Mathematics - Quadratic Equation Question with Solution | TestHub

Let $f(x)=\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx\Rightarrow f(0)=0$ and $f(1)=\frac{a}{3}+\frac{b}{2}+c=\frac{2a+3b+6c}{6}=0$ Also $f(x)$ is continuous and differentiable in $[0,1]$ and $[0,1]$. So by Rolle's theorem, $f^{\prime }(x)=0$. i.e. $a{x}^2+bx+c=0$ has at least one root in $[0,1]$