Multiple choice

In each of the following question, two equations are I and II given. You have to solve them and give answer $I. (2x^2 – 23x + 65 = 0) (II. 2y^2 – 3y = 27)$

  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
A Correct answer
Explanation

Equation I: 2x² - 23x + 65 = 0 gives x = (23 ± √(529-520))/4 = (23 ± 3)/4, so x = 5 or 6.5. Equation II: 2y² - 3y - 27 = 0 gives y = (3 ± √(9+216))/4 = (3 ± 15)/4, so y = 4.5 or -3. Comparing minimum values: x = 5 > y = 4.5. Therefore x > y is always true.

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