The set of values of p for which the roots of the equation 3x^2 + 2x + p(p - 1) = 0 are of opposite sign is A. (-inf, 0) B. (0, 1) C. (1, inf) D. (0, inf)

GATE 2013 · General Aptitude · Quadratic Equations · medium

Answer: p in (0, 1).

1. Set up the product-of-roots condition: Here a = 3, c = p(p-1). Product = p(p-1)/3. For opposite signs, need p(p-1)/3 < 0, i.e., p(p-1) < 0.
2. Solve p(p-1) < 0: p(p-1) < 0 means p and (p-1) have opposite signs. Case 1: p > 0 and p - 1 < 0 => 0 < p < 1. Case 2: p < 0 and p - 1 > 0 => impossible. So solution is 0 < p < 1, i.e., p in (0, 1).