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Cardano's method: help needed

  1. Jan 15, 2007 #1
    i am trying to solve [tex]x^3 + \frac{5}{3}x - \frac{200}{27} = 0[/tex] using Cardano's method. i got one solution,

    [tex]x = \sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}[/tex]

    now, using a calculator, i can see that x = 5/3. but how can i determine that without using a calculator, i.e., how can i simplify the above expression?

    thanks in advance to anybody who can help.
     
    Last edited: Jan 15, 2007
  2. jcsd
  3. Jan 15, 2007 #2

    Gib Z

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    Well since you have one factor, divide the original expression by that to get a quadratic equation, use quadratic forumula to solve for remaining roots.
     
  4. Jan 15, 2007 #3
    that's not what i asked. i know how i can solve for the remaining roots. i wanted to know how i can simplify the following expression:
    [tex]\sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}[/tex]

    using a calculator i found that it is equal to 5/3. but how can i find that without using a calculator?
     
    Last edited: Jan 15, 2007
  5. Jan 15, 2007 #4

    HallsofIvy

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    Your equation is the same as 27x3+ 45x- 200= 0. Any rational root must have numerator a factor of 200= 2852 and denominator a factor of 27= 33. Try combinations of those until one works. Other than that, there is no "elementary" method to reduce [tex]x = \sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}[/tex] to 5/3.
     
  6. Jan 15, 2007 #5

    dextercioby

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    Sure it is

    [tex]x = \sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}=\frac{1}{3}\left(\sqrt[3]{100+45\sqrt{5}} +\sqrt[3]{100-45\sqrt{5}}\right) [/tex]

    Denote by "a" the number inside the round bracket. You wish to show that "a=5".

    Consider [itex] a^{3} [/itex]. By an elementary application of the formula

    [itex] (x+y)^{3} \equiv x^{3} +y^{3} +3x^{2}y+3xy^{2} [/itex]

    one finds out that

    [tex] a^{3}=200-15 a [/tex]

    The only viable solution to our initial purpose is a=5.

    QED.

    Daniel.
     
  7. Jan 15, 2007 #6
    how does one find that out? it involves solving another cubic equation using cardano's method.
     
  8. Jan 15, 2007 #7

    dextercioby

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    No need for Cardano method. You can see very clearly that a=5 is the viable solution by looking at the graph intersection for the curve y=x^3 and y=200-15 x.

    They intersect in one point only. x=5 and y=125.

    Daniel.
     
  9. Jan 15, 2007 #8

    Hurkyl

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    Apply the rational root theorem: it always gives you all possible rational roots to an integer polynomial. (HoI applied it to your problem)
     
  10. Jan 15, 2007 #9

    uart

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    Well it's easy enough to verify that 5 is the only real solution to that equation which is in effect the same thing as verifying that the original solution is equal to 5/3.
     
  11. Jan 15, 2007 #10
    thanks a lot. i can of course use the rational root theorem. i just wanted to know if i used Cardano's method, whether there is a way to simplify
    [tex]\sqrt[3]{\frac{100}{27} + \frac{5\sqrt{5}}{3}} + \sqrt[3]{\frac{100}{27} - \frac{5\sqrt{5}}{3}}[/tex] to [tex]\frac{5}{3}[/tex] directly.

    anyway, thanks a lot guys.
     
    Last edited: Jan 15, 2007
  12. Jan 15, 2007 #11

    HallsofIvy

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    Very nice. Thank you
     
  13. Jan 16, 2007 #12
    Y=[tex]\sqrt[3]{2+\sqrt5} +\sqrt[3]{2-\sqrt5 }[/tex] Niederhoffer, writing in "The Education of a Speculator," got this problem on a high school math honor course. He was to chose a simple answer for this. Among the choices, he just guessed 1, which was correct.

    The dextercioby method works well on that. Just set it equal to Y and cube. Then replace the middle part by -Y. The result is Y^3 = 4-3Y, making 1 an easy choice.
     
    Last edited: Jan 16, 2007
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