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

Question

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

Accepted Answer

Let $g(x)=a\dfrac{x^3}{3}+b\dfrac{x^2}{2}+cx$; continuous and differentiable everywhere, $g(0)=0$, $g(1)=\dfrac a3+\dfrac b2+c=\dfrac{2a+3b+6c}{6}=0$. By Rolle's theorem, $g'(x)=ax^2+bx+c=0$ has a root in $(0,1)$.

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