${\beta ^2} - 5\beta + 3 = 0$ …..$(ii)$
From $(i) -(ii),$
==> $({\alpha ^2} - {\beta ^2}) - 5\alpha + 5\beta = 0$
==> ${\alpha ^2} - {\beta ^2} = 5(\alpha - \beta )$$ \Rightarrow \alpha + \beta = 5$
From $(i) + (ii),$
==> $({\alpha ^2} + {\beta ^2}) - 5(\alpha + \beta ) + 6 = 0$
==> $({\alpha ^2} + {\beta ^2}) - 5.5 + 6 = 0$ ==> ${\alpha ^2} + {\beta ^2} = 19$
Then ${(\alpha + \beta )^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta $
$ \Rightarrow 25 = 19 + 2\alpha \beta $ $ \Rightarrow \alpha \beta = 3$
then the equation, whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$, is
${x^2} - x\left( {\frac{\alpha }{\beta } + \frac{\beta }{\alpha }} \right) + \frac{\alpha }{\beta }.\frac{\beta }{\alpha } = 0$
==> ${x^2} - x\left( {\frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}} \right) + 1 = 0$
$ \Rightarrow {x^2} - x.\frac{{19}}{3} + 1 = 0$
==> $3{x^2} - 19x + 3 = 0$.
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