Question

If ${\alpha }^2=5\alpha -3$ and ${\beta }^2=5\beta -3$ then the value of $\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$ is

Answer 1

${\alpha }^2=5\alpha -3-\left(1\right)$

${\beta }^2=5\beta -3-\left(2\right)$

$Addingeq1\&2$

${\alpha }^2+{\beta }^2=5\left(\alpha +\beta \right)-6-\left(3\right)$

$subtracting2from1$

${\alpha }^2-{\beta }^2=5\left(\alpha -\beta \right)$

$\alpha +\beta =5$

$from3$

${\alpha }^2+{\beta }^2=5\left(\alpha +\beta \right)-6$

${\alpha }^2+{\beta }^2=5\times 5-6$

${\alpha }^2+{\beta }^2=19$

$now$

${\alpha }^2+{\beta }^2=19$

$\left(\alpha +\beta \right)-2\alpha \beta =19$

${\left(5\right)}^2-2\alpha \beta =19$

$2\alpha \beta =6$

$\alpha \beta =3$

$so,$

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }$

$=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{19}{3}$