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Thread: formula question

  1. #1
    Newbie yuriythebest's Avatar
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    formula question

    hi! is this correct???

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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi

    No, it's wrong, unless * is the symbol you use for the sum and + the one you use for the product...
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    Not it isn't.

    $\displaystyle a^{{\color{red}n}*k}$ (I suppose it's n, not a) is equal to $\displaystyle (a^n)^k=(a^k)^n$

    The only thing you can do is to factor by $\displaystyle a^k$, if $\displaystyle k<n$ :

    $\displaystyle a^n+a^k=a^k \left(a^{n-k}+1 \right)$
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by yuriythebest View Post
    hi! is this correct???

    no.....the only manipulation I could see would be this $\displaystyle a^n+a^k=a^n(1+a^{k-m})$

    and to test your hypothesis let a=2 and k=1 and n=2
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  5. #5
    Newbie yuriythebest's Avatar
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    wow thanks for the lighting-quick responses. But it seems I am still doing something terribly wrong:

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  6. #6
    Moo
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    I'd make a substitution :

    $\displaystyle u=(x^3+8)^{1/4}$

    Then the equation is now uČ+u=6
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  7. #7
    Newbie yuriythebest's Avatar
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    ok I tried it that way but it seems I am going around in circles...

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  8. #8
    Super Member flyingsquirrel's Avatar
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    No, you're not "going around in circles". Once you have found the two solutions $\displaystyle u_1,\,u_2$ of $\displaystyle u^2+u-6=0$, you can solve for $\displaystyle x$ : $\displaystyle \sqrt[4]{x^3+8}=u_1 \Rightarrow x^3+8=u_1^4$ and so on...

    Be careful about $\displaystyle \Delta=b^2-4ac=1^2-4\times 1\times (-6)\neq 40 $
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  9. #9
    Newbie yuriythebest's Avatar
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    ok I tried that and got 2 negative roots:

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  10. #10
    Super Member flyingsquirrel's Avatar
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    OK you got the idea, now, read again the post #8. (especially the last sentence "Be careful...")
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  11. #11
    Newbie yuriythebest's Avatar
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    thank you flyingsquirrel! answer iz correct!
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  12. #12
    Super Member
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    Quote Originally Posted by yuriythebest View Post
    thank you flyingsquirrel! answer iz correct!
    I'm referring to post #7: I don't understand where you've got the results from:

    $\displaystyle u^2+u-6=0~\implies~u = -3~\vee~u=2$

    And therefore

    $\displaystyle \sqrt[4]{x^3+8} = -3$ that means there doesn't exist a real x to satisfy this equation

    or

    $\displaystyle \sqrt[4]{x^3+8} = 2~\implies~x^3+8=16~\implies~x^3=8~\implies~\boxed {x=2}$
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  13. #13
    Newbie yuriythebest's Avatar
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    yeah I understood that. cause of my spelling I mistook one of my u's for a 4 and therefore the incorrect answer. With your help I redid it and everything is correct.
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