Question

If $a+b+c=0,$ then the quadratic equation $3a{x}^2+2bx+c=0$ has

• A. imaginary roots
• B. at least one real root in $[0,1]$
• C. one root in $[2,3]$ and the other in $[3,6]$
• D. none of these

Answer 1

Answer: (B)

at least one real root in $[0,1]$

Let $f\left(x\right)=a{x}^3+b{x}^2+cx$

$\therefore f^{\prime }\left(x\right)=3a{x}^2+2bx+c$

Clearly $f\left(x\right)$ is continuous in $[0,1],$ derivable in $(0,1)$ and $f\left(0\right)=0=f\left(1\right)$

$\because f\left(1\right)=a+b+c=0$ (given)

$\therefore$ By Rolle's theorem, there exists at least one real $x\in (0,1)$ such that $f^{\prime }\left(x\right)=0$

$\Rightarrow 3a{x}^2+2bx+c=0$ exist

Hence there exist at least one real root of $3a{x}^2+abx+c=0$ in $(0,1)$