It's Just Roots

Let the first term and the common difference of the arithmetic progression be $a_1$ and $d$ respectively. Then the four roots are $\\ a_1$ , $a_1+d$ , $a_1+2d$ , and $a_1+3d$ . By Vieta's formula we have:

$a_1 + (a_1 + d) + (a_1 + 2d) + (a_1+3d) = 4a_1+6d = 14 \implies 2a_1 + 3d = 7 \quad ...(1)$

$a_1(a_1+d) + a_1(a_1+2d) + a_1(a_1+3d) + (a_1+d)(a_1+2d) + (a_1+d)(a_1+3d) + (a_1+2d)(a_1+3d) \\ = 6a_1^2 + 18ad_1 + 11d^2 = 51 \quad ...(2)$

$\begin{aligned} (1)^2: \ \ \ 4a_1^2 + 12a_1d + 9d^2 & = 49 & ...(3) \\ (2) - (3): \ \ \ \ \ 2a_1^2 + 6a_1d + 2d^2 & = 2 & ...(4) \\ (4) \times 2: \ \ \ 4a_1^2 + 12a_1d + 4d^2 & = 4 & ...(5) \\ (3)-(5): \ \ \ \ \ \qquad \qquad \qquad 5 d^2 & = 45 \\ \implies d & = 3 \\ (1): \ \ \ \ \ \qquad \qquad \implies a_1 & = \frac {7-3(3)}2 = -1 \end{aligned}$

Then the four roots are $-1$ , $2$ , $5$ , and $8$ . And $a$ and $b$ are:

$\begin{aligned} a & = - ((-1)(2)(5)+(-1)(5)(8)+(-1)(8)(2) + (2)(5)(8)) = -14 \\ b & = (-1)(2)(5)(8) = - 80 \\ \implies a+b & = \boxed{-94} \end{aligned}$

Let the roots of this quartic equation be $A - \frac {3D}2, A - \frac D2 , A + \frac D2, A + \frac {3D}2$ , where $D> 0$ .

By Vieta's formula , $\left( A - \frac {3D}2\right) + \left( A - \frac D2 \right) + \left( A + \frac D2 \right) + \left(A+ \frac {3D}2 \right) = 14 \quad \implies \quad A = \frac 72$ And the $2^\text{nd}$ symmetric sum of these 4 numbers is equal to 51. For simplicities sake, let $A_1, A_2, A_3, A_4$ be these numbers. $A_1 A_2 + A_1 A_3 + A_1 A_4 + A_2 A_3 + A_2 A_4 + A_3 A_4 = 51$ Grouping them as such: $A_1 (A_2 + A_3) + A_4 (A_2 +A_3) + A_1 A_4 + A_2 A_3 = (A_1 + A_4)(A_2 + A_3) + A_1 A_4 + A_2 A_3$ Then substitute: $2A \cdot 2A + \left( A^2 - \frac{9D^2}{4} \right) + \left( A^2 - \frac {D^2}4 \right) = 51$ with $A = \frac72$ , we get $D = 3$ .

So the 4 roots are $-1,2,5,8$ , thus the quartic equation is $(x+1)(x-2)(x-5)(x-8) = x^4 -14x^3 + 51x^2 - 14x-80 = 0$ with $(a,b) = (-14,-80) \implies a+b= \boxed{-94} .$

Let the four roots be in arithmetic progression: $r, r+ \delta, r + 2\delta, r + 3\delta$ such that by Vieta's Formulae:

$\Sigma_{i=1}^{4} r_{i} = 4r+ 6\delta = 14;$

$\Sigma_{i=1}^{3} \Sigma_{j=i+1}^{4} r_{i}r_{j} = 6r^2 + 18\delta r +11\delta^{2} = 51$

which yields $(r,\delta) = (-1,3); (8,-3)$ , and the four roots are $-1, 2, 5, 8$ . Now we calculate:

$a = -[\Sigma_{i=1}^{2} \Sigma_{j=i+1}^{3} \Sigma_{k=j+1}^{4} r_{i}r_{j}r_{k} ]= -14$ ;

$b = \Pi_{i=1}^{4} r_{i} = -80$

Hence, $a+b = \boxed{-94}.$