Question

Let $f(x)=p{x}^2+qx+r$ , where $p,q,r$ are constants and $p\ne 0$ . If $f(5)=-3f(2)$ and $f(-4)=0$ , then the other root of $f$ is

• A. 3
• B. -7
• C. -2
• D. 2
• E. 6

Answer 1

Answer: (A)

$f(x)=p{x}^2+qx+r$
$f(-4)=0$
$\Rightarrow 16p-4q+r=0$ ...(i)
One root is $x=-4$
and $f(5)=-3f(2)$
$25p+5q+r=-3(4p+2q+r)$
$\Rightarrow 37p+11q+4r=0$ ...(ii)
Eq. (ii) - Eq. (i), we get
$\Rightarrow -27p+27q=0$
$\Rightarrow p=0$
Then, equation is $p{x}^2+qx+r=0$
Roots $=-4,\alpha$
Sum of roots $=-4x+\alpha =-\frac{p}{q}=-1$
So, another root $\alpha =3$.