Question

If $\alpha ,\beta$ be the roots of the equation $x^2+px+q=0,$ then the equation whose roots are $\frac{1}{\alpha +\beta }$ and $\frac{1}{\alpha }+\frac{1}{\beta }$ will be:

• A. None of the above
• B. $pq{x}^2+(p^2+q)x+p=0$
• C. $pq{x}^2+(p^2+q^2)x+p=0$
• D. $pq{x}^2+(q^2+p)x+p=0$

Answer 1

The correct option is B $pq{x}^2+(p^2+q)x+p=0$

Since, $\alpha ,\beta$ are the roots of $x^2+px+q=0$

$\therefore \alpha +\beta =-p$ and $\alpha .\beta =q$

Let, $\gamma =\frac{1}{\alpha +\beta }=\frac{1}{-p}$

And, $\delta =\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha .\beta }=\frac{-p}{q}$

So, Quadratic equation whose roots are $\gamma$ and $\delta ,$

$x^2-(\gamma +\delta )x+\gamma \delta =0$

$\Rightarrow x^2-(\frac{1}{-p}+\frac{-p}{q})x+\frac{1}{-p}\times \frac{-p}{q}=0$
Upon Simplification we will get the equation as:
$pq{x}^2+(p^2+q)x+p=0$