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Thread: Inequalities - due tomorrow!

  1. #1
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    Inequalities - due tomorrow!

    Hey, this assignment is due tomorrow and we never focused on inequalities before, so I need a hand with these couple problems. Also, I don't really quite know how to use most of the symbols, so please bear with me.

    These have to be solved and the answer stated in interval notation.

    a. 45x^4 - 20x^2 greater than 0

    b. -2/(x+5) greater than 0

    c. (x^2 + 8x + 15)/(x^2 - 1) greater than or equal to 0

    d. X^3 - 63 less than or equal to 1

    I know a little bit about key numbers, number lines, etc. The main purpose for this is to confirm the answers i have already, especially since she doesn't give partial credit!
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  2. #2
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    Hello, leviathanwave!

    Here are some of them . . .


    $\displaystyle a)\;\;45x^4 - 20x^2 \:> \:0$

    Factor: .$\displaystyle 5x^2(9x^2 - 4) \:> \:0$

    For $\displaystyle x \neq 0,\;5x^2$ is always positive.
    Hence: $\displaystyle 9x^2 - 4$ must be positive.

    We have: .$\displaystyle 9x^2 - 4 \:>\:0\quad\Rightarrow\quad x^2 \:> \:\frac{4}{9}\quad\Rightarrow\quad|x| > \frac{2}{3}$

    Solution: .$\displaystyle \left(\text{-}\infty,\,\text{-}\frac{2}{3}\right) \:\cup \:\left(\frac{2}{3},\,\infty\right)$



    $\displaystyle b)\;\;\frac{-2}{x+5} \:> \:0$

    The numerator is already negative.
    For the fraction to be positive, its denominator must also be negative.

    We have: .$\displaystyle x + 5 \:< \:0\quad\Rightarrow\quad x \:< \:-5$

    Solution: .$\displaystyle (\text{-}\infty,\,\text{-}5)$



    $\displaystyle c)\;\;\frac{x^2 + 8x + 15}{x^2 - 1} \:\geq \:0$

    This is a messy one . . . I'll let someone else explain it.



    $\displaystyle d)\;\;x^3 - 63\:\leq \:1$

    We have: .$\displaystyle x^3 \:\leq \:64\quad\Rightarrow\quad x \:\leq \:4$

    Solution: .$\displaystyle (-\infty,\,4] $

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  3. #3
    TD!
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    For c: determine where the numerator is positive, zero and negative and do the same for the denominator.
    The inequality then holds wherever the numerator (only) is 0 or where numerator and denominator have the same sign.
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  4. #4
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    Quote Originally Posted by leviathanwave View Post
    c. (x^2 + 8x + 15)/(x^2 - 1) greater than or equal to 0
    First we exclude $\displaystyle x=\pm 1$.

    There are 2 possibilities,
    $\displaystyle x^2-1<0$
    $\displaystyle x^2-1>0$

    Let us first assume the first case,
    $\displaystyle x^2-1<0$
    $\displaystyle x^2<1$
    $\displaystyle |x|<1$ *IMPORTANT*
    Multiplying both sides by the denominator will flip the inequality sign because it is negative,
    $\displaystyle x^2+8x+15\leq 0$
    $\displaystyle x^2+8x \leq -15$
    $\displaystyle x^2+8x+16 \leq -15+16$
    $\displaystyle (x+4)^2 \leq 1$
    $\displaystyle -1\leq x+4 \leq 1$
    $\displaystyle 3\leq x\leq 5$
    Look at the important message, we assumed that $\displaystyle |x|<1$ the only ones that work are $\displaystyle -1<x<1$. Thus, that is one solution,
    $\displaystyle x=(-1,1)$

    Let us assume the second case,
    $\displaystyle x^2-1>0$
    $\displaystyle x^2>1$
    $\displaystyle x>1 \mbox{ or }x<-1$ *IMPORTANT*
    Multiplying both sides by the denominator will not change the inequality sign for it is positive,
    $\displaystyle x^2+8x+15\geq 0$
    $\displaystyle x^2+8x+16\geq 1$
    $\displaystyle (x+4)^2\geq 1$
    $\displaystyle x+4 \geq 1 \mbox{ or }x+4\leq -1$
    $\displaystyle x \geq -3 \mbox{ or }x\leq -5$
    It needs to agree with the important interval.
    Thus, $\displaystyle -3\leq x< -1 \mbox { or } x>1 \mbox{ or } x\leq -5 $
    The solution is hence,
    $\displaystyle x=[-3,-1)\cup (1,\infty) \cup (-\infty, -5]$

    Thus, the full solution (from first problem) is,
    $\displaystyle x=[-3,-1)\cup (1,\infty) \cup (-\infty, -5] \cup (-1,1)$
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  5. #5
    TD!
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    Quote Originally Posted by ThePerfectHacker View Post
    Let us first assume the first case,
    $\displaystyle x^2-1<0$
    $\displaystyle x^2<1$
    $\displaystyle |x|<1$ *IMPORTANT*
    Multiplying both sides by the denominator will flip the inequality sign because it is negative,
    $\displaystyle x^2+8x+15\leq 0$
    $\displaystyle x^2+8x \leq -15$
    $\displaystyle x^2+8x+16 \leq -15+16$
    $\displaystyle (x+4)^2 \leq 1$
    $\displaystyle -1\leq x+4 \leq 1$
    $\displaystyle 3\leq x\leq 5$
    Look at the important message, we assumed that $\displaystyle |x|<1$ the only ones that work are $\displaystyle -1<x<1$. Thus, that is one solution,
    $\displaystyle x=(-1,1)$
    If you find that x has to be in [3,5] when you're in the case |x| < 1, then there are no solutions.
    So in your final solution, the interval (-1,1) shouldn't be there. The rest is correct.
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  6. #6
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    Hey guys, thanks so much for the help. Yeah, b and c were the ones i didnt actually have the right answers for. Thanks again for the help, you saved my life!
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