NIMCET 2021 Question Paper with solutions

Given ${\mathrm{\alpha }}^2=5\mathrm{\alpha }-3$
and ${\mathrm{\beta }}^2=5\mathrm{\beta }-3$
hence, $\mathrm{\alpha }$ and $\mathrm{\beta }$ satisfies ${\mathit{x}}^2=5\mathit{x}-3$
${\mathit{x}}^2-5\mathit{x}+3=0$ .......(1)
sum of roots $\mathrm{\alpha }+\mathrm{\beta }=-\mathit{b}∕\mathit{a}=5$
product of roots $\mathrm{\alpha }.\mathrm{\beta }=\mathit{c}∕\mathit{a}=3$
quadratic equation whose roots are $\frac{\mathrm{\alpha }}{\mathrm{\beta }},\frac{\mathrm{\beta }}{\mathrm{\alpha }}$ is given by
${\mathit{x}}^2-\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)\mathit{x}+\left(\frac{\mathrm{\alpha }}{\mathrm{\beta }}\times \frac{\mathrm{\beta }}{\mathrm{\alpha }}\right)=0$ ......(2)
Now, $\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}=\frac{{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2}{\mathrm{\alpha }\mathrm{\beta }}$
$\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}=\frac{{\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^2-2\mathrm{\alpha }\mathrm{\beta }}{\mathrm{\alpha }\mathrm{\beta }}$
$\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}=\frac{25-2\times 3}{3}$
$\frac{\mathrm{\alpha }}{\mathrm{\beta }}+\frac{\mathrm{\beta }}{\mathrm{\alpha }}=\frac{19}{3}$
And $\frac{\mathrm{\alpha }}{\mathrm{\beta }}\times \frac{\mathrm{\beta }}{\mathrm{\alpha }}=1$
On putting this value in equation (2), we get
${\mathit{x}}^2-\left(\frac{19}{3}\right)\mathit{x}+1=0$
$3{\mathit{x}}^2-19\mathit{x}+3=0$