Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equations and choose the correct answer. $I. (3x^2 + 8x + 5 = 0) II. (2y^2 + y - 1 = 0)$

  1. $\(x > y\)$
  2. $\(x < y\)$
  3. $\(x \le y\)$
  4. $\(x \ge y\)$

  5. x = y or the relationship cannot be established
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Solving equation I: 3x² + 8x + 5 = 0 gives (3x + 5)(x + 1) = 0, so x = -5/3 or x = -1. Solving equation II: 2y² + y - 1 = 0 gives (2y - 1)(y + 1) = 0, so y = 1/2 or y = -1. Comparing values: x = -5/3 < y = 1/2, and x = -5/3 < y = -1. But x = -1 = y = -1. Since x is sometimes less than y and sometimes equal to y, the relationship is x ≤ y.

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