Question

If $\frac{\alpha -1}{\alpha }$ and $\frac{\beta -1}{\beta }$ are the roots of the quadratic equation $x^2-2px-1=0,p\in R$, then find the minimum value of $\left(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}\right)$.

Answer 1

Answer: 3

$\frac{\alpha -1}{\alpha }+\frac{\beta -1}{\beta }=2p\Rightarrow 1-\frac{1}{\alpha }+1-\frac{1}{\beta }=2p\Rightarrow 2-2p=\frac{1}{\alpha }+\frac{1}{\beta }\dots \dots .\text{ (i) }$
$\left(1-\frac{1}{\alpha }\right)\left(1-\frac{1}{\beta }\right)=1\Rightarrow 1-\frac{1}{\alpha }-\frac{1}{\beta }+\frac{1}{\alpha \beta }=1$....(ii)
$1-(2-2p)+\frac{1}{\alpha \beta }=-1\Rightarrow \frac{1}{\alpha \beta }=-2p$
$\frac{1}{{\alpha }^2+{\beta }^2}={\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)}^2-\frac{2}{\alpha \beta }=(2-2p{)}^2+4p=4\left(p^2-p+1\right)=4\left[{\left(p-\frac{1}{2}\right)}^2+\frac{3}{4}\right]$
${\therefore \left(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}\right)|}_{\min .}=3$