Real roots added

It would be tempting to use Vieta's here, but since we are just looking for the sum of all real roots I'll try a different approach.

Let $y = x + 4.$ Then the equation becomes

$(y - 1)^{4} + (y + 1)^{4} = 20 \Longrightarrow 2y^{4} + 12y^{2} + 2 = 20 \Longrightarrow y^{4} + 6y^{2} - 9 = 0.$

This is a quadratic in $y^{2}$ and has solutions $y^{2} = -3 \pm 3\sqrt{2}.$

Now as we are looking for just real solutions here, we take $y^{2} = 3(\sqrt{2} - 1).$

Thus $x + 4 = y = \pm \sqrt{3(\sqrt{2} - 1)} \Longrightarrow x = -4 \pm \sqrt{3(\sqrt{2} - 1)}$ ,

and so the sum of the two real roots is just $\boxed{-8}.$

Graph of LHS is symmetric about x= -4,

also it is inceasing to its right and decreasing at its left,

So it cuts y=20 only twice,

so sum of real roots is 2(-4) = -8

to understand why i did the last step, do this, if the graph was symmetric about origin, roots would have been perhaps t and -t, then upon shifting the whole graph by 4 units left, the positive root becomes t-4 and the negative ones become -t-4, adding them we get -8

Let ${y}$ = ${x}$ +4, such that the equation becomes: $(y-1)^4 + (y+1)^4 = 2y^4 +12y^2 +2 = 20 \Rightarrow\ y^4 +6y^2 -9 = 0$ Now completing the square and taking positive values only (as $y^{2}$ $\geq$ 0 ): $(y^2+3)^2 = 18 \Rightarrow\ y^2 = -3 +3\sqrt{2}$ Now from our original substitution we have: $y^2 = (x+4)^2 = -3 +3\sqrt{2} \Rightarrow\ x^2 +8x +(19 - 3\sqrt{2} ) = 0$ $Now \ b^2 - 4ac > 0 \ so \ we \ can \ use \ vieta \ sums:$ $\therefore\sum_{}^{} roots \ = -\frac{b}{a} = \boxed{-8}$

Let $n=x+4$ , for real number $x$ . The transformed equation is:

$(n-1)^{ 4 }+(n+1)^{ 4 }=20\Rightarrow { 2n }^{ 4 }+12{ n }^{ 2 }+2=20\Rightarrow { n }^{ 4 }+6{ n }^{ 2 }-9=0$ $\Rightarrow { n }^{ 2 }=-3\pm \sqrt { 18 }$

Because, n is real, so: $n=\pm \sqrt { \sqrt { 18 } -3 } \Rightarrow x=-4\pm \sqrt { \sqrt { 18 } -3 }$ , therefore the sum of 2 real roots is $-8$ .

Ok so u may take derivatives of the function given above.....The 3rd derivative is linear from there u csn extract the coefficients of x^4 (16) and x^3 (2) easily.The reqd ans. then is -8...from sum of roots theorem.

How is this a level 5 problem? It took me less than 10 seconds to solve it. Based on the diagram, all the roots are part of the roots of unity (so they are symmetric), so instead of using -b/a as per Vieta's, use -b/2a = -8.

Eliminates difficulty of solving quartic polynomial equation,

x^4 + 16 x^3 + 102 x^2 + 304 x + 343 = 0

-2.8852620545082 + -5.1147379454918 = - 8

Other two are non-real numbers.

Put $y = x + 4$

You get some value of $y$

ie $y = \pm a [\text{for some } a]$

Adding, $\begin{aligned}x &= y - 4\\&=\pm a - 4\\ \Rightarrow \sum x &= 2\times -4 \\&= -8\end{aligned}$

$\Huge \displaystyle \therefore \boxed{Sum = -8}$

put a=x+4 , rearrange the eqution
expand by binomial theorem solve quadratic to get roots