Symmetry!

$x^4+x^3+x^2+x+1=0$ The coefficients of this quartic equation are symmetrical, so we can do a neat trick by factoring out an $x^2$ and getting everything in terms of $\left(x+x^{-1}\right)$ : $x^2\left[x^2+x+1+x^{-1}+x^{-2}\right]=0\\ x^2\left[\left(x^2+2+x^{-2}\right)+\left(x+x^{-1}\right)-1\right]=0\\ x^2\left[\left(x+x^{-1}\right)^2+\left(x+x^{-1}\right)-1\right]=0$ Here, we can use the quadratic formula to get the roots and factorise. Most textbook questions would give tidy results, but unfortunately this ends up fairly messy: $\begin{aligned}\left(x+x^{-1}\right)&=\frac{-(1)\pm\sqrt{1^2+4(1)(1)}}{2(1)}\\ &=\frac{-1\pm\sqrt5}{2}\end{aligned}$ $x^2\left(x+x^{-1}-\frac{-1-\sqrt5}{2}\right)\left(x+x^{-1}-\frac{-1+\sqrt5}{2}\right)=0\\ x^2\left(x+\frac{1+\sqrt5}{2}+x^{-1}\right)\left(x+\frac{1-\sqrt5}{2}+x^{-1}\right)=0$ Then we can redistribute one half of the $x^2$ into the first bracket and the other into the second, resulting in two quadratic equations: $\left(x^2+\frac{1+\sqrt5}{2}x+1\right)\left(x^2+\frac{1-\sqrt5}{2}x+1\right)=0$ We could have substituted $x$ for $\mathrm{cis}\theta$ and simplified this with trigonometry, but looking at the answer required, quadratic formula is probably more direct.

I will multiply everything by 2 to remove annoying fractions, then use quadratic formula one more time: $x^2+\frac{1\pm\sqrt5}{2}x+1=0\\ 2x^2+\left(1\pm\sqrt5\right)x+2=0\\ \begin{aligned}x&=\frac{-\left(1\pm\sqrt5\right)\pm\sqrt{\left(1\pm\sqrt5\right)^2-4(2)(2)}}{2(2)}\\ &=\frac{-1\pm\sqrt5\pm\sqrt{1\pm2\sqrt5+5-16}}{4}\\ &=\frac{-1\pm\sqrt5\pm\sqrt{-10\pm2\sqrt5}}{4}\\ &=\frac{-1\pm\sqrt5\pm i\sqrt{10\pm2\sqrt5}}{4}\end{aligned}$ So we get $a=-1,\ b=5,\ c=10,\ d=2,\ e=5,\ f=4$ . This means our final answer is: $\sqrt{-1+5+10+2+5+4}\\ \begin{aligned}&=\sqrt{25}\\ &=\boxed5\end{aligned}$

I used Descartes's method, which can be used to solve any general quartic equation $x^4+ax^3+bx^2+cx+d=0$ . First obtain the depressed quartic by making the substitution $x=y-\frac{a}{4}$ . (A depressed quartic is a quartic with no cube term.) For this problem, this results in $y^4+\frac{5}{8}y^2+\frac{5}{8}y+\frac{205}{256}=0$ . With no cube term, the quartic can be factorized into the form $(y^2+ky+m)(y^2-ky+n)$ . Equating the coefficients leads to the system of equations: $m+n-k^2=\frac{5}{8}$ $k(n-m)=\frac{5}{8}$ $mn=\frac{205}{256}$ I solved the system in the following way: Divide by $k$ in the second equation, then add it to the first. This results in: $2n=k^2+\frac{5}{8}\Big(1+\frac{1}{k}\Big)$ Then divide by $k$ again in the second equation and subtract the first and second to obtain: $2m=k^2+\frac{5}{8}\Big(1-\frac{1}{k}\Big)$ Then multiply the two resulting equations: $4mn=\frac{205}{64}=\Big((k^2+\frac{5}{8}\Big(1-\frac{1}{k}\Big)\Big)\Big(k^2+\frac{5}{8}\Big(1+\frac{1}{k}\Big)\Big)=k^4+\frac{5}{4}k^2+\frac{25}{64}\Big(1-\frac{1}{k^2}\Big)$ $\Rightarrow 64k^6+80k^4-180k^2-25=0$ Using the rational zeros theorem, one can find $k^2=\frac{5}{4}$ as a possible solution. I will use $k=\frac{\sqrt{5}}{2}$ . This leads to $n=\frac{15}{16}+\frac{\sqrt{5}}{8}$ and $m=\frac{15}{16}-\frac{\sqrt{5}}{8}$ . Applying the quadratic formula on the two equations: $y^2+\frac{\sqrt{5}}{2}y+\frac{15}{16}-\frac{\sqrt{5}}{8}=0$ $y^2-\frac{\sqrt{5}}{2}y+\frac{15}{16}+\frac{\sqrt{5}}{8}=0$ one gets $y=\frac{-\sqrt{5}\pm i\sqrt{10+2\sqrt{5}}}{4}$ and $y=\frac{\sqrt{5}\pm i\sqrt{10-2\sqrt{5}}}{4}$ . Then remember to subtract $\frac{1}{4}$ (since $x=y-\frac{1}{4}$ ) to get the final answer: $x=\frac{-1-\sqrt{5}\pm i\sqrt{10+2\sqrt{5}}}{4}, \frac{-1+\sqrt{5}\pm i\sqrt{10-2\sqrt{5}}}{4}$ And, therefore, $\sqrt{a+b+c+d+e+f}=\sqrt{-1+5+10+2+5+4}=\sqrt{25}=5$ . One might also want to note that these are the fifth roots of unity (other than 1).