3x2 + 8x + 4 = 0; 4y2 - 19y + 12 = 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.

3x² + 8x + 4 = 0; 4y² - 19y + 12 = 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: C

3x² + 8x + 4 = 0
⇒ 3x² + 6x + 2x + 4 = 0
⇒ 3x(x + 2) + 2(x + 2) = 0
⇒ (x + 2) (3x + 2) = 0
∴ x = - 2, - 2/3
and 4y² - 19y + 12 = 0
⇒ 4y² - 16y - 3y + 12 = 0
⇒ 4y(y - 4) -3(y - 4) = 0
⇒ (y - 4) (4y - 3) = 0
∴ y = 4, 3/4
Hence, y > x or x < y