# Thread: How to solve this inequality?

1. ## How to solve this inequality?

I'm grateful for all and any replies.

I attached the inequality and the purported solution in an image, as I determined it'd be easier than figuring out the formatting in this site.

I'm really at a loss. I've tried solving it numerous times, thought I had it a couple, and haven't actually gotten anywhere.

2. ## Re: How to solve this inequality?

Hola

Esa inecuación es equivalente a:

$\displaystyle \Leftrightarrow\dfrac{(x+2)^2(x-8)(x+3)^3(x-10)^2(x+5)}{(x-5)^3(x-7)^4}\leq 0,\qquadx\neq 0$

YO
particularmente lo resuelvo así:

$\displaystyle \mathrm{C.S.=\left[5,\;3\right]\cup<5,\;7>\cup<7; 8]}$

PS: Lo siento, resolví el problema para $\displaystyle \geq 0$, como es $\displaystyle >0$, simplemente no toma el -5, -3 y -8, así que es abierto en esos puntos:

$\displaystyle \mathrm{C.S.=<5,\;3>\cup<5,\;7>\cup<7; 8>}$

3. ## Re: How to solve this inequality?

This can be done qualitatively. First lets identify our domain. Where is the denominator equal to 0 ?, at points 0, 5, 7. So I got D = $\displaystyle (-\infty, 0) \cup (0, 5) \cup (5, 7) \cup (7, \infty$ . Now lets since the inequality > 0, the numerator cannot be zero also. So where is the numerator 0? I got -2, 8, -3, 10, -5. So our set of possible solutions consists of $\displaystyle (-\infty, -5) \cup \(-5, -3) \cup (-3, -2) \cup (-2, 0) \cup (0, 5) \cup (5, 7) \cup (7, 8) \cup (8, 10) \cup (10, \infty) = S$.

Now consider any $\displaystyle (x \in (-\infty, 0)$. We see that the denominator > 0. So, the numerator must also be > 0, So only consider odd power terms in numerator, since even power gives positive no matter what. So take $\displaystyle (x-8)(x+3)^3(x+5) > 0$, since we see $\displaystyle (x-8) < 0 for x \in (-\infty, 0)$, it must be that $\displaystyle (x+3)^3(x+5) < 0$(negative * negative = positive) . Which means either $\displaystyle x - 3 < 0 AND x + 5 > 0$ OR $\displaystyle x - 3 > 0 AND x + 5 < 0$ which means $\displaystyle x < 3 AND x > -5$ OR $\displaystyle x > 3 AND x < -5$., the second is impossible, x cannot be > 3 and < -5 at the same time.. So x \in $\displaystyle (-5, 3) \cap S$

Now we solved what happens when $\displaystyle x \in (-\infty, 0))$ we got for $\displaystyle x \in (-5, -3) \cap S = (-5, -3)$ the inequality holds. Now assume $\displaystyle x \in (0, \infty)$ We only need to consider $\displaystyle \frac{x-8}{5-x}^3$ because every other term evaluates to positive. So for $\displaystyle \frac{x-8}{5-x}^3 > 0$ One option is for both num and denom to be positive. which gives $\displaystyle x > 8 AND x < 5$ which is not possible. So it must be that both num and denom are negative. so that means $\displaystyle x - 8 < 0 AND 5 - x < 0$ which gives us $\displaystyle x < 8 AND x > 5$ which gives us the interval $\displaystyle (5, 8)$. So for $\displaystyle x \in (0, \infty)$ the inequality holds in $\displaystyle (5, 8) \cap S) = (5, 7) \cup (7,8)$. So our soln is $\displaystyle (-5, -3) \cup (5, 7) \cup (7, 8)$