Let f(x) = $x^2+ax+b,a,b \in R$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is .....
Solution 1:
We have three real terms summing up to $0$.
They can't be all zero as a quadratic has at most two zeroes.
Hence at least one term is positive and at least one term is negative, hence the roots must be distinct real roots.
Solution 2:
Brute force,
\begin{align} \Delta &= a^2-4b \\ &=a ^2-4\left( -\frac{14+6a}{3} \right) \\ &= a^2+8a \color{red}{+16} +\frac{56}{3} \color{red}{-16} \\ &= (a+4)^2+\frac{8}{3} \\ &> 0 \end{align}