Question

If reciprocal of every root of the equation $2x^3-7x^2+px-q=0$, (where $p,q$ are positive integers) is also a root then $(2p-q)$ is divisible by

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (A), (B), (C)

$2x^3-7x^2+px+q=0$
Let the root are $\alpha ,\frac{1}{\alpha },1$
$\therefore \alpha +\frac{1}{\alpha }+1=\frac{7}{2}$
$\Rightarrow 2\left({\alpha }^2+\alpha +1\right)=7\alpha$
$\Rightarrow 2{\alpha }^2-5\alpha +2\Rightarrow \alpha =2\text{ or }\frac{1}{2}$
$\therefore \text{ roots are }2,\frac{1}{2},1$
$\frac{q}{2}=1\Rightarrow q=2$
$\text{ and }p-q=5\Rightarrow p=7$
$\therefore 2p-q=14-2=12$