3y2 + 7y + 1 = 0

(vii) 3y² + 7y + 1 = 0

Sol.

3y² + 7y + 1 = 0

∴ 3y² + 7y = –1

Dividing both the sides by 3 we get,

y² + (⁷⁄₃)y = ^-1⁄₃ .....(i)

Third term = ( ½ × coefficient of y)²

= ( ½ × ⁷⁄₃)²

= (⁷⁄₆)²

= ⁴⁹⁄₃₆

Adding ⁴⁹⁄₃₆ to both sides of (i) we get,

y² + (⁷⁄₃)y + ⁴⁹⁄₃₆ = ^-1⁄₃ + ⁴⁹⁄₃₆

∴ (y + ⁷⁄₆)² = ³⁷⁄₃₆

∴ y + ⁷⁄₆ = ± √(³⁷⁄₃₆)

∴ y + ⁷⁄₆ = ± ^√37⁄₆

∴ y = -⁷⁄₆ ± ^√37⁄₆

∴ y = ^{(-7 ± √37)}⁄₆

∴ y = ^{(-7 + √37)}⁄₆  or  y = ^{(-7 - √37)}⁄₆