Consider a quadratic equation ax^2+bx+c=0, where 2a+3b+6c=0, and let g(x)=ax^33+bx^22+cx.

Question

Consider a quadratic equation $ax^2+bx+c=0$, where $2a+3b+6c=0$, and let $g(x)=a\dfrac{x^3}{3}+b\dfrac{x^2}{2}+cx$.

Statement 1: The quadratic equation has at least one root in $(0,1)$.

Statement 2: Rolle's theorem is applicable to $g(x)$ on $[0,1]$.

Accepted Answer

$g'(x)=ax^2+bx+c$. $g(0)=0$, $g(1)=\dfrac a3+\dfrac b2+c=\dfrac{2a+3b+6c}{6}=0=g(0)$. Since $g$ is a polynomial (continuous, differentiable everywhere), Rolle's theorem applies on $[0,1]$, giving a root of $g'(x)=ax^2+bx+c=0$ in $(0,1)$. This directly proves Statement 1, so Statement 2 correctly explains Statement 1.

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