<s>[INST] $xy + ax^2 + bx + c = 0 , x+y$ has relative max or min if... Options :
$b^2-4ac > 0$
$b^2/4ac >0$
$b/(c-1) >0$
$c/(a-1)>0$
$a/(b-1)>0$
Relative max or min is local max or min, means max or min at certain open interval..
$xy + ax^2 + bx + c = 0 $ 
$ax^2 + (y+b)x + c = 0 $  if its a qudratic fuction.
Then the max or min point will be (-b/2a , f(-b/2a))
$y = (-ax^2 - bx - c)/x$
But what does it mean by $x+y$ ?
When $x+y$ will have relative max or min?
 [/INST] Get $y$ by itself so that $$y = \frac{-ax^{2}-bx-c}{x}$$ 
Then substitute this for $y$ in $f(x) = x+y$ so we have 
$$f(x) = \frac{-ax^{2}-bx-c}{x} + x$$ 
and we want to know if $f(x)$ will have a relative minimum or maximum. The cases where it will not have a minimum or maximum are when the derivative has no real zeroes. Lets find the derivative using the product rule: 
$$f'(x) = \frac{1}{x}(-2ax-b)-\frac{1}{x^{2}}(-ax^{2}-bx-c)+1$$
Simplify fractions
$$f'(x) = \frac{x(-2ax-b)-(-ax^{2}-bx-c)+x^{2}}{x^{2}}$$
Simplify
$$f'(x) = \frac{(-a+1)x^{2} + c}{x^{2}}$$
We wish to know when the derivative will be $0$, and we can multiply away the denominator to get
$$0 = (-a+1)x^{2} + c$$
$$x^{2} = \frac{-c}{-a+1} = \frac{c}{a-1}$$
Notice that we need real roots in order to have a relative minimum or maximum, and we are taking a square root, thus
$$\frac{c}{a-1}>0$$
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