How do you solve $x/(x^2 - 16)>0$ ?

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Answer 1 · 2016-06-17T00:24:11

$x/(x^2-16) > 0 iff x in (-4,0)uu(4,oo)$

An expression $f(x)/g(x)$ will be greater than $0$ when either both $f(x)$ and $g(x)$ are positive or both $f(x)$ and $g(x)$ are negative.

Looking at the signs of the given functions, we have

$x < 0 if x in (-oo,0)$
$x>0 if x in (0,oo)$

and

$x^2-16 < 0 if x in (-4,4)$
$x^2-16 > 0 if x in (-oo,-4)uu(4,oo)$

If we compare the signs, then, we have both being negative on $(-4,0)$ and both being positive on $(4,oo)$ .

Thus, we have $x/(x^2-16) > 0 iff x in (-4,0)uu(4,oo)$

How do you solve x/(x^2 - 16)>0x/(x^2 - 16)>0?