TITLE: Why can't I simply use algebra to solve this inequality?
QUESTION [7 upvotes]: Consider the inequality:
$\frac{(x+3)(x-5)}{x(x+2)}\geq 0$
Why can't I simply multiply both sides by $x(x+2)$ and get $(x+3)(x-5)\geq 0$ ?
Which would yield: $x^2-2x-15\geq 0$ and I could then use the quadratic formula to derive the answer..?
This seems algebraically correct but I assume that is because I am misunderstanding something fundamental about inequalities.

REPLY [14 votes]: You're making your life more difficult by expanding the numerator out and suggesting the use of the quadratic formula. In the expression $$Y = \frac{AB}{CD},$$ the sign of $Y$ is determined by the signs of $A, B, C, D$, or whether any of them are zero. Specifically, if either of $C$ or $D$ are zero, then $Y$ is undefined and has no sign. Otherwise, if either of $A$ or $B$ are zero, then $Y$ is zero. If all quantities in the numerator and denominator are nonzero, then $Y$ is positive if and only if there are an even number of negative quantities among $A, B, C, D$, and negative otherwise.
With that in mind, have a look at your expression, $$\frac{(x+3)(x-5)}{x(x+2)}.$$
Each monomial will change sign when it crosses zero, which are the points $-3, 5, 0, -2$, which in sorted order is $-3, -2, 0, 5$. So, with the above rules in mind, you just need to look at what happens on the intervals and points $(-\infty, -3), -3, (-3, -2), -2, (-2, 0), 0, (0, 5), 5,$ and $(5, \infty)$.