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Math Help - stuck....

  1. #1
    Newbie yuriythebest's Avatar
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    stuck....

    ok I got this far, everything seems to be nice and neat, no crazy numbers, but what do I do now??

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  2. #2
    MHF Contributor
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    You need to learn Latex my friend. I think I can follow it though.

    I see a mistake from the 3rd line to the 4th line when you're FOILing. x^{\frac{1}{3}} \times x^{\frac{1}{3}} = x^{\frac{2}{3}} since you add the exponents in this case. You wrote 1/6 and I think it should be 2/3.

    After this I'm losing what you're doing. The right hand side was 2, but now it looks like you're dividing by 4. Where'd that come from?

    Anyway, first couple of things I noticed. I'll come back later and give this a more thorough answer if no one has done so. I gotta head out for now.
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  3. #3
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    Hello
    As you know now you must to solve
    equations system:


    x^1/6-5x^1/3+6=0(1)
    5x^1/3+10<>0 (2)

    in (1) equation x^1/6=-1 no solvation
    only x^1/6=6/5

    in (2) equation x^1/3=-2 no solvation so 5x^1/3+10 are allways > 0

    and we have only one solvation:Ats.: x=46656/15625

    Good luck
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  4. #4
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    oh i`m so sorry, because i looked only in the last line
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  5. #5
    Newbie yuriythebest's Avatar
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    ok made some more progress. Now though I have what seems like a quadratic equation but it is made of fractions? what to do??

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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by yuriythebest View Post
    ok made some more progress. Now though I have what seems like a quadratic equation but it is made of fractions? what to do??
    Susbstitute u = x^{\frac13}
    Now the last equation looks like this:
    u^2 - 5u + 22 = 0
    You will not be very happy. It has complex roots I think
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  7. #7
    Super Member
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    Quote Originally Posted by yuriythebest View Post
    ok I got this far, everything seems to be nice and neat, no crazy numbers, but what do I do now??
    Let's start from the beginning:

    \frac4{\sqrt[3]{x}+2}+\frac{\sqrt[3]{x}+3}5=2

    Multiply both sides of the equation by 5(\sqrt[3]{x}+2) to get rid of all fractions:

    20+(\sqrt[3]{x}+3)(\sqrt[3]{x}+2) = 10(\sqrt[3]{x}+2)

    20+(\sqrt[3]{x})^2+5 \cdot \sqrt[3]{x} + 6=10\sqrt[3]{x} + 20

    Now substitute y = \sqrt[3]{x}:

    y^2-5y+6=0~\iff~y = 2~\vee~y=3

    And therefore
    2 = \sqrt[3]{x}~\implies~x=8
    3 = \sqrt[3]{x}~\implies~x = 27
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  8. #8
    Newbie yuriythebest's Avatar
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    ok thanks!
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