1. Tardigrade
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  4. For x, y such that x>y>0, x y=4, find the minimum value of the expression E=((x+y)2/x-y).

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Q. For $x, y$ such that $x>y>0, x y=4$, find the minimum value of the expression $E=\frac{(x+y)^2}{x-y}$.

Complex Numbers and Quadratic Equations

Solution:

$E=\frac{(x-y)^2+4 x y}{x-y}=\frac{(x-y)^2+16}{x-y}=(x-y)+\frac{16}{x-y}$
$=\left(\sqrt{x-y}-\frac{4}{\sqrt{x-y}}\right)^2+8$
$\therefore E \geq 8 $
$\therefore E_{\min }=8$.