Ariana Welch

2022-11-26

An equation which one's root is $x=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and coefficients are integer.
How many equation we can write for it?

Damion Ray

Expert

One way is :
$x=\sqrt{2}+\sqrt{3}+\sqrt{5}\phantom{\rule{0ex}{0ex}}x-\sqrt{5}=\sqrt{2}+\sqrt{3}\phantom{\rule{0ex}{0ex}}\left(x-9\sqrt{5}=\sqrt{2}+\sqrt{3}{\right)}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}+5-2\sqrt{5}x=2+3+2\sqrt{6}\phantom{\rule{0ex}{0ex}}{x}^{2}-2\sqrt{5}x=2\sqrt{6}\phantom{\rule{0ex}{0ex}}\left({x}^{2}-2\sqrt{5}x=2\sqrt{6}{\right)}^{2}\phantom{\rule{0ex}{0ex}}{x}^{4}+20{x}^{2}-4\sqrt{5}{x}^{3}=24\phantom{\rule{0ex}{0ex}}\left({x}^{4}+20{x}^{2}-24{\right)}^{2}=\left(4\sqrt{5}{x}^{3}{\right)}^{2}$

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