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

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$

Let $\mathrm{F}\left(\mathrm{x}\right)=\int \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{a}}{3}{\mathrm{x}}^3+\frac{\mathrm{b}}{2}{\mathrm{x}}^2+\mathrm{cx}$

Clearly $\mathrm{F}\left(0\right)=0 \mathrm{and} \mathrm{F}\left(1\right)=\frac{\mathrm{a}}{3}+\frac{\mathrm{b}}{2}+\mathrm{c}$

$=\frac{2a+3b+6c}{6}=0$

$\Rightarrow \mathrm{F}\left(0\right)=\mathrm{F}\left(1\right)=0$

There exist at least one point c in between 0 and 1 such that ${\mathrm{F}}^{\text{'}}\left(\mathrm{x}\right)=0$ or $a{x}^2+bx+c=0$ for some $x\in (0, 1)$.