<p>If \(a(p+q)^2 + 2bpq + c = 0\) and \(a(p+r)^2 + 2bpr + c = 0\) \((a \neq 0)\), then</p>
Step-by-Step Solution
Key Concept: Recognize that p satisfies both equations simultaneously, making them identical expressions in different variables q and r. This means q and r must be roots of the same quadratic equation in a new variable.
<p><strong>Step 1:</strong> Rewrite the given equations as quadratics in the variable 'p':</p><p>First equation: a(p+q)² + 2bpq + c = 0</p><p>Expanding: ap² + 2apq + aq² + 2bpq + c = 0</p><p>ap² + (2aq + 2bq)p + (aq² + c) = 0</p><p><strong>Step 2:</strong> Similarly, the second equation is:</p><p>ap² + (2ar + 2br)p + (ar² + c) = 0</p><p><strong>Step 3:</strong> Since both equations equal zero, they represent the same quadratic in p. Therefore, coefficients of like powers must be equal:</p><p>Coefficient of p: 2aq + 2bq = 2ar + 2br</p><p>2q(a + b) = 2r(a + b)</p><p><strong>Step 4:</strong> If a + b ≠ 0, then q = r (trivial case)</p><p>If a + b = 0, then b = -a, which gives the non-trivial relationship.</p><p>Constant term: aq² + c = ar² + c</p><p>This gives: aq² = ar², so q = ±r</p><p><strong>Step 5:</strong> From the structure of these matching equations with p as the common root: <strong>b = -a</strong> is the key relationship, or equivalently <strong>a + b = 0</strong></p><p>∴ Answer: B</p>
Correct Answer: B