Multiple choice

In each of the following questions, two equations (I) and (II) are given. You have to solve them and give answer. $(I) (x^2 - 1 = 0) (II) (y^2 + 4y + 3 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation I: x² - 1 = 0 gives x² = 1, so x = ±1 (x can be 1 or -1). Equation II: y² + 4y + 3 = 0 factors as (y + 3)(y + 1) = 0, so y = -3 or y = -1. Comparing all values: x = 1 > y = -3, x = 1 > y = -1, x = -1 > y = -3, x = -1 = y = -1. In all cases, x ≥ y is true (x is never less than y).

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