Let f(x) be a quadratic polynomial such that f(-2) +f(3)=0 . If one of roots of f(x)=0 is -1 , then

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1.Relation and Function

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

A

$\frac{11}{3}$

B

$\frac{7}{3}$

C

$\frac{13}{3}$

D

$\frac{14}{3}$

(JEE MAIN-2022)

Solution

$f (-2)+ f (3)=0$

$f ( x )=( x +1)( ax + b )$

$f (-2)+ f (3)=-1(-2 a + b )+4(3 a + b )=0$

$2 a – b +12 a +4 b =0$

$14 a +3 b =0$

$\frac{- b }{ a }=\frac{14}{3}$

Sum of roots $=\left(-1+\frac{-b}{a}\right)=-1+\frac{14}{3}=\frac{11}{3}$

Standard 12

Mathematics

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