Multiple choice

In the following question two equation number I and II are given you have to solve both equation and (I. x^2 + 29x + 210 = 0 )(II. 2y^2 + 12y + 16 = 0)

  1. x > y

  2. x ≥ y

  3. x ≤ y

  4. x < y

  5. x = y or the relationship cannot be established
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A Correct answer
Explanation

Solving equation I: x² + 29x + 210 = 0 gives (x+14)(x+15) = 0, so x = -14 or x = -15. Solving equation II: 2y² + 12y + 16 = 0. Divide by 2: y² + 6y + 8 = 0 gives (y+2)(y+4) = 0, so y = -2 or y = -4. Comparing: Both values of x (-14, -15) are less than both values of y (-2, -4). So x < y is false, x > y is true for all combinations.

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