The function g is defined by \(\mathrm{g(x) = px^2 + qx + r}\), where p, q, and r are constants....

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{g(x) = px^2 + qx + r}\) (quadratic function)
  • Sum of roots = 2
  • Product of roots = -15
  • p is a prime number
  • Need to find p + q

2. TRANSLATE using Vieta's formulas

• For quadratic \(\mathrm{px^2 + qx + r = 0}\), Vieta's formulas tell us:
  • Sum of roots = \(\mathrm{-q/p}\)
  • Product of roots = \(\mathrm{r/p}\)
• This gives us the equations:
  • \(\mathrm{-q/p = 2}\)
  • \(\mathrm{r/p = -15}\)

3. SIMPLIFY to find relationships

• From sum equation: \(\mathrm{-q/p = 2}\)
  • Multiply both sides by p: \(\mathrm{-q = 2p}\)
  • Therefore: \(\mathrm{q = -2p}\)
• From product equation: \(\mathrm{r/p = -15}\)
  • Multiply both sides by p: \(\mathrm{r = -15p}\)

4. INFER what we need to find

• We want p + q
• Since \(\mathrm{q = -2p}\), we have:
  \(\mathrm{p + q = p + (-2p)}\)
  \(\mathrm{= p - 2p}\)
  \(\mathrm{= -p}\)
• So \(\mathrm{p + q = -p}\), which means we need to find the value of -p

5. CONSIDER ALL CASES for prime values of p

• Since p is prime, test small prime numbers:
  • If \(\mathrm{p = 2}\): \(\mathrm{p + q = -2}\) ← This matches choice (C)
  • If \(\mathrm{p = 3}\): \(\mathrm{p + q = -3}\) ← Not among the choices
  • If \(\mathrm{p = 5}\): \(\mathrm{p + q = -5}\) ← Not among the choices
  • If \(\mathrm{p = 7}\): \(\mathrm{p + q = -7}\) ← Not among the choices

6. APPLY CONSTRAINTS to select the answer

• Only \(\mathrm{p = 2}\) gives a result that appears in the answer choices
• Therefore: \(\mathrm{p + q = -2}\)

Answer: C (-2)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may not recognize that they need to use Vieta's formulas to connect the given root information to the coefficients. Instead, they might try to solve the quadratic directly or get confused about how to use the sum and product of roots.

This leads to confusion and random guessing among the answer choices.

Second Most Common Error:

Poor CONSIDER ALL CASES execution: Students might correctly establish that \(\mathrm{p + q = -p}\) but then either forget that p must be prime or fail to systematically test prime values against the answer choices. They might test non-prime values or stop after the first prime doesn't work.

This may lead them to select an incorrect choice or abandon the systematic approach and guess.

The Bottom Line:

This problem requires students to bridge the gap between abstract root properties and concrete coefficient relationships using Vieta's formulas, then systematically apply the constraint that p is prime. The key insight is recognizing that only certain prime values will yield answer choices that actually appear.