Question

If $\alpha ,\beta$ are the zeroes of $x^2+px+q$, find the value of $(\frac{\alpha }{\beta }+2).(\frac{\beta }{\alpha }+2)$.

Answer 1

Equation $\to x^2+px+q$

$\alpha$ and $\beta$ are its zeroes

$\alpha +\beta =-p$ and $\alpha \beta =q$

$(\frac{\alpha }{\beta }+2).(\frac{\beta }{\alpha }+2)=1+\frac{2\alpha }{\beta }+\frac{2\beta }{\alpha }+4$

$=5+2(\frac{\alpha }{\beta }+\frac{\beta }{\alpha })$

$=5+2\frac{({\alpha }^2+{\beta }^2)}{\alpha \beta }$

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

$5+2\frac{[p^2-2q]}{q}$

$=\frac{5q+2p^2-4q}{q}$

$=\frac{2p^2+q}{q}$