Question

Let $f\left(x\right)$ be a quadratic polynomial such that $f(-2)+f\left(3\right)=0.$ If one of the roots of $f\left(x\right)=0$ is $-1,$ then the sum of the roots of $f\left(x\right)=0$ is equal to:

Answer 1

$\because x=-1$ be the roots of $f\left(x\right)=0$
$\therefore$ let $f\left(x\right)=A(x+1)(x–b)$ …(i)
Now, $f(–2)+f\left(3\right)=0$
$\Rightarrow A[-1(-2-b)+4(3-b\left)\right]=0$
$b=\frac{14}{3}$
$\therefore$ Second root of $f\left(x\right)=0$ will be $\frac{14}{3}$
$\therefore$ Sum of roots $=\frac{14}{3}-1=\frac{11}{3}$