<p>Let \( p(x) = 51x^2 + mx + c \) and \( q(x) = 3x^2 + bx + a \) are two quadratic polynomials with integer coefficients such that \( p(r) = q(r) = 0 \). If \( r \) is an irrational number, then the value of \( \dfrac{c}{a} \) is:</p>
Step-by-Step Solution
Key Concept: If p(x) and q(x) share an irrational root r with integer coefficients, then r must also be a root of their linear combination. Since p(x) = 51q(x) + (m-153b)x + (c-51a), the coefficients force specific divisibility relationships between a and c.
<p><strong>Step 1:</strong> Since p(r) = q(r) = 0, we have:</p><p>51r² + mr + c = 0 ... (1)</p><p>3r² + br + a = 0 ... (2)</p><p><strong>Step 2:</strong> Multiply equation (2) by 17:</p><p>51r² + 17br + 17a = 0 ... (3)</p><p><strong>Step 3:</strong> Subtract (3) from (1):</p><p>(m - 17b)r + (c - 17a) = 0</p><p><strong>Step 4:</strong> Since r is irrational and this equation must hold, both the rational and irrational parts must separately equal zero:</p><p>m - 17b = 0 ⟹ m = 17b</p><p>c - 17a = 0 ⟹ c = 17a</p><p><strong>Step 5:</strong> Therefore:</p><p>c/a = 17a/a = 17</p><p>∴ Answer: B</p>
Correct Answer: B