Ⅰ. 6x2 + 41x + 63 = 0 Ⅱ. 4y2 + 8y + 3 = 0

Direction: In this question 2 equations numbered I and II are given. You have to solve both the equations and find out the correct option.

1. Relationship between x and y cannot be established
2. x ≥ y
3. x < y
4. x > y
5. x ≤ y

According to question ,
From equation Ⅰ. 6x² + 41x + 63 = 0
⇒ 6x² + 27x + 14x + 63 = 0
⇒ 3x(2x + 9) + 7(2x + 9) = 0
⇒ (3x + 7)(2x + 9) = 0

∴ x = -	3	, -	2
	7		9

From equation Ⅱ. 4y² + 8y + 3 = 0
⇒ 4y² + 6y + 2y + 3 = 0
⇒ 2y(2y + 3) + 1(2y + 3)
⇒ (2y + 1)(2y + 3)

∴ y = -	1	, -	3
	2		2

Hence, required answer is x < y