Question

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

Answer 1

$\alpha ,\beta satisfy{x}^2=5x-3i.e.x^2-5x+3=0$
$\therefore \alpha +\beta =5,\alpha \beta =3$
$\Rightarrow$ Sum$=\frac{a}{\beta }+\frac{\beta }{a}=\frac{a^2+b^2}{\alpha \beta }=\frac{19}{3}$
and product = 1
$\therefore$ Equation is $x^2-S.x+P=0$
$\Rightarrow x^2-\frac{19}{3}x+1=0$ or $3x^2-19x+3=0$