x2 - x - 12 = 0; y2 + 5y + 6 = 0

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Direction: In each of the following question, there are two equations. you have to solve both equations and give answer.

x² - x - 12 = 0; y² + 5y + 6 = 0

1. If x > y
2. If x ≥ y
3. If x < y
4. If x ≤ y
5. If x = y or relation cannot be established

Correct Option: B

x² - x - 12 = 0
⇒ x² - 4x + 3x - 12 = 0
⇒ x(x - 4) + 3 (x - 4) = 0
⇒ (x - 4) (x + 3) = 0
∴ x = - 3, 4
and y² + 5y + 6 = 0
⇒ y² + 3y + 2y + 6 = 0
⇒ y(y + 3) + 2(y + 3) = 0
⇒ (y + 3) (y + 2) = 0
⇒ y = - 3, - 2
∴ x ≥ y
[∵ x = - 3 and y = - 3, so x = y and x = 4 and y = - 2, hence x > y]