Multiple choice

In each of the following questions, two equations are given. You have to solve them and $I. (x^2 + 3x - 18 = 0) II. (y^2 + 9y + 18 = 0)$

  1. If x > y

  2. If x ≥ y

  3. If x < y

  4. If x ≤ y

  5. If x = y or relationship can not be established
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

I: x² + 3x - 18 = 0 gives (x + 6)(x - 3) = 0, so x = -6 or 3. II: y² + 9y + 18 = 0 gives (y + 6)(y + 3) = 0, so y = -6 or -3. For x = 3: x > y (both -6, -3). For x = -6: x = y when y = -6, and x > y when y = -3. Therefore x ≥ y is always true.

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