Multiple choice

In each question two equations numbered I and II are given, you have to solve both the equation and choose the correct answer. $I. (3x^2 - 15x + 12 = 0) II. (3y^2 - 19y + 20 = 0)$

  1. X > Y

  2. X < Y

  3. X ≤ Y

  4. X ≥ Y

  5. X = Y or the relationship can’t be established
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 3x² - 15x + 12 = 0. Simplifying: x² - 5x + 4 = 0, so (x-1)(x-4) = 0. Values: x = 1 or x = 4. Equation II: 3y² - 19y + 20 = 0. Solving: y = [19 ± √(361 - 240)]/6 = [19 ± √121]/6 = [19 ± 11]/6. So y = 30/6 = 5 or y = 8/6 = 4/3. Comparing: When x=1, y=5 gives xy. The relationship varies, so it cannot be established.

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