A rectangle has a length of x units and a width of \(\mathrm{(x - 15)}\) units. If the rectangle has...

Step-by-Step Solution

1. TRANSLATE the problem information

• Given information:
  • Rectangle length: \(\mathrm{x}\) units
  • Rectangle width: \(\mathrm{(x - 15)}\) units
  • Rectangle area: \(\mathrm{76}\) square units
  • Need to find: value of \(\mathrm{x}\)

2. TRANSLATE the area relationship into an equation

• Using Area = length × width:
  Area = \(\mathrm{x(x - 15) = 76}\)
• This gives us the equation: \(\mathrm{x(x - 15) = 76}\)

3. SIMPLIFY by expanding and rearranging

• Expand the left side: \(\mathrm{x^2 - 15x = 76}\)
• Move all terms to one side: \(\mathrm{x^2 - 15x - 76 = 0}\)
• Now we have a quadratic equation in standard form

4. SIMPLIFY by factoring the quadratic

• We need two numbers that multiply to \(\mathrm{-76}\) and add to \(\mathrm{-15}\)
• Those numbers are \(\mathrm{-19}\) and \(\mathrm{+4}\): \(\mathrm{(-19)(4) = -76}\) and \(\mathrm{(-19) + 4 = -15}\)
• Factor: \(\mathrm{(x - 19)(x + 4) = 0}\)

5. INFER the solutions using zero product property

• If \(\mathrm{(x - 19)(x + 4) = 0}\), then either:
  • \(\mathrm{x - 19 = 0}\), so \(\mathrm{x = 19}\), or
  • \(\mathrm{x + 4 = 0}\), so \(\mathrm{x = -4}\)

6. APPLY CONSTRAINTS to select the valid answer

• Since \(\mathrm{x}\) represents length, it must be positive
• Therefore \(\mathrm{x = 19}\) (rejecting \(\mathrm{x = -4}\))

Answer: B. 19

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Incorrectly factoring \(\mathrm{x^2 - 15x - 76 = 0}\)

Students often struggle to find the correct factor pair for \(\mathrm{-76}\) that also adds to \(\mathrm{-15}\). They might try factors like \(\mathrm{(-38, 2)}\) or \(\mathrm{(-2, 38)}\), leading to incorrect factorizations such as \(\mathrm{(x - 38)(x + 2) = 0}\) or \(\mathrm{(x - 2)(x + 38) = 0}\). This produces wrong solutions that don't match any answer choices, causing confusion and guessing.

Second Most Common Error:

Conceptual confusion: Misidentifying what the question asks for

After correctly solving for \(\mathrm{x = 19}\), some students select Choice A (4) because they calculate the width as \(\mathrm{19 - 15 = 4}\) and mistakenly think this is what was asked. Others might select Choice D (76) by confusing the area value with the answer to "what is x?"

The Bottom Line:

This problem requires strong algebraic manipulation skills, particularly factoring quadratics. The key challenge is systematically finding the correct factor pair while keeping track of what quantity the question actually asks for.