Let `f(x)` be a quadratic expression such that `f(-1)+f(2)=0`. If one root of `f(x)=0` is `3`, then the other root of `f(x)=0` lies in (A) `(-oo,-3)` (B) `(-3,oo)` (C) `(0,5)` (D) `(5,oo)`

Options

A. `(-oo,-3)`
B. `(-3,oo)`
C. `(0,5)`
D. `(5,oo)`

Answer

B

Explanation

`(b)` Since one root is `3`, let
`f(x)=(x-3)(ax+b)`
`implies f(x)=ax^(2)+(b-3a)x-3b`……….`(1)`
Given, `f(-1)+f(2)=0`
`implies (a+(3a-b)-3b)+(4a+2b-6a-3b)=0`
`implies 2a-5b=0`
`implies (b)/(a)=(2)/(5)` ………..`(2)`
Now, product of roots `=(-3b)/(a)=-3((2)/(5))`
since one root of `f(x)=0` is `3`, the oher root `=(-2)/(5) in (-3,0)`