If $\alpha, \beta$ are the roots of the equation $x^{2}+p x+q=0$ then $-\frac{1}{\alpha}+\frac{1}{\beta}$ are the roots of the equation
(a) $x^{2}-p x+q=0$
(b) $x^{2}+p x+q=0$
(c) $q x^{2}+p x+1=0$
(d) $q x^{2}-p x+1=0$
(d) $q x^{2}-p x+1=0$
Given equation: $x^{2}+p x+q=0$
Also, $\alpha$ and $\beta$ are the roots of the given equation.
Then, sum of the roots $=\alpha+\beta=-p$
Product of the roots $=\alpha \beta=q$
Now, for roots $-\frac{1}{\alpha},-\frac{1}{\beta}$, we have:
Sum of the roots $=-\frac{1}{\alpha}-\frac{1}{\beta}=-\frac{\alpha+\beta}{\alpha \beta}=-\left(\frac{-p}{q}\right)=\frac{p}{q}$
Product of the roots $=\frac{1}{\alpha \beta}=\frac{1}{q}$
Hence, the equation involving the roots $-\frac{1}{\alpha},-\frac{1}{\beta}$ is as follows:
$x^{2}-(\alpha+\beta) x+\alpha \beta=0$
$\Rightarrow x^{2}-\frac{p}{q} x+\frac{1}{q}=0$
$\Rightarrow q x^{2}-p x+1=0$