y + k = x + 26y - k = x^2 - 5xIn the given system of equations, k is...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{y + k = x + 26}\)
  • \(\mathrm{y - k = x^2 - 5x}\)
  • The system has exactly one distinct real solution
• What this tells us: We need to find the value of k that makes this condition true

2. INFER the solution strategy

• 'Exactly one distinct real solution' is key - this means when we eliminate one variable, we'll get a quadratic equation whose discriminant equals zero
• Strategy: Eliminate y to get a quadratic in x, then use discriminant = 0

3. SIMPLIFY to eliminate y

• From first equation: \(\mathrm{y = x + 26 - k}\)
• From second equation: \(\mathrm{y = x^2 - 5x + k}\)
• Set equal: \(\mathrm{x + 26 - k = x^2 - 5x + k}\)
• Collect terms: \(\mathrm{x + 26 - 2k = x^2 - 5x}\)
• Rearrange to standard form: \(\mathrm{x^2 - 6x + (2k - 26) = 0}\)

4. INFER the discriminant condition

• For exactly one real solution: discriminant = 0
• For \(\mathrm{ax^2 + bx + c = 0}\): discriminant = \(\mathrm{b^2 - 4ac}\)
• Here: \(\mathrm{a = 1}\), \(\mathrm{b = -6}\), \(\mathrm{c = (2k - 26)}\)

5. SIMPLIFY the discriminant equation

• Discriminant = \(\mathrm{(-6)^2 - 4(1)(2k - 26) = 0}\)
• \(\mathrm{36 - 4(2k - 26) = 0}\)
• \(\mathrm{36 - 8k + 104 = 0}\)
• \(\mathrm{140 - 8k = 0}\)
• \(\mathrm{k = 140/8 = 17.5}\) (use calculator)

Answer: 17.5 or 35/2

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't connect 'exactly one distinct real solution' to the discriminant condition. Instead, they try to solve the system directly using substitution or elimination, not realizing they need to find when the resulting quadratic has a repeated root.

This leads to confusion because they can't determine a unique value of k from direct substitution, causing them to get stuck and guess randomly.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the need for discriminant = 0 but make arithmetic errors when expanding \(\mathrm{(-6)^2 - 4(1)(2k - 26)}\) or when solving \(\mathrm{140 - 8k = 0}\).

Common mistakes include getting \(\mathrm{36 + 8k - 104 = 0}\) (sign error) or \(\mathrm{k = 140/-8 = -17.5}\) (division error), leading them to incorrect negative values.

The Bottom Line:

The key insight is recognizing that the constraint 'exactly one distinct real solution' translates to a discriminant condition on the quadratic equation obtained after eliminating y. Students who miss this connection often get lost trying to solve the system conventionally.