Sum of roots?

Let $a=\sqrt[4]{57-x}$ and $b=\sqrt[4]{x+40}$ . Then, we have:

$\begin{aligned} a + b & = 5 \\ (a+b)^4 & = 5^4 \\ a^4+4a^3b+6a^2b^2+4ab^3+b^4 & = 625 \\ {\color{#3D99F6}a^4+b^4}+4ab(a^2 +b^2)+6a^2b^2 & = 625 & \small \color{#3D99F6} \text{Note that } a^4+b^4 = 57-x+x+40 = 97 \\ {\color{#3D99F6}97}+4ab(({\color{#D61F06}a+b})^2 -2ab)+6a^2b^2 & = 625 & \small \color{#D61F06} \text{Note that } a+b = 5 \\ 4ab({\color{#D61F06}25} -2ab)+6a^2b^2 & = 528 \\ 100ab - 8a^2b^2 + 6a^2b^2 & = 528 \\ (ab)^2 - 50ab + 254 & = 0 \\ (ab-6)(ab-44) & = 0 \ \\ \implies ab & = k = \begin{cases} 6 \\ 44 \end{cases} \\ (ab)^4 & = k^4 \\ (57-x)(x+40) & = k^4 \\ 2280 + 17x - x^2 & = k^4 \\ x^2 -{\color{#3D99F6}17}x - 2280+k^4 & = 0 \end{aligned}$

By Vieta's formula , the sum of real solutions is $\color{#3D99F6}\boxed{17}$ . The solutions are 41 and -24.

Let $u = \sqrt[4]{x+40} \implies \sqrt[4]{57-x} = \sqrt[4]{97-u^4}$ , and so we have

$\begin{aligned} & \sqrt[4]{57-x} + \sqrt[4]{x+40} = 5 \\ \implies & \sqrt[4]{97-u^4} + u = 5 \\ \implies & \sqrt[4]{97-u^4} = 5 - u & \small {\color{#3D99F6} \text{Raising both sides to the power of } 4} \\ \implies & 97-u^4 = {\left(5-u\right)}^4 \\ \implies & 97-u^4 = 625 - 500u + 150u^2 - 20u^3 + u^4 \\ \implies & 2u^4 - 20u^3 + 150u^2 -500u+528 =0 \\ \implies & u^4 - 10u^3 + 75u^2 -250u + 264 = 0 \\ \implies & (u-2)(u-3)(u^2 - 5u + 44) = 0 \end{aligned}$

Now $u^2 -5u + 44 =0$ gives no real roots for $u$ , thus we have

$\begin{cases} u = 2 \implies \sqrt[4]{x+40} = 2 \implies x= -24 \\ u = 3 \implies \sqrt[4]{x+40} = 3 \implies x= 41 \end{cases}$

Thus sum of real values of $x$ is $(-24)+41 = \boxed{17}$ .

Since $\sqrt [ 4 ]{ 57-x }+\sqrt [ 4 ]{ x+40 }=\sqrt [ 4 ]{ 48.5-(x-8.5) }+\sqrt [ 4 ]{ 48.5+(1-8.5) }$ . This means that the graph of $f(x)=\sqrt [ 4 ]{ 57-x }+\sqrt [ 4 ]{ x+40 }$ is symmetry about $x=8.5$ . By sketching the graph of $y=f(x)$ , we know that there are two solutions to $f(x)=5$ . Then the solution is of $x_1=8.5+\alpha$ and $x_2=8.5-\alpha$ for some $\alpha>0$ . So the sum of roots is $\boxed{17}$ .

We know that neither fourth root can have a negative number inside it, so the domain of x is [-40,57]. We also know that both fourth roots have to work out to be integers, since the sum of two irrational numbers won't be rational in this case (since neither is less than zero). So, 5 will be expressed as either 5+0, 4+1, 3+2, 2+3, 1+4, or 0+5. Due to our restricted domain, we see that we will be unable to get one of the fourth roots to reduce to 5 or 4, since 625 and 256 are larger than the max value of what is inside the fourth roots (97). So, we know that the solutions will work out to 3+2 and 2+3. This means that the solutions will yield 57-x=81 and x+40=16, and 57-x=16 and x+40=81. For the first one, x=-24; for the second one, x=41. Lastly, 41-24=17.