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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.

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    $\begingroup$ By the way, when you are confronted with a problem of the form $(x\pm a)(x\pm b)=0$, the correct approach is to divide by each monomial, not use quadratic formula. The quadratic formula is only for when you can't find the linear factors. $\endgroup$ – AJMansfield Nov 29 '13 at 1:55
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If $x(x+2)$ is negative, then you need to switch the direction of the inequality.

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  • $\begingroup$ So would the correct way be to first assume that x > 0 and solve the inequality, and then assume x < 0 and solve the inequality? So this is only true if it's true for both cases? $\endgroup$ – Cruncher Nov 28 '13 at 20:06
  • $\begingroup$ @Cruncher The denominator is negative for $-2<x<0$ only. I think the easiest way is to use a sign chart. Write down for what values the numerator is positive and for which it is negative, and do the same for the denominator. Then, when both are positive or when both are negative, that is your solution (plus when the numerator is 0). $\endgroup$ – hejseb Nov 28 '13 at 20:15
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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 $(5, \infty)$.

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    $\begingroup$ +1 for explaining how to deal with the problems presented by the effects of multiplying by x(x+2). $\endgroup$ – Kevin Nov 28 '13 at 19:46
  • $\begingroup$ Thanks! I thought the quadratic formula remark needed addressing. $\endgroup$ – Dustan Levenstein Nov 28 '13 at 21:04
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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.

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    $\begingroup$ You need to be sure that x≠0 and (x+2)≠0 I'm not sure if that's much of an issue here. If x is 0 or x+2 is 0, then this question is already botched. It's not a matter of having to be careful at this point. If you got to this expression without considering the x=0 or x=-2 case, then the mistake is already made $\endgroup$ – Cruncher Nov 28 '13 at 20:03
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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 $$B>0\to A\le BC,~~~B<0\to A\ge BC$$

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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 $$\frac{f(x)}{g(x)}\geq 0 \iff f(x)g(x)\geq 0 \text{ and } g(x) \neq 0$$

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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.

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