mathrm{(x + 3)^2}\) or when factoring \(\mathrm{x^2 + 3x - 108 = 0}\). Common mistakes include getting the expansion wrong (forgetting the \(\mathrm{6x}\) term) or not finding the correct factors of \(\mathrm{-108}\). This leads to an incorrect quadratic equation and ultimately a wrong answer.

The Bottom Line:

This problem tests whether students can connect geometry (rectangle properties) with algebra (quadratic equations). The key insight is recognizing that a rectangle's diagonal problem is really a disguised Pythagorean theorem problem.