If $a+b+c = 0$ then the quadratic equation $3ax^{2}+2bx +c=0$ has atleast one root in _________?

Solution 1:

Rolle's theorem states that, if a function is continuous in $[a,b]$ , differentiable

in $(a,b)$, and $f(a)=f(b)$ then $f^{\prime}(c)=0$, $\forall\,c\in(a,b)$

This implies there is at least one root of $f^{\prime}(x)$ lying in $(a,b)$

We have to find the root of $\,3ax^{2}+2bx+c=0$

So, according to Rolle's theorem,

Let $\,f(x)=a{x}^{3}+b{x}^{2}+c{x}$

Now, $\,\,f(0)=0$

And, $\,\,f(1)=a+b+c=0\,\,\,\,[\mathsf{given}]$

Since $\,\,f(x)$ is a polynomial function, so, it is continuous in $\,[0,1]$ ,

differentiable in $\,\,(0,1)$ and $f(0)=f(1)$

Hence, there is at least one root of $f^{\prime}(x)=3a{x}^{2}+2b{x}+c$ lying in between $\,\,(0,1)$