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Q.
If $x$ and $y$ are real numbers with $x>y$ and $x y=1$ then the minimum possible value of $\frac{x^2+y^2}{x-y}$ is
Complex Numbers and Quadratic Equations
Solution:
$Z =\frac{ x ^2+\frac{1}{ x ^2}}{ x -\frac{1}{ x }}=\frac{\left( x -\frac{1}{ x }\right)^2+2}{ x -\frac{1}{ x }}= x -\frac{1}{ x }+\frac{2}{ x -\frac{1}{ x }}= t +\frac{2}{ t }( t >0)$
$=\left(\sqrt{t}-\sqrt{\frac{2}{t}}\right)^2+2 \sqrt{2} $
$\therefore Z_{\min }=2 \sqrt{2} \text { when } t=\sqrt{2} \text { i.e. } x-\frac{1}{x}=\sqrt{2}$