If $2 a+3 b+6 c=0(a, b, c \in R)$ then the quadratic equation $a x^2+b x+c=0$ has

Question

Aicharya · Accepted Answer

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

If $2 a+3 b+6 c=0(a, b, c \in R)$ then the quadratic equation $a x^2+b x+c=0$ has

Solution