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Find a polynomial with integer coefficients with sqrt(2) + sqrt(3) as a root.

 May 23, 2020
 #1
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I just muddled my way through this.  

Maybe there is a better way to do it.

 

 

\((x-(\sqrt2+\sqrt3))\quad \text{is a factor}\\consider\\~\\ (x-(\sqrt2+\sqrt3))(x+(\sqrt2+\sqrt3))\\ =x^2-(\sqrt2+\sqrt3)^2\\ =x^2-(2+3+2\sqrt2\sqrt3)\\ =x^2-(5+2\sqrt6)\\ =((x^2-5)-2\sqrt6)\\~\\ Now\\ ((x^2-5)-2\sqrt6)((x^2-5)+2\sqrt6)\\ =(x^2-5)^2-(2\sqrt6)^2\\ =x^4-10x^2+25-4*6\\ =x^4-10x^2+1\\~\\ so\\ x^4-10x^2+1 \quad has \quad \sqrt2+\sqrt3\quad \text{ as a root} \)

 

 

 

 

LaTex coding:

(x-(\sqrt2+\sqrt3))\quad \text{is a factor}\\consider\\~\\
(x-(\sqrt2+\sqrt3))(x+(\sqrt2+\sqrt3))\\
=x^2-(\sqrt2+\sqrt3)^2\\
=x^2-(2+3+2\sqrt2\sqrt3)\\
=x^2-(5+2\sqrt6)\\
=((x^2-5)-2\sqrt6)\\~\\
Now\\
((x^2-5)-2\sqrt6)((x^2-5)+2\sqrt6)\\
=(x^2-5)^2-(2\sqrt6)^2\\
=x^4-10x^2+25-4*6\\
=x^4-10x^2+1\\~\\
so\\
x^4-10x^2+1 \quad has \quad \sqrt2+\sqrt3\quad \text{ as a root}

 May 24, 2020

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