<p>A quadratic equation with integral coefficients has two different prime numbers as its roots. If the sum of the coefficients of the equation is prime, then the sum of the roots is</p>
Step-by-Step Solution
Key Concept: If a quadratic with integer coefficients has prime roots p and q, then the equation is a(x² - (p+q)x + pq) = 0. The sum of coefficients equals a(1 - (p+q) + pq), which must be prime. This constraint severely limits which pairs of primes work.
<p><strong>Step 1:</strong> Let the two different prime roots be p and q. The quadratic equation is a(x² - (p+q)x + pq) = 0 where a ≠ 0 is an integer.</p><p><strong>Step 2:</strong> Sum of coefficients = a(1 - (p+q) + pq) = a((1-p)(1-q)). This must be prime.</p><p><strong>Step 3:</strong> For this to be prime, we need |a · (1-p)(1-q)| to be prime. Since a is a non-zero integer and (1-p)(1-q) is an integer, one of them must be ±1.</p><p><strong>Step 4:</strong> If p and q are both odd primes, then (1-p) and (1-q) are both even, so (1-p)(1-q) is divisible by 4. This cannot equal ±1.</p><p><strong>Step 5:</strong> One prime must be 2. Let p = 2 and q be an odd prime. Then (1-2)(1-q) = (-1)(1-q) = q-1. For a(q-1) to be prime, we need a = ±1 and q-1 = ±p' (a prime).</p><p><strong>Step 6:</strong> If a = 1: q - 1 must be prime. Try q = 3: then q-1 = 2 (prime) ✓. Sum of coefficients = 1·2 = 2 (prime) ✓</p><p><strong>Step 7:</strong> Verify: Equation is x² - 5x + 6 = 0 with roots 2 and 3. Sum of coefficients: 1 - 5 + 6 = 2 (prime) ✓</p><p>∴ <strong>Sum of roots = 2 + 3 = 5</strong></p>
Correct Answer: B