If 2a+3b+6c=0, (a,b,cR) then the quadratic equation ax^2+bx+c=0 has

Question

If $2a+3b+6c=0$, $(a,b,c\in\mathbb{R})$ then the quadratic equation $ax^2+bx+c=0$ has

Accepted Answer

Let $g(x)=a\dfrac{x^3}{3}+b\dfrac{x^2}{2}+cx$; $g(0)=0$ and $g(1)=\dfrac a3+\dfrac b2+c=\dfrac{2a+3b+6c}{6}=0$. Since $g$ is continuous and differentiable everywhere (polynomial), by Rolle's theorem $g'(x)=ax^2+bx+c=0$ has at least one root in $(0,1)\subset[0,1]$.

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