Question
If $$\alpha ,\beta $$ be the roots of the equation $$x^2 - px + q = 0$$ and $${\alpha_1} , {\beta_1} $$ the roots of the equation $$x^ 2 - qx + p = 0 ,$$ then the equation whose roots are $$\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\,{\text{and}}\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}{\text{is}}$$
A.
$$pq{x^2} - pqx + {p^2} + {q^2} + 4pq = 0$$
B.
$${p^2}{q^2}{x^2} - {p^2}{q^2}x + {p^3} + {q^3} - 4pq = 0$$
C.
$${p^3}{q^3}{x^3} - {p^3}{q^3}x + {p^4} + {q^4} - 4{p^2}{q^2} = 0$$
D.
$$\left( {p + q} \right){x^2} - \left( {p + q} \right)x + {p^2} + {q^2} + pq = 0$$
Answer :
$${p^2}{q^2}{x^2} - {p^2}{q^2}x + {p^3} + {q^3} - 4pq = 0$$
Solution :
Here, $$\alpha + \beta = p,\alpha \beta = q$$
$${\alpha _1} + {\beta _1} = q,{\alpha _1}{\beta _1} = p$$
Sum of given roots
$$\eqalign{
& = \left( {\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}} \right)\,\left( {\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}} \right) \cr
& = \frac{{\alpha {\beta _1} + {\alpha _1}\beta + \beta {\beta _1} + \alpha {\alpha _1}}}{{\alpha \beta {\alpha _1}{\beta _1}}} \cr
& = \frac{{\left( {\alpha + \beta } \right)\left( {{\alpha _1} + {\beta _1}} \right)}}{{\left( {\alpha \beta } \right)\left( {{\alpha _1}{\beta _1}} \right)}} = \frac{{pq}}{{qp}} = 1 \cr} $$
Product of given roots
$$\eqalign{
& = \left( {\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}} \right)\left( {\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}} \right) \cr
& = - \frac{{\left( {\alpha {\beta _1} + {\alpha _1}\beta } \right)\left( {\alpha {\alpha _1} + \beta {\beta _1}} \right)}}{{{\alpha ^2}{\beta ^2}\alpha _1^2\beta _1^2}} \cr
& = \frac{{\alpha \beta \left( {\alpha _1^2 + \beta _1^2} \right) + {\alpha _1}{\beta _1}\left( {{\alpha ^2} + {\beta ^2}} \right)}}{{{\alpha ^2}{\beta ^2}\alpha _1^2\beta _1^2}} \cr
& = \frac{{\alpha \beta \left[ {{{\left( {{\alpha _1} + {\beta _1}} \right)}^2} - 2{\alpha _1}{\beta _1}} \right] + {\alpha _1}{\beta _1}\left[ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right]}}{{{{\left( {\alpha \beta } \right)}^2}{{\left( {{\alpha _1}{\beta _1}} \right)}^2}}} \cr
& = \frac{{q\left( {{q^2} - 2p} \right) + p\left( {{p^2} - 2q} \right)}}{{{q^2}{p^2}}} \cr
& = \,\frac{{{p^3} + {q^3} - 4pq}}{{{p^2}{q^2}}} \cr} $$
Hence, the required equation is
$$\left( {{p^2}{q^2}} \right){x^2} - \left( {{p^2}{q^2}} \right)x + {p^3} + {q^3} - 4pq = 0$$