A right triangle has legs of length x and x+1. If the hypotenuse has a length of 4, what is...

1. TRANSLATE the problem information

• Given information:
  • Right triangle with legs of length \(\mathrm{x}\) and \(\mathrm{x+1}\)
  • Hypotenuse has length 4
  • Need to find the value of \(\mathrm{x}\)

2. INFER the approach

• Since we have a right triangle with known relationships, use the Pythagorean theorem
• Set up: \(\mathrm{x^2 + (x+1)^2 = 4^2}\)

3. SIMPLIFY the equation to standard form

• Expand \(\mathrm{(x+1)^2}\): \(\mathrm{x^2 + x^2 + 2x + 1 = 16}\)
• Combine like terms: \(\mathrm{2x^2 + 2x + 1 = 16}\)
• Move everything to one side: \(\mathrm{2x^2 + 2x - 15 = 0}\)

4. INFER the solution method

• This quadratic doesn't factor easily, so use the quadratic formula
• Identify \(\mathrm{a = 2}\), \(\mathrm{b = 2}\), \(\mathrm{c = -15}\)

5. SIMPLIFY using the quadratic formula

• \(\mathrm{x = \frac{-2 \pm \sqrt{4 - 4(2)(-15)}}{2 \times 2}}\)
• \(\mathrm{x = \frac{-2 \pm \sqrt{4 + 120}}{4}}\)
• \(\mathrm{x = \frac{-2 \pm \sqrt{124}}{4}}\)
• Simplify \(\mathrm{\sqrt{124} = \sqrt{4 \times 31} = 2\sqrt{31}}\)
• \(\mathrm{x = \frac{-2 \pm 2\sqrt{31}}{4} = \frac{-1 \pm \sqrt{31}}{2}}\)

6. APPLY CONSTRAINTS to select the valid solution

• Two solutions: \(\mathrm{x = \frac{-1 + \sqrt{31}}{2}}\) and \(\mathrm{x = \frac{-1 - \sqrt{31}}{2}}\)
• Since \(\mathrm{x}\) represents a triangle side length, \(\mathrm{x}\) must be positive
• Since \(\mathrm{\sqrt{31} \approx 5.57}\) (use calculator), \(\mathrm{\frac{-1 + \sqrt{31}}{2} \approx 2.29 \gt 0}\) ✓
• The negative solution \(\mathrm{\frac{-1 - \sqrt{31}}{2} \approx -2.29 \lt 0}\) ✗

Answer: C. \(\mathrm{\frac{-1 + \sqrt{31}}{2}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make algebraic errors when expanding \(\mathrm{(x+1)^2}\) or setting up the quadratic equation correctly.

Many students write \(\mathrm{x^2 + (x+1)^2 = 16}\) as \(\mathrm{x^2 + x + 1 = 16}\), forgetting to square the entire binomial \(\mathrm{(x+1)}\). This leads to \(\mathrm{2x^2 + x - 15 = 0}\) instead of \(\mathrm{2x^2 + 2x - 15 = 0}\), giving completely different solutions that don't match any answer choice. This leads to confusion and guessing.

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students correctly solve the quadratic but select the negative root without considering the physical constraint.

They get both solutions \(\mathrm{x = \frac{-1 \pm \sqrt{31}}{2}}\) but don't recognize that triangle side lengths must be positive. This may lead them to select Choice A \(\mathrm{\left(\frac{-1 + \sqrt{31}}{4}\right)}\) if they make an additional simplification error, or to guess between the positive and negative forms of the answer.

The Bottom Line:

This problem tests whether students can properly set up and solve a quadratic equation from a geometric context, then apply real-world constraints to select the meaningful solution.