Is there any simple way to find a polynomial with integer coefficients so that ($x=\sqrt{2} +\sqrt{3}$) is one of its roots? I know one way is to get rid of all the square roots in the equation to be left with rational numbers, but is there any other simple way?
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"I know one way": should we guess what your way is ? – Aug 12 '18 at 12:09
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And yet another failure of "trusted" users to search. – Jyrki Lahtonen Aug 12 '18 at 12:10
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Mind you, I didn't downvote the question. Newbies get some slack. The answerers... I don't know. My finger is twitching. OTOH this isn't calculus or such. – Jyrki Lahtonen Aug 12 '18 at 12:15
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Is this simple? $$x=\sqrt2+\sqrt3$$ $$x^2=2+3+2\sqrt2\sqrt3$$ $$x^2-5=2\sqrt6$$ $$(x^2-5)^2=24$$ $$x^4-10x^2+1=0$$
Angina Seng
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Square it $$x^2 =2+2\sqrt{6}+3$$ so $$(x^2-5)^2 = 4\cdot 6$$ so $$ p(x) = x^4 -10x^2+1$$
nonuser
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We have $x_0=\sqrt2+\sqrt3$. If we take another root to be $x_1=\sqrt2-\sqrt3$, by Vieta's formulas we get the equation
$$x^2-2\sqrt2x-1=0$$ and we can get rid of the square root by writing
$$(x^2-1)^2=8x^2.$$
Alternatively,
$$x-\sqrt2=\sqrt3\to x^2-2\sqrt2x+2=3\to(x^2-1)^2=8x^2.$$