(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 (- 7 - √37)/6
