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Find all real roots of the following fifth-degree equation

$$x^5+10x^3+20x-4=0.$$

In fact, the only one root is $\sqrt[5]8-\sqrt[5]4$. How to get that?

P..
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ziang chen
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  • What have you done so far? – AlexR Oct 30 '13 at 21:12
  • According to Matematica there is only one and it appears to be irrational. $x=0.19620865573750382...$ – Wintermute Oct 30 '13 at 21:16
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    This equation has one real root, and it's not very nice. Are you trying to find an approximation? – vadim123 Oct 30 '13 at 21:17
  • The result is $x=0.196209$ – 1190 Oct 30 '13 at 21:17
  • Maple gives $x_0 \approx 0.1962086557375038229732578300768051056478145297847229605316974461169176963085374475908223396134722920$ If you want more digits, tell me. But this looks nothing like fun to me. – AlexR Oct 30 '13 at 21:17
  • @B11b please add the ellipsis. What you wrote is just wrong. – AlexR Oct 30 '13 at 21:18
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    I think I have seen this exact equation on this site before, but I can't recall where. Putting $x=t-\frac{2}{t}$ gives a quadratic in $t^5$ if I remember correctly, which is then easily solved. – Ivan Loh Oct 30 '13 at 21:19
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    Aha I found it. http://math.stackexchange.com/questions/545651/how-solve-this-equation – Ivan Loh Oct 30 '13 at 21:20
  • Note that it is at least easy to prove that there is only one root. It's derivative is strictly positive and clearly contains no horizontal asymptotes: from this it is simple to prove there can only be one real root. After that I can only imagine doing some nasty magic. – Tyler Holden Oct 30 '13 at 21:37
  • @TylerHolden: Or you can use Descartes' rule of signs. – P.. Oct 30 '13 at 21:42

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