How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$
I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing.
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3I think the most simple way is to notice that $x = -\frac{1}{\sqrt{2}}$ is root. After that you need to divide polynomial to $x + \frac{1}{\sqrt{2}}$ and get the polynomial of 4-th degree. Than you can apply https://en.wikipedia.org/wiki/Quartic_function#General_formula_for_roots – Eugene Sirkiza Jul 05 '19 at 12:12
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This is so embarrassing :( – nonuser Jul 05 '19 at 12:15
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1https://www.wolframalpha.com/input/?i=solve(2x%5E5%2B5%5Csqrt%7B2%7Dx%5E4%2B20x%5E3%2B20%5Csqrt%7B2%7Dx%5E2%2B20x%2B4%5Csqrt%7B2%7D) – Martin R Jul 05 '19 at 12:21
3 Answers
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Substitute $x = \sqrt{2}y$ and divide by $\sqrt{2}^5$. The resulting equation is
$$y^5 + (y+1)^5 = 0.$$
Now substitute $y = -1/(z+1)$ and multiply by $(z+1)^5$ and you get $$z^5 - 1 = 0.$$
Magma
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The only real root is $$ \alpha=-\frac{1}{\sqrt{2}}=- 0.707106781187\cdots $$
The polynomial divided by $x-\alpha$ then has degree $4$ and has roots $\alpha\pm i\beta$, $\alpha\pm i \gamma$.
Dietrich Burde
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@Zacky. With all these powers of $2$ and two occurences of $\sqrt 2$, it looks "natural" (at least to me). – Claude Leibovici Jul 05 '19 at 12:53
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We had one root $x=-\frac{1}{\sqrt{2}}$ as already said in answers and comments. After long division remains $$2 x^4+4 \sqrt{2} x^3+16 x^2+12 \sqrt{2} x+8=0$$ Make $x=y-\frac{1}{\sqrt{2}}$ to get $$2 y^4+10 y^2+\frac{5}{2}=0$$ which is a quadratic in $y^2$.
Claude Leibovici
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@Aqua. Where is the problem ? But thnks for your comment. Cheers :-) – Claude Leibovici Jul 05 '19 at 12:51