Reciprocal Sum Equations!

Wow! Loved solving it!

$\frac { 1 }{ x } +\frac { 1 }{ x+2 } -\frac { 1 }{ x+4 } -\frac { 1 }{ x+6 } -\frac { 1 }{ x+8 } -\frac { 1 }{ x+10 } +\frac { 1 }{ x+12 } +\frac { 1 }{ x+14 } =0\\ \\ (\frac { 1 }{ x } +\frac { 1 }{ x+14 } )\quad +\quad (\frac { 1 }{ x+12 } +\frac { 1 }{ x+2 } )\quad =\quad (\frac { 1 }{ x+4 } +\frac { 1 }{ x+10 } )\quad +\quad (\frac { 1 }{ x+6 } +\frac { 1 }{ x+8 } )\\ \\ \frac { 2x+14 }{ x(x+14) } +\frac { 2x+14 }{ (x+12)(x+2) } \quad =\quad \frac { 2x+14 }{ (x+4)(x+10) } +\frac { 2x+14 }{ (x+6)(x+8) }$

Clearly, we can see that $x=-7$ is a root of the equation...So, there's our $A = -7$
Now, the next part is tricky...By looking at the sense of the next 4 roots, we can see that we surely need to form a quadratic equation somewhere so as to use the quadratic formula...Ok!

$\frac { 1 }{ x(x+14) } -\frac { 1 }{ (x+6)(x+8) } \quad =\quad \frac { 1 }{ (x+4)(x+10) } -\frac { 1 }{ (x+12)(x+2) } \\ \\ \frac { 1 }{ { x }^{ 2 }+14x } -\frac { 1 }{ { x }^{ 2 }+14x+48 } \quad =\quad \frac { 1 }{ { x }^{ 2 }+14x+40 } -\frac { 1 }{ { x }^{ 2 }+14x+24 } \\ \\ \frac { 48 }{ x(x+14)(x+6)(x+8) } \quad =\quad \frac { -16 }{ (x+4)(x+10)(x+2)(x+12) } \\$
Phew!

$-3(x+4)(x+10)(x+2)(x+12)\quad =\quad x(x+14)(x+6)(x+8)\\ \\ x(x+14)(x+6)(x+8)\quad +\quad 3(x+4)(x+10)(x+2)(x+12)\quad =\quad 0\\ \\ ({ x }^{ 2 }+14x)\quad ({ x }^{ 2 }+14x+48)\quad +\quad 3({ x }^{ 2 }+14x+40)({ x }^{ 2 }+14x+24)\quad =\quad 0\\ \\ { ((x+7) }^{ 2 }-49)({ (x+7) }^{ 2 }-1)\quad +\quad 3({ (x+7) }^{ 2 }-9)({ (x+7) }^{ 2 }-25)\quad =\quad 0$

Now, there we have our quadratic... Put ${ (x+7) }^{ 2 }=y$ and we get

${ (y }-49)(y-1)\quad +\quad 3(y-25)(y-9)\quad =\quad 0\\ \\ { y }^{ 2 }-50y+49+3{ y }^{ 2 }-102y+675\quad =\quad 0\\ \\ { 4y }^{ 2 }-152y+724\quad =\quad 0\\ \\ { y }^{ 2 }-38y+181\quad =\quad 0$

Now, using the quadratic formula:

$y\quad =\quad \frac { 38\pm \sqrt { 720 } }{ 2 } \\ { (x+7) }^{ 2 }\quad =\quad 19\pm 6\sqrt { 5 } \\ \\ x+7\quad =\quad \pm \sqrt { 19\pm 6\sqrt { 5 } } \\ \\ x\quad =\quad -7\pm \sqrt { 19\pm 6\sqrt { 5 } }$
Comparing.... $A=-7,\quad \alpha =7,\quad \beta =19,\quad \gamma =6,\quad \delta =5\\ A+\alpha +\beta +\gamma +\delta \quad =\quad 30$

TaDa!!!! :-p

first, substitute $x=y-7$ ,then it becomes $\dfrac{1}{y-7}+\dfrac{1}{y-5}-\dfrac{1}{y-3}-\dfrac{1}{y-1}-\dfrac{1}{y+1}-\dfrac{1}{y+3}+\dfrac{1}{y-5}+\dfrac{1}{y+7}=0$ $\dfrac{2y}{y^2-7^2}+\dfrac{2y}{y^2-5^2}-\dfrac{2y}{y^2-3^2}-\dfrac{2y}{y^2-1^2}=0$ we see that one solution is $y=0\rightarrow A=x=-7$ after dividing the eq by 2y, $\dfrac{1}{y^2-49}+\dfrac{1}{y^2-25}-\dfrac{1}{y^2-9}-\dfrac{1}{y^2-1}=0$ put $z=y^2$ : $\dfrac{1}{z-49}+\dfrac{1}{z-25}-\dfrac{1}{z-9}-\dfrac{1}{z-1}=0$ $64z^2-2432z+11584=0,z≠49,25,9,1$ now, we can solve to get $\begin{aligned} z = 19\pm 6\sqrt{5}\\ y^2=19\pm 6\sqrt{5}\\ y=\pm \sqrt{19\pm 6\sqrt{5}}\\ x=-7\pm \sqrt{19\pm 6\sqrt{5}} \end{aligned}$ so $\begin{cases} \alpha =7\\ \beta = 19\\ \gamma=6\\ \delta = 5\\ A=-7 \end{cases}$ and $7+19+6+5-7=\boxed{30}$

Using Excel for Brute Force way,

$x^5 + 35 x^4 + 452 x^3 + 2632 x^2 + 6600 x + 5040 = 0$ is the equation.

Excel's figures correlated using 18 S.F. solver:

$\alpha$ = -1.306459110096890 $\rightarrow -7 + \sqrt{19 + 6 \sqrt5}$

$\beta$ = -4.637037424121732 $\rightarrow -7 + \sqrt{19 - 6 \sqrt5}$

$\gamma$ = -7.00000000000000

$\delta$ = -9.362962575878268 $\rightarrow -7 - \sqrt{19 - 6 \sqrt5}$

$\epsilon$ = -12.69354088990311 $\rightarrow -7 - \sqrt{19 + 6 \sqrt5}$

Found by matching.

-7 + 7 + 19 + 6 + 5 = 30

Answer: $\boxed{30}$