Question

If $\alpha$, $\beta$ are the roots of the equation $x^2-px+q=0$ and ${\alpha }_1,{\beta }_1$ be the roots of the equation $x^2-qx+p=0$, then form the quadratic equation whose roots are $\frac{1}{{\alpha }_1\beta }+\frac{1}{\alpha {\beta }_1}$ and $\frac{1}{\alpha {\alpha }_1}+\frac{1}{\beta {\beta }_1}$

Answer 1

Given, $\alpha$ and $\beta$ roots of $x^2-px+q=0$

$\therefore \alpha +\beta =p$

$\alpha \beta =q$

And $\alpha$, and $\beta$ are roots of $x^2-qx+p=0$

$\therefore {\alpha }_1+{\beta }_1=q$

${\alpha }_1.{\beta }_1=p$

We have to find quadratic equation whose whose roots are

$\frac{1}{{\alpha }_1\beta }+\frac{1}{\alpha {\beta }_1}$ and $\frac{1}{\alpha {\alpha }_1}+\frac{1}{\beta {\beta }_1}$

Now,

$\frac{1}{{\alpha }_1\beta }+\frac{1}{\alpha {\beta }_1}+\frac{1}{\alpha {\alpha }_1}+\frac{1}{\beta {\beta }_1}=\frac{1}{{\alpha }_1}[\frac{1}{\alpha }+\frac{1}{\beta }]+\frac{1}{{\beta }_1}[\frac{1}{\alpha }+\frac{1}{\beta }]$

$=\frac{\alpha +\beta }{\alpha \beta }\times \frac{{\alpha }_1+{\beta }_1}{{\alpha }_1{\beta }_1}=\frac{p}{q}\times \frac{q}{p}=1$

and

$(\frac{1}{\alpha \beta }+\frac{1}{\alpha {\beta }_1})\times (\frac{1}{{\alpha }_1\alpha }+\frac{1}{{\beta }_1\beta })=\frac{1}{{\alpha }_1^2\alpha \beta }+\frac{1}{{\alpha }_1{\beta }_1{\beta }^2}+\frac{1}{{\beta }_1{\alpha }_1{\alpha }^2}+\frac{1}{{\beta }_1^2\alpha \beta }$

$=\frac{1}{\alpha \beta }[\frac{1}{{\alpha }_1^2}+\frac{1}{{\beta }_1^2}]+\frac{1}{{\alpha }_1{\beta }_1}[\frac{1}{{\beta }^2}+\frac{1}{{\alpha }^2}]$

$=\frac{1}{q}\left[\frac{{\beta }_1^2+{\alpha }_1^2}{{\alpha }_1^2.{\beta }_1^2}\right]+\frac{1}{p}\left[\frac{{\alpha }^2+{\beta }^2}{{\alpha }^2{\beta }^2}\right]$

$=\frac{1}{q.p^2}\left[\right({\alpha }_1+{\beta }_1{)}^2-2{\alpha }_1{\beta }_1]+\frac{1}{p{q}^2}[(\alpha +\beta {)}^2-2\alpha \beta ]$

$=\frac{1}{q.p^2}[q^2-2p]+\frac{1}{p{q}^2}[p^2-2q]$

$=\frac{q^3+p^3-pq}{p^2{q}^2}$

Therefore Equation is

$x^2-x+\frac{q^3+{\beta }^2-4pq}{p^2{q}^2}$