Multiple choice

In each of the following questions, two equations are I and II given. You have to solve them and give the answer. $I. (12x^2 - x - 1 = 0) II. (18y^2 - 9y + 1 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

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

Solving equation I: 12x^2 - x - 1 = 0 factors to (4x + 1)(3x - 1) = 0, giving x = -1/4 or x = 1/3. Solving equation II: 18y^2 - 9y + 1 = 0 factors to (6y - 1)(3y - 1) = 0, giving y = 1/6 or y = 1/3. Comparing: when x = 1/3, it can equal y = 1/3 or be greater than y = 1/6; when x = -1/4, it's less than both y values. The relationship cannot be established.

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