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Math Help - Help on this inequation

  1. #1
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    Help on this inequation

    \sqrt{-x-2}-\sqrt[3]{x+5}<3 In this inequation what i'd usually do is bring -x-2 and x+5 to the "same root" that's 6. But i don't think i'm going anywhere because of that 3 on the right. What would you suggest me to do here?

    the condition is x\in(-\infty,-2]
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    Have you tried graphing the functions y = \sqrt{-x-2} - \sqrt[3]{x + 5} and y = 3 and determining the x values for which this inequality holds true?
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    i need to prepare for an exam, i don't think doing graphs will help me if i have to solve a similar inequation
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    Solve the equation \sqrt{-2-x}+ \sqrt[3]{x+ 5}= 3. The points where those are equal or where the roots are not defined (x must be less than -2, of course) separate ">" from "<".
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    Quote Originally Posted by Utherr View Post
    i need to prepare for an exam, i don't think doing graphs will help me if i have to solve a similar inequation
    On the contrary, you should NEVER try to solve any equations or inequations without picturing what you are actually dealing with (i.e. graphing). To gain a deep understanding of function analysis, you need to understand and be able to connect the numerical, graphical and algebraic representations of your function.
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  6. #6
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    Quote Originally Posted by Utherr View Post
    \sqrt{-x-2}-\sqrt[3]{x+5}<3 In this inequation what i'd usually do is bring -x-2 and x+5 to the "same root" that's 6. But i don't think i'm going anywhere because of that 3 on the right. What would you suggest me to do here?


    The condititon is x\in(-\infty,-2]
    If you want to use the 6th roots

    \sqrt{-x-2}=(-x-2)^{\frac{3}{6}}=\left(\sqrt[6]{-x-2}\right)^3

    \sqrt[3]{x+5}=(x+5)^{\frac{2}{6}}=\left(\sqrt[6]{x+5}\right)^2

    Instead...

    x=-2\ \Rightarrow\ \sqrt{0}-\sqrt[3]{3}<3

    x=-3\ \Rightarrow\ \sqrt{1}-\sqrt[3]{2}<3

    x=-4\ \Rightarrow\ \sqrt{2}-\sqrt[3]{1}<3

    x=-5\ \Rightarrow\ \sqrt{3}-\sqrt[3]{0}<3

    x=-6\ \Rightarrow\ \sqrt{4}-\sqrt[3]{-1}=3
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