Question

Let $a,b,c$ be three real number such that $2a+3b+6c=0$ Then the quadratic equation $a{x}^2+bx+c=0$ has

• A. imaginary roots
• B. at least one roots in $(0,1)$
• C. at least one roots in $(-1,0)$
• D. both roots in $(1,2)$

Answer 1

The correct option is B at least one roots in $(0,1)$

Given that $2a+3b+6c=0$
consider, $f\left(x\right)=\frac{a{x}^3}{3}+\frac{{bx}^2}{2}+cx+d$
Since, $f\left(0\right)=f\left(1\right)$
Therefore, $f^{{}^{\prime }}\left(x\right)=0\Rightarrow a{x}^2+bx+c=0$ at least once in $(0,1)$ [ mean value theorem ]

Ans: B