Positive reals $x, y$ satisfy $x \neq y$ and $\frac{x^2 + y^2}{xy} = k$. If replacing $x$ by $x + y$ and $y$ by $|x - y|$ leaves the value of $k$ unchanged, then $k$ equals

Given $\dfrac{x^2 + y^2}{xy} = k$.

After replacement, the new $k$ is:

$\dfrac{(x+y)^2 + (x-y)^2}{(x+y)|x-y|} = \dfrac{2(x^2 + y^2)}{(x+y)|x-y|}$

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Since $x, y$ are positive reals, $(x+y)|x-y| = |x^2 - y^2|$.

Set the new $k$ equal to the old $k$:

$\dfrac{2(x^2 + y^2)}{|x^2 - y^2|} = \dfrac{x^2 + y^2}{xy}$

Cancel $(x^2 + y^2)$, which is positive:

$\dfrac{2}{|x^2 - y^2|} = \dfrac{1}{xy}$

$|x^2 - y^2| = 2xy$

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Divide by $xy$:

$\left|\dfrac{x}{y} - \dfrac{y}{x}\right| = 2$

Let $r = \dfrac{x}{y} + \dfrac{y}{x} = k$ (this is the original $k$).

Using the identity $\left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2 - \left(\dfrac{x}{y} - \dfrac{y}{x}\right)^2 = 4$:

$k^2 - 4 = 4$

$k^2 = 8$, so $k = 2\sqrt{2}$ (positive since $x, y > 0$).

Answer $= 2\sqrt{2}$