Let α(a) and β(a) be the roots of the equation (3√(1+a))x² + (√(1+a))x + (6√(1+a)) = 0 where a > -1. Then lima→0+α(a) and lima→0+β(a) are

Let 1+a = y. The equation becomes (y1/3)x² + (y1/2)x + (y1/6) = 0
Dividing by y1/6, we get (y1/6)x² + (y1/3)x + 1 = 0
When a → 0, y → 1. Hence, taking limy→1 on both the sides,
→ limy→1((y1/6)x² + (y1/3)x + 1) = 0
→ x² + x + 1 = 0
Solving the quadratic equation, we get
→ x = -1/2 ± i√3/2
Since α(a) and β(a) are roots of the given equation, we get
lima→0+α(a) = -1/2 + i√3/2 and lima→0+β(a) = -1/2 - i√3/2