another roots

$\begin{aligned} (x^2+x+4)^2 + 3x (x^2+x+4) + 2x^2 & = 0 \\ \left((x^2+x+4) + 2x\right) \left((x^2+x+4) + x\right) & = 0 \\ \left(x^2+3x+4 \right) \left(x^2+2x+4 \right) & = 0 \\ \end{aligned}$

$\implies \begin{cases} x^2+3x+4 = 0 & \small \color{#3D99F6} \text{Let the roots be }\alpha, \beta \\ x^2+2x+4 = 0 & \small \color{#3D99F6} \text{Let the roots be }\gamma, \delta \end{cases}$

Since the coefficients of the two equations are real and their roots are complex, the roots are conjugates. And we have:

$\implies \begin{cases} |\alpha| = |\beta| = \sqrt{|\alpha \beta|} = \sqrt 4 = 2 \\ |\gamma| = |\delta| = \sqrt{|\gamma \delta|} = \sqrt 4 = 2 \end{cases}$

$\implies |\alpha| + |\beta| + |\gamma| + |\delta| = 2+2+2+2 = \boxed{8}$

First let us find a formula for $|\alpha|$ if it is a complex root of a quadratic equation:-

For a quadratic equation $ax^2+bx+c=0$ with $b^2<4ac$ , It will have complex roots given by $x=\frac{-b\pm i\sqrt{4ac-b^2}}{2a}$

Taking $|x|$ we have $|x|=\frac{\sqrt{b^2+4ac-b^2}}{2a}=\sqrt{\frac{c}{a}}$

Now , We see that if $x^2+x+4$ is replaced by $y$ then the equation becomes $y^2+3xy+2x^2=0$ Factorizing this equation, we get $y^2+2xy+xy+2x^2=0$ $y(y+2x)+x(y+2x)=0$ $(y+x)(y+2x)=0$ Resubstituting y as $x^2+x+4$ $(x^2+2x+4)(x^2+3x+4)$ For both quadratic equations, $b^2<4ac$ so we have imaginary roots. We also see that for both equations, $a=1$ and $c=4$ .

Thus $|\alpha|=|\beta|=|\gamma|=|\delta|=\sqrt{\frac{4}{1}}=2$

And answer is $2+2+2+2=8$