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Solve in R

  1. Jan 27, 2008 #1
    solve in R

    [tex](x^2+2)^{1/3}+(4x^2+3x-2)^{1/3}=(3x^2+x+5)^{1/3}+(2x^2+2x-5)^{1/3}[/tex]
     
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  3. Jan 27, 2008 #2

    EnumaElish

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    Any thoughts?
     
  4. Jan 28, 2008 #3

    VietDao29

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    You should notice that:
    (x2 + 2) + (4x2 + 3x - 2) = (3x2 + x + 5) + (2x2 + 2x - 5) = 5x2 + 3x.

    So, if you let: [tex]\alpha = x ^ 2 + 2[/tex]
    [tex]\beta = 3x ^ 2 + x + 5[/tex]
    and [tex]\gamma = 5x ^ 2 + 3x[/tex]

    Then, your original equation will becomes:

    [tex]\sqrt[3]{\alpha} + \sqrt[3]{\gamma - \alpha} = \sqrt[3]{\beta} + \sqrt[3]{\gamma - \beta}[/tex]

    When [tex]\gamma = 0[/tex], both sides equal 0, so this is one obvious solution. You'll get 2 (of a toal 4) solutions when setting [tex]\gamma = 0[/tex], and solve for x.

    On graphing it, I can see there are actually 4 solutions.. Not sure how you can obtain another 2 solutions. :frown:
     
  5. Jan 28, 2008 #4
    it has got only one solution i.e 0
     
  6. Jan 28, 2008 #5

    CompuChip

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    Are you sure that
    [tex]\sqrt[3]{\alpha} + \sqrt[3]{-\alpha} = \sqrt[3]{\alpha - \alpha} = 0[/tex]
     
  7. Jan 28, 2008 #6

    VietDao29

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    We are working in the reals, aren't we?

    Btw, it's not like that, since: [tex]\sqrt[3]{a} + \sqrt[3]{b} \neq \sqrt[3]{a + b}[/tex].

    It should've read:

    [tex]\sqrt[3]{\alpha} + \sqrt[3]{-\alpha} = \sqrt[3]{\alpha} - \sqrt[3]{\alpha} = 0[/tex][/QUOTE]

    :)
     
  8. Jan 28, 2008 #7

    CompuChip

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    Yes, we are working in the reals, and the identity I gave obviously doesn't hold.. that's what confused me.
    I still don't really get what you try to do. Setting [itex]\gamma = 0[/itex] means that
    [tex]\gamma = 5 x^2 + 3 x = x(5 x + 3) = 0[/tex]
    so x = 0 or x = -3/5.
    Taking x = 0 the formula becomes
    [tex]\sqrt[3]{2} + \sqrt[3]{-2} = \sqrt[3]{5} + \sqrt[3]{-5}[/tex]
    which, numerically, is something like
    [tex]1.89 + 1.09 i = 2.56 + 1.48 i[/tex]

    So it appears I didn't get what you were trying to say...

    Instead, there appear to be two real (non-zero) solutions and two complex solutions (which we are not interested in). I have no idea how to find them though.
     
    Last edited: Jan 28, 2008
  9. Jan 28, 2008 #8

    VietDao29

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    Well, no. We are not working in the complex. "i" belongs to the complex, not the reals. In the complex, [tex]\sqrt[3]{a}[/tex] has 3 different values (for a <> 0), so actually [tex]\sqrt[3]{a} + \sqrt[3]{b}[/tex] has at most 6 different values.

    Whereas, in the reals, [tex]\sqrt[3]{a}[/tex] only has one value, i.e, the real number whose cubed is a.

    E.g, we have: [tex]\sqrt[3]{8} + \sqrt[3]{-8} = 2 + (-2) = 0[/tex] (since (-2)3 = -8)

    In general, we have: [tex]\sqrt[3]{-a} = -\sqrt[3]{a}[/tex].

    Proof:

    Let [tex]b = \sqrt[3]{a}[/tex], b is unique, since we are in the reals.
    [tex]\Rightarrow b ^ 3 = a[/tex] (the definition of cube root)
    [tex]\Rightarrow - b ^ 3 = -a[/tex] (take the additive inverse of both sides)
    [tex]\Rightarrow (- b) ^ 3 = -a[/tex]
    [tex]\Rightarrow - b = \sqrt[3]{-a}[/tex] (again, the definition of cube root).
    So we have [tex]\sqrt[3]{-a} = -\sqrt[3]{a}[/tex] (Q.E.D)

    And so, that means:

    [tex]\sqrt[3]{2} + \sqrt[3]{-2} = \sqrt[3]{2} - \sqrt[3]{2} = 0 = \sqrt[3]{5} + \sqrt[3]{-5}[/tex]

    Is everything clear now? :)
     
    Last edited: Jan 28, 2008
  10. Jan 29, 2008 #9
    good work but

    let [tex]$ \sqrt[3]{x^2+2}=a, \ \sqrt[3]{4x^2+3x-2}=b, \ \sqrt[3]{3x^2+x+5}=c, \ \sqrt[3]{2x^2+2x-5}=d.[/tex]

    these may be help
     
  11. Jan 29, 2008 #10

    VietDao29

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    Then, how will you go from there?? :confused: :confused:
     
  12. Jan 29, 2008 #11

    NateTG

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    [tex]\alpha=\beta[/tex]
    and
    [tex]\gamma=\alpha + \beta[/tex]

    Are obviously solutions - no idea if that gets you anything.

    Edit: Looks like those are all imaginary solutions.
     
    Last edited: Jan 29, 2008
  13. Jan 29, 2008 #12

    VietDao29

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    Well, ya, why didn't I think of it before. Damn it. >.<

    [tex]\alpha = \beta[/tex] will gives you 2 solutions in the complex, whereas [tex]\gamma = \alpha + \beta[/tex] will give you another 2 real solutions as wishes.

    If you are allowed to graph, and then solve for x, then it's done. Woohoo. But if you are not allowed to graph :(, then you must first prove that it has at most 4 solutions, then point out all the 4 solutions.

    But well, I think, in this problem, you are allowed to use graph. :biggrin:
     
  14. Feb 5, 2008 #13

    Let

    [tex]\alpha=x^2+2,\,\beta=4x^2+3x-2\,\gamma=3x^2+x+5,\,\delta=2x^2+2x-5[/tex]

    with [itex]\alpha\neq 0,\,\beta\neq 0,\,\gamma \neq 0,\delta\neq 0[/itex] since the roots of these equations does not satisfy the original equation. Then

    [tex]\alpha^{1/3}+\beta^{1/3}=\gamma^{1/3}+\delta^{1/3}\neq 0 \quad (1)[/tex] and
    [tex]\alpha+\beta=\gamma+\delta \quad (2)[/tex]

    Raising (1) to the 3rd power

    [tex]\alpha+3\,\alpha^{2/3}\,\beta^{1/3}+3\,\alpha^{1/3}\,\beta^{2/3}+\beta=\gamma+3\,\gamma^{2/3}\,\delta^{1/3}+3\,\gamma^{1/3}\,\delta^{2/3}+\delta\Rightarrow[/tex]

    [tex]3\,\alpha^{2/3}\,\beta^{1/3}+3\,\alpha^{1/3}\,\beta^{2/3}=3\,\gamma^{2/3}\,\delta^{1/3}+3\,\gamma^{1/3}\,\delta^{2/3}\Rightarrow \quad \text{from (2)} [/tex]

    [tex]\alpha^{1/3}\,\beta^{1/3}\,(\alpha^{1/3}+\beta^{1/3})=\gamma^{1/3}\,\delta^{1/3}\,(\gamma^{1/3}+\delta^{1/3})\Rightarrow[/tex]

    [tex]\alpha^{1/3}\,\beta^{1/3}=\gamma^{1/3}\,\delta^{1/3}\Rightarrow \quad \text{from (1)}[/tex]

    [tex]\alpha\,\beta=\gamma\,\delta \quad (3)[/tex]

    Thus we have to solve the system of (1), (2), (3) which is rather simple. From (3) we have

    [tex]\frac{\alpha}{\gamma}=\frac{\delta}{\beta}=k[/tex]

    which makes (2)

    [tex](k-1)\,\gamma=(k-1)\,\beta[/tex]

    with [itex]k\neq 1[/itex] because leads to inconsistency the original equation. Thus [itex]\gamma=\beta[/itex] and from (2) [itex]\alpha=\delta[/itex], with the final answer

    [tex]x=-1-2\,\sqrt{2},\, x=-1+2\,\sqrt{2}[/tex]​

    P.S. I hope this time the mentors would not delete this post, after the conversation we had about santa's threads.
     
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