Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and Give answer $I. (4x^2 - 20x + 21 = 0)$ $II. (2y^2 - 13y + 20 = 0)$

  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

Solve equation I: 4x² - 20x + 21 = 0 factors as (2x-3)(2x-7), giving x = 3/2 or x = 7/2. Solve equation II: 2y² - 13y + 20 = 0 factors as (2y-5)(y-4), giving y = 5/2 or y = 4. When x = 1.5, it's less than y = 2.5 but greater than y = 4. When x = 3.5, it's greater than y = 2.5 but less than y = 4. Since the relationship varies, it cannot be established.

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