Multiple choice

In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer- $I. (4x^2 + 12x + 5 = 0) II. (6y^2 + 27y + 12 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or the relation can't be determined
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Equation I: 4x² + 12x + 5 = 0. Discriminant = 144 - 80 = 64. Roots: x = (-12 ± 8)/8, so x = -1/2 or x = -5/2. Equation II: 6y² + 27y + 12 = 0. Discriminant = 729 - 288 = 441. Roots: y = (-27 ± 21)/12, so y = -1/2 or y = -4. Comparing: x can equal y (both -1/2) or x can be > y or < y. So no unique relation can be established.

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