# If α, β are the roots of the equation x

Question:

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) $q x^{2}-p x+1=0$

Solution:

(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:

$x^{2}-(\alpha+\beta) x+\alpha \beta=0$

$\Rightarrow x^{2}-\frac{p}{q} x+\frac{1}{q}=0$

$\Rightarrow q x^{2}-p x+1=0$