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Let r be a root of the equation x^2 + 2x + 6 = 0. Then the value of the expression (r + 2)(r + 3)(r + 4)(r + 5) is: A. 51 B. -51 C. 126 D. -126

GATE 2022 · General Aptitude · Quadratic Equations · medium

Answer: (r+2)(r+3)(r+4)(r+5) = -126 (option D)

  1. Establish the root substitution rule: r is a root of x^2 + 2x + 6 = 0, so r^2 = -2r - 6.
  2. Compute (r + 2)(r + 5): (r+2)(r+5) = r^2 + 7r + 10. Substitute r^2 = -2r-6: = (-2r-6) + 7r + 10 = 5r + 4.
  3. Compute (r + 3)(r + 4): (r+3)(r+4) = r^2 + 7r + 12. Substitute r^2 = -2r-6: = (-2r-6) + 7r + 12 = 5r + 6.
  4. Multiply the two linear results: (5r+4)(5r+6) = 25r^2 + 30r + 20r + 24 = 25r^2 + 50r + 24. Now substitute r^2 = -2r-6: = 25(-2r-6) + 50r + 24 = -50r - 150 + 50r + 24 = -126.