The system of equations y = x^2 + px + q and y = 2x + r, where p, q,...

1. TRANSLATE the problem information

• Given information:
  • System: \(\mathrm{y = x^2 + px + q}\) and \(\mathrm{y = 2x + r}\)
  • System has exactly one solution
  • \(\mathrm{p = 6}\) and \(\mathrm{r = 8}\)
  • Need to find minimum possible value of \(\mathrm{2q}\)

2. INFER the mathematical meaning

• "Exactly one solution" means the parabola and line intersect at exactly one point
• This happens when they're tangent to each other
• For a quadratic equation, exactly one solution occurs when the discriminant equals zero

3. Set up the intersection equation

• Since both expressions equal y, set them equal to each other:
  \(\mathrm{x^2 + px + q = 2x + r}\)
• Rearrange to standard quadratic form:
  \(\mathrm{x^2 + (p - 2)x + (q - r) = 0}\)

4. SIMPLIFY by substituting known values

• Substitute \(\mathrm{p = 6}\) and \(\mathrm{r = 8}\):
  \(\mathrm{x^2 + (6 - 2)x + (q - 8) = 0}\)
  \(\mathrm{x^2 + 4x + (q - 8) = 0}\)

5. APPLY the discriminant condition

• For exactly one solution, discriminant \(\mathrm{Δ = 0}\):
  \(\mathrm{Δ = b^2 - 4ac = 4^2 - 4(1)(q - 8) = 0}\)

6. SIMPLIFY to solve for q

• Expand: \(\mathrm{16 - 4(q - 8) = 0}\)
• Distribute: \(\mathrm{16 - 4q + 32 = 0}\)
• Combine: \(\mathrm{48 - 4q = 0}\)
• Solve: \(\mathrm{4q = 48}\), so \(\mathrm{q = 12}\)

7. Find the final answer

• \(\mathrm{2q = 2(12) = 24}\)

Answer: 24

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't connect "exactly one solution" to the discriminant condition. They might try to solve the system directly or get confused about what "exactly one solution" means mathematically.

Without this key insight, they may attempt to find specific x and y values or use elimination/substitution methods inappropriately, leading to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when working with the discriminant equation, such as sign errors when distributing \(\mathrm{-4(q - 8)}\) or combining like terms incorrectly.

For example, they might get \(\mathrm{16 - 4q - 32 = 0}\) instead of \(\mathrm{16 - 4q + 32 = 0}\), leading to \(\mathrm{q = -4}\) and \(\mathrm{2q = -8}\), causing them to get stuck since this doesn't match typical answer patterns.

The Bottom Line:

This problem requires students to translate the geometric concept of "exactly one solution" into the algebraic condition of zero discriminant, then execute careful algebraic manipulation without computational errors.