Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations 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 relation cannot be established.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solve Equation I: x² - 1 = 0 gives x = ±1. Solve Equation II: y² + 4y + 3 = 0 factors to (y+3)(y+1) = 0, so y = -3 or y = -1. When x = 1, x > y (both -3 and -1). When x = -1, x = y (= -1). Therefore x ≥ y always holds. Option C is correct.

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