general math MCQ #1320

I. a² + 8a + 16 = 0, II. b² - 4b + 3 = 0 to solve both the equations to find the values of a and b?

general math MCQ #1320

  1. Question 1

    Q1. I. a² + 8a + 16 = 0, II. b² - 4b + 3 = 0 to solve both the equations to find the values of a and b?

    • A) If a < b
    • B) If a ≤ b
    • C) If the relationship between a and b cannot be established
    • D) If a > b

    Answer: If a < b

    Explanation: (a+4)² = 0 gives a = −4; (b−1)(b−3) = 0 gives b = 1 or 3. Since −4 < 1 or 3, a < b.