If 2a + 3b + 6c = 0, then at least one root of the equation $ax^2+ bx+ c= 0$ lies in the interval:

(0, 1)

Let $f(x)=2ax^3+ 3bx^2+ 6cx= 0$ 

Clearly f(x) is continuous on [0, 1], derivable on (0, 1) and f(0) = 0, f = (1) = 2a + 3b + 6c = 0

Also, $f'(x)=6ax^2+ 6bx+ 6c$

∴ By Rolle’s theorem $∀\, α ∈ (0, 1)$ such that $f'(a) = 0⇒6aα^2+6bα+6c=0⇒aα^2+2bα+c=0$

$∴ax^2+ bx+ c= 0$ has a root in (0, 1)