Question

If $x=\sqrt{2}+\sqrt{3}+\sqrt{6}$ is a root of $x^4+a{x}^3+b{x}^2+cx+d=0$ where $a,b,c,d$ are integers, what is the value of $|a+b+c+d|$ ?

Answer 1

$x=\sqrt{2}+\sqrt{3}+\sqrt{6}$ is a root of equation:
$x^4+a{x}^3+b{x}^2+cx+d=0$; where $a,b,c,d\in Z$

Now, $x-\sqrt{2}=\sqrt{3}+\sqrt{6}$
Squaring both sides, we get
$(x-\sqrt{2}{)}^2=(\sqrt{3}+\sqrt{6}{)}^2$
$\Rightarrow x^2-2\sqrt{2x}+2=9+6\sqrt{2}$
$\Rightarrow x^2-7=2\sqrt{2}(3+x)$

Squaring again both sides, we get
$\Rightarrow x^4-14x^2+49=8(x^2+6x+9)$
$\Rightarrow x^4-22x^2-48x-23=0$

$\therefore a=0,b=-22,c=-48$ and $d=-23$

$\therefore |a+b+c+d|=93$