# Let f(x) be a quadratic polynomial such that

Question:

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

1. (1) $(-1,0)$

2. (2) $(1,3)$

3. (3) $(-3,-1)$

4. (4) $(0,1)$

Correct Option: 1

Solution:

Let $f(x)=a x^{2}+b x+c$

Given: $f(-1)+f(2)=0$

$a-b+c+4 a+2 b+c=0$

$\Rightarrow 5 a+b+2 c=0 \quad \ldots$ (i)

and $f(3)=0 \Rightarrow 9 a+3 b+c=0$...(ii)

From equations (i) and (ii),

$\frac{a}{1-6}=\frac{b}{18-5}=\frac{c}{15-9} \Rightarrow \frac{a}{-5}=\frac{b}{13}=\frac{c}{6}$

Product of roots, $\alpha \beta=\frac{c}{a}=\frac{-6}{5}$ and $\alpha=3$

$\Rightarrow \beta=\frac{-2}{5} \in(-1,0)$