JEE MAIN - Mathematics (2002 - No. 39)
If $$\alpha \ne \beta $$ but $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3$$ then the equation having $$\alpha /\beta $$ and $$\beta /\alpha \,\,$$ as its roots is
$$3{x^2} - 19x + 3 = 0$$
$$3{x^2} + 19x - 3 = 0$$
$$3{x^2} - 19x - 3 = 0$$
$${x^2} - 5x + 3 = 0$$
Explanation
We have $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3;$$
$$ \Rightarrow \alpha \,\,\& \,\,\beta $$ are roots of
equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$
$$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$
Thus, the equation having $${\alpha \over \beta }\,\,\& \,\,{\beta \over \alpha }$$ as its roots is
$${x^2} - x\left( {{\alpha \over \beta } + {\beta \over \alpha }} \right) + {{\alpha \beta } \over {\alpha \beta }} = 0$$
$$ \Rightarrow {x^2} - x\left( {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right) + 1 = 0$$
or $$3{x^2} - 19x + 3 = 0$$
$$ \Rightarrow \alpha \,\,\& \,\,\beta $$ are roots of
equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$
$$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$
Thus, the equation having $${\alpha \over \beta }\,\,\& \,\,{\beta \over \alpha }$$ as its roots is
$${x^2} - x\left( {{\alpha \over \beta } + {\beta \over \alpha }} \right) + {{\alpha \beta } \over {\alpha \beta }} = 0$$
$$ \Rightarrow {x^2} - x\left( {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right) + 1 = 0$$
or $$3{x^2} - 19x + 3 = 0$$
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