Find the real root of $$x^5+5x^3+5x-1$$
I have tried with squaring but the numbers that I got is not correct, any help ?
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Setting $x=y-\frac 1y$, equation is $y^{10}-y^5-1=0$
And so $$y=\sqrt[5]{\frac{1+\sqrt 5}2}$$ or $$y=\sqrt[5]{\frac{1-\sqrt 5}2}$$
And in both cases, we get $${x=\sqrt[5]{\frac{\sqrt 5+1}2}-\sqrt[5]{\frac{\sqrt 5-1}2}}$$
AND WE'RE DONE
Oiue
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4+1 but Is there any technique you use to recongnize wether or not to use that method? – Oussama Boussif Aug 12 '15 at 18:21
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6@OussamaBoussif: That technique can be used for polynomials that are symmetric in their coefficients: i.e. if you reverse the coefficients you get the same polynomial. The original question is equivalent to $$x^5+5x^3+5x+1=2$$ and the polynomials on both sides satisfy that requirement. – Rory Daulton Aug 12 '15 at 18:25
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1@OussamaBoussif Yes there is. – Oiue Aug 12 '15 at 18:25
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Ah thanks to both of you ^^ – Oussama Boussif Aug 12 '15 at 18:26