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Thread: How to solve this inequality?

  1. #1
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    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.
    Attached Thumbnails Attached Thumbnails How to solve this inequality?-math.png  
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  2. #2
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    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:

    $\displaystyle \mathrm{C.S.=<5,\;3>\cup<5,\;7>\cup<7; 8>}$
    Last edited by darthjavier; Nov 13th 2012 at 03:36 AM. Reason: misread problem xD
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  3. #3
    Senior Member jakncoke's Avatar
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    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) $
    Last edited by jakncoke; Nov 12th 2012 at 07:44 PM.
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