<s>[INST] Horizontal line(s) that intersect $f(x)=x-2+\frac{5}{x}$ in two points. Compute exactly the value(s) of $q$ for which the horizontal line $y = q$
intersects the graph of $f(x)$ in two points that are located on a distance $4$ from each other.
While I found the two lines as $y = 4$ and $y = -8$
, I do not know how to find it by any mathematical means.
 [/INST] We want to values of $x$ (of distance $4$ apart) such that $x-2 + \frac{5}{x} = q$, but this is equivalent to $x^2 - 2x + 5 = qx$ by multiplying throughout by $x$, more simply:  $x^2 - (2+q)x + 5 = 0$. 
Now say that the smaller point is $\alpha$, then the other one is necessarily $\alpha + 4$. Hence $2\alpha + 4= 2 +q$ and $\alpha(\alpha + 4) = 5$ using the relationship between roots and coefficients. 
Isolating $\alpha$ from the first equation and plugging it into the second gives $(q-2)(q+6) = 20$ and equivalently $(q-4)(q+8) = 0.$
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