<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 two polynomials with integer coefficients share an irrational root r, then r and its conjugate must both satisfy both polynomials. This forces a specific ratio relationship between the leading coefficients and constant terms through the common root structure.
<p><strong>Step 1: Recognize the constraint.</strong> If p(r) = q(r) = 0 where r is irrational and both polynomials have integer coefficients, then the conjugate root must also satisfy both equations (since minimal polynomials of algebraic numbers are unique).</p><p><strong>Step 2: Establish root structure.</strong> Let r satisfy the irreducible polynomial with integer coefficients. Since both q(x) and p(x) have r as a root with integer coefficients, p(x) and q(x) must share a common factor corresponding to r.</p><p><strong>Step 3: Use divisibility.</strong> Since gcd of degrees implies q(x) divides p(x) or they share a quadratic factor. Given deg(q) = 2 and deg(p) = 2, if they share an irrational root r, they must be scalar multiples of the same minimal polynomial.</p><p><strong>Step 4: Compare coefficients.</strong> If p(x) = k·q(x) for some rational k, then:
51x² + mx + c = k(3x² + bx + a)
Comparing leading coefficients: 51 = 3k, so k = 17</p><p><strong>Step 5: Find the ratio.</strong> Then c = 17a, giving us:
$$\frac{c}{a} = 17$$</p><p>∴ Answer: B</p>
Correct Answer: B