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  (- 7 -  √37)/6


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