Question

Let $\left(p+\frac{1}{p}\right)\left(q+\frac{1}{q}\right)\left(r+\frac{1}{r}\right)<0$ where $p,q,r\in R-\{0\}$ and $\alpha =\frac{|p|}{p}+\frac{|q|}{q}+\frac{|r|}{r}$. If the equation $x^2+(m-2)x-n=0$ is satisfied by distinct values of ' $\alpha$ ' then find the value of $(m+n)$.

Answer 1

Answer: 7

$\left(p+\frac{1}{p}\right)\left(q+\frac{1}{q}\right)\left(r+\frac{1}{r}\right)<0$
$\Rightarrow p,q,r$ all three are negative or exactly one of then is negative
$\therefore \alpha =-3$ or 1

$-(m-2)=-2\Rightarrow m=4$
$-n=3\Rightarrow n=3$
$\therefore m+n=7.$