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 :

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 : (A) (-3,-1) (B) (1,3) (C) (-1,0) (D) (0,1)

Tardigrade Admin · Accepted Answer

f(x)=a(x-3)(x-α) f(2)=a(α-2) f(-1)=4 a(1+α) f(-1)+f(2)=0 ⇒ a(α-2+4+4 α)=0 a ≠ 0 ⇒ 5 α=-2 α=-(2/5)=-0.4 α ∈(-1,0)

Q. 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 :

$(- 3, - 1)$

$(1, 3)$

$(- 1, 0)$

$(0, 1)$

$f (x) = a (x - 3) (x - α)$
$f (2) = a (α - 2)$
$f (- 1) = 4 a (1 + α)$
$f (- 1) + f (2) = 0 \Rightarrow a (α - 2 + 4 + 4 α) = 0$
$a \neq = 0 \Rightarrow 5 α = - 2$
$α = - \frac{2}{5} = - 0.4$
$α \in (- 1, 0)$

Q. 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 :

(−3,−1)

(1,3)

(−1,0)

(0,1)