Why can't I simply use algebra to solve this inequality? 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.
 A: Note that in Calculus whenever you have something like $A/B=C$ and you want to eliminate the denominator by multiplication, you should assume that $B\neq 0$. Moreover if $$A/B\le C, ~B\neq 0$$ then $$B>0\to A\le BC,~~~B<0\to A\ge BC$$
A: When multiplying both sides by negative number the sign between both sides change to opposite, as for example $1 < 2$, but $-1 = 1\cdot(-1) > 2\cdot(-1) = -2$ and you would need to consider different cases depending on the sign of denominator. So first you would need to solve $x(x+2) \geq 0$ which gives $x\in\left[-\infty, -2\right]\cup\left[0, \infty\right]$. For those numbers we have $(x+3)(x-5) \geq 0$, and for $x\in(-2, 0)$ we have $(x+3)(x-5) \leq 0$ which is a little bit of work. Instead it's better to just use the fact that
$$\frac{f(x)}{g(x)}\geq 0 \iff f(x)g(x)\geq 0 \text{ and } g(x) \neq 0$$
A: if we assume that $x \gt 0$ then $(x+2) \gt 0$ so
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \ge 0$$
but if $x \cdot (x+2) < 0$
$$\frac{(x+3)(x-5)}{x(x+2)} \ge 0 \Rightarrow (x+3)(x-5) \le 0$$
and if $x = 0$ or $x = -2$ then $\frac{(x+3)(x-5)}{x(x+2)}$ is undefined.
A: If $x(x+2)$ is negative, then you need to switch the direction of the inequality.
A: 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 $(5, \infty)$.
A: You need to be sure that $x\neq 0$ and $(x+2)\neq 0$. Moreover, multiplying both sides of the inequality by $x(x+2)$ has the effect of changing the inequality if $x(x+2)<0$. In other words, you need to study also the sign of the denominator of the given fraction.
