If $ x=\sqrt{2}+\sqrt{3}+\sqrt{6} $ is root of $ x^{4}+a x^{3}+b x^{2}+c x+d=0 $ where $a, b, c, d$ are integers, what is the value of |$a+b+c+d$|?A. 52
B. 90
C. 21
D.93

Given:

Given equation is \( x^{4}+a x^{3}+b x^{2}+c x+d=0 \).

\( x=\sqrt{2}+\sqrt{3}+\sqrt{6} \) is root of \( x^{4}+a x^{3}+b x^{2}+c x+d=0 \) where $a, b, c, d$ are integers.

To do:

We have to find the value of |$a+b+c+d$|.

Solution:

$x^{4}+a x^{3}+b x^{2}+c x+d=0$

$x=\sqrt{2}+\sqrt{3}+\sqrt{6}$

$\Rightarrow x-\sqrt6=\sqrt2+\sqrt3$

Squaring on both sides, we get,

$(x-\sqrt6)^2=(\sqrt2+\sqrt3)^2$

$x^2-2(x)(\sqrt6)+(\sqrt6)^2=(\sqrt2)^2+2(\sqrt2)(\sqrt3)+(\sqrt3)^2$

$x^2-2\sqrt6 x+6=2+2\sqrt6+3$

$x^2+6-5=2\sqrt6+2\sqrt6 x$

$x^2+1=2\sqrt6(1+x)$

Squaring on both sides,

$(x^2+1)^2=(2\sqrt6(1+x))^2$

$x^4+2(1)(x^2)+1=4(6)[1+2(1)(x)+x^2]$

$x^4+2x^2+1=24(x^2+2x+1)$

$x^4+2x^2+1=24x^2+48x+24$

$x^4+x^2(2-24)-48x+1-24=0$

$x^4-22x^2-48x-23=0$

Comparing it with the given equation, we get,

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

The value of |$a+b+c+d$|$=0+(-22)+(-48)+(-23)$

$=$|$-93$|

$=93$.

The value of |$a+b+c+d$| is 93.

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Updated on: 10-Oct-2022

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