If $\mathrm{\alpha }\ne \mathrm{\beta } \mathrm{but} {\mathrm{\alpha }}^2=5\mathrm{\alpha }-3$ and ${\mathrm{\beta }}^2=5\mathrm{\beta }-3$, then the equation whose roots are $\mathrm{\alpha }/\mathrm{\beta }$ and $\mathrm{\beta }/\mathrm{\alpha }$ is

${\mathrm{\alpha }}^2-5\mathrm{\alpha }+3=0 ...\left(\mathrm{i}\right)$

${\mathrm{\beta }}^2-5\mathrm{\beta }+3=0 ....\left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)-\left(\mathrm{ii}\right),$

$\Rightarrow ({\mathrm{\alpha }}^2-{\mathrm{\beta }}^2)-5\mathrm{\alpha }+5\mathrm{\beta }=0$

$\Rightarrow {\mathrm{\alpha }}^2-{\mathrm{\beta }}^2=5(\mathrm{\alpha }-\mathrm{\beta })\Rightarrow \mathrm{\alpha }+\mathrm{\beta }=5$

From $\left(\mathrm{i}\right)+\left(\mathrm{ii}\right),$

$\Rightarrow ({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2)-5(\mathrm{\alpha }+\mathrm{\beta })+6=0$

$\Rightarrow ({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2)-5.5+6=0\Rightarrow {\mathrm{\alpha }}^2+{\mathrm{\beta }}^2=19$

Then ${(\mathrm{\alpha }+\mathrm{\beta })}^2={\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+2\mathrm{\alpha \beta }$

$\Rightarrow 25=19+2\mathrm{\alpha \beta }\Rightarrow \mathrm{\alpha \beta }=3$

then the equation, whose roots are $\frac{\mathrm{\alpha }}{\mathrm{\beta }} \mathrm{and} \frac{\mathrm{\beta }}{\mathrm{\alpha }}$, is

${\mathrm{x}}^2-\mathrm{x}(\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }})+\frac{\mathrm{\alpha }}{\mathrm{\beta }}.\frac{\mathrm{\beta }}{\mathrm{\alpha }}=0$

$\Rightarrow {\mathrm{x}}^2-\mathrm{x}\left(\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha \beta }}\right)+1=0$

$\Rightarrow {\mathrm{x}}^2-\mathrm{x}.\frac{19}{3}+1=0\Rightarrow 3{\mathrm{x}}^2-19\mathrm{x}+3=0$