Question:
Let $f(x)$ be a quadratic polynomial such that $\mathrm{f}(-1)+\mathrm{f}(2)=0$. If one of the roots of $\mathrm{f}(\mathrm{x})=0$ is 3 , then its other root lies in :
Correct Option: , 3
Solution:
$f(x)=a(x-3)(x-\alpha)$
$f(2)=a(\alpha-2)$
$f(-1)=4 a(1+\alpha)$
$f(-1)+f(2)=0 \Rightarrow a(\alpha-2+4+4 \alpha)=0$
$a \neq 0 \Rightarrow 5 \alpha=-2$
$\alpha=-\frac{2}{5}=-0.4$
$\alpha \in(-1,0)$
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