Let \(\alpha, \beta\) be the roots of the equation \(x^2-|a| x-|b|=0\) such that \(|\alpha| < |\beta|\). If…
Let \(\alpha, \beta\) be the roots of the equation \(x^2-|a| x-|b|=0\) such that \(|\alpha| < |\beta|\). If \(|a| < \beta-1\), then the positive root of \(\log _{\mid \text {여 }}\left(\frac{x^2}{\beta^2}\right)-1=0\), is
\( < |\alpha|\)
\( < \alpha\)
\( < \beta\)
\( > \beta\)
Solution
Given, \(\log _{|\alpha|}\left(\frac{x^2}{\beta^2}\right)=1\)
\(\Rightarrow \quad \frac{x^2}{\beta^2}=|\alpha| \Rightarrow x^2=\beta^2|\alpha|\)
and \(\alpha, \beta\) are roots of equation
\(x^2-|a| x-|b|=0\)...(i)
so, \(\alpha+\beta=-|a| < 0\)
so, \(\beta < 0\) as \(|\alpha| < |\beta|\)
From Eq. (i), we get \(x= \pm|\beta \| \alpha|^{1 / 2}\)
So, the positive root is
\(\left.|\beta \| \alpha|^{1 / 2} > \beta \quad \text { [as } \beta < 0\right]\)