Question

Let $f\left(x\right)=a{x}^2+bx+c$, where a, b and c are certain constants and a $\ne 0$. It is known that $f\left(5\right)=-3f\left(2\right)$ and that 3 is a root of $f\left(x\right)=0$.
What is the other root of f(x) = 0?

• A. -7
• B. -4
• C. 2
• D. 6
• E. cannot be determined

Answer 1

Answer: (A), (C)

The correct options are
A -7
C 2

$f\left(x\right)=a{x}^2+bx+c$
Given $f\left(5\right)=-3f\left(2\right)\Rightarrow 25a+5b+c=-3(4a+2b+c)\to 37a+11b+4c=0$ ---(1)
Also 3 is a root of $f\left(x\right)=0\Rightarrow f\left(3\right)=0\Rightarrow 9a+3b+c=0$ ---(2)
On solving (1) and (2) we get, $b=a,c=-12a$ ---(3)
Substituting (3) in $f\left(x\right)=0\Rightarrow a{x}^2+ax-12a=0\Rightarrow a(x2+x-12)=0$
Given that a $\ne 0\Rightarrow (x-3)(x+4)=0\Rightarrow x=3\left(or\right)-4$
x=3 root is already given hence the other root is -4
Answer option for first question is (2)