Question

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

Answer 1

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