<s>[INST] Sum of absolute values of coefficients of an equation? If $x =√2+√5$ is a root of $kx^4+mx^3+nx^2+l = 0$ where $k, m, n, l$ are integers
find the value of $|k| + |m| + |n| + |l|$.
Now as all coefficients are integers, so conjugate of x is also a root of the above.
So putting $x=√2+√5$ and $x=√2-√5$ in the above eq one by one and then subtracting the two equations we got.
We get $k.52.√10 + m.22.√5+ n.4.√10+l=0$
Now $m=0$ and $l=0$ for it to become integer.
and $k.52.√10+ n.4.√10=0$
$\implies n=-13k$
Now putting all values in the original equation
we get $kx^4-13kx^2=0 \implies x^2(kx^2-13k)=0 $ and for x not equal to 0
we get $kx^2-13k=0\implies k=0$ is the only integer solution
Hence $k=l=m=n=0$
hence final answer becomes $0$
Is this the correct way to deduce the problem?
One other way is below
$x =√2+√5 $
$⇒ x−√2 = √5 $
$⇒ (x−√2)^2 = (√5)^2 $
$⇒ x^2+2−2√2x =5 $
$⇒ (x^2 − 3) = 2√2x $
$⇒ (x^2 − 3)^2 = (2√2x)^2 $
$⇒ x^4 − 6x^2 + 9 = 8x^2 $
$⇒x^4 − 14x^2 + 9 = 0 $
$⇒ k = 1, m = 0, n = −14, l = 9 $
$⇒ |k| + |m| + |n| + |l| =1 + 0 + 14 + 9 = 24
$
 [/INST] $x=\sqrt{2}+\sqrt{5}$ is a root of the equation
$$Nx^4-14Nx^2+9N=0$$
for any integer $N\neq 0$. Because, $(x^2-7)^2-40= x^4-14x+9=0.$
So, the sum asked can be $24|N|=24M$ for some positive integer $M$.
It can not be zero.
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