Consider a quadratic equation a x^2+b x+c=0 , where 2 a+3 b+6 c=0...

Let $g(x)=\frac{a x^3}{3}+b \cdot \frac{x^2}{2}+c x$ \mathrmg^{\prime}(x)=a x²+b x+c Given: $a x^2+b x+c=0 and 2 a+3 b+6 c=0$ Statement-2: (i) g(0)=0 and $g(1)$$\begin{aligned}& =\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6}\\ & =\frac{0}{6}=0 \\ & \Rightarrow g(0)=g(1) \end{aligned}$$(ii)$g$is continuous on$[0,1] and differentiable on $(0,1)\therefore$ By Rolle's theorem $\exists k \in(0,1)$ such that $g^{\prime}(k)=0$ This holds the statement 2. Also, from statement-2, we can say $a x^2+b x+c=0$ has at least one root in $(0,1)$. Thus statement-1 and 2 both are true and statement-2 is a correct explanation for statement-1.