Question

Let $f\left(x\right)$ be a quadratic polynomial such that $f(-1)+f\left(2\right)=0$. If one of the roots of $f\left(x\right)=0$ is $3$, then its other roots lies in

• A. $\left(0,1\right)$
• B. $\left(1,3\right)$
• C. $\left(-1,0\right)$
• D. $\left(-3,-1\right)$

Answer 1

The correct option is C

$\left(-1,0\right)$

Explanation for the correct option:

Finding the other root:

let $f=a\left(x-\alpha \right)\left(x-\beta \right)$

$f(-1)+f\left(2\right)=0$

$$\begin{gathered} a\left[\left(-1-\alpha \right)\left(-1-\beta \right)+\left(2-\alpha \right)\left(2-\beta \right)\right]=0 \\ \Rightarrow a\left[5\alpha +2\right]=0 \\ \Rightarrow \alpha =\frac{-2}{5}\in \left(-1,0\right) \end{gathered}$$

Hence option (C) is the correct answer