LIC AAO Model Paper 10

I. $12{\mathit{x}}^2$ + 7x + 1 = 0
⇒ $12{\mathit{x}}^2$ + 4x + 3x + 1 = 0
⇒ 4x(3x+ 1) + 1 (3x+ 1) = 0
⇒ (3x+ 1) (4x+ 1) = 0
⇒ x = $-\frac{1}{3}$ or , $-\frac{1}{4}$
II. $6{\mathit{y}}^2+5\mathit{y}+1=0$
⇒ $6{\mathit{y}}^2$ + 3y + 2y + 1 = 0
⇒ 3y (2y + 1) + 1 (2y + 1) = 0
⇒ (3y + 1) (2y+ 1) = 0
⇒ y = $-\frac{1}{3}$ or, $-\frac{1}{2}$
Clearly x ≥ y