USAMO is filled with Vieta

For sake of variety, here is mine. Let us factor this into two quadratics. Hence they will be $(x^2+ax-32)$ $(x^2+bx+62)$ because we are given $-32$ as the product of two roots, so the other two must have product $62$ as we know the product of all four is $-1984$ by Vieta's formulae. Now comparing coefficients after expanding, we get $-18$ , the desired $x^3$ coefficient, equal to $a+b$ . Also 200, the desired $x$ coefficient equal to $62a-32b$ . Now we have a system. Solving for a and b get get $a=-4, b=-14$ . Expand the two quadratics with variables $a$ and $b$ inserted and it can be seen $k$ is equal to $86$ .

Let $a$ , $b$ , $c$ and $d$ be the zeros of $x^4-18x^3+kx^2+200x-1984=0$ .

By Vieta's Formulas, we have:

$\begin{cases} a+b+c+d & = 18 &... (1) \\ ab+ac+ad+bc+bd+cd & = k &... (2) \\ abc+abd+acd+bcd & = -200 &... (3) \\ abcd & = -1984 &... (4) \end{cases}$

Let $cd=-32$ . From equation 4: $\quad \Rightarrow ab = \dfrac {-1984}{-32} = 62$

From equation 1: $\quad \Rightarrow c+d = 18-(a+b)\quad ...(1a)$

From equation 3:

$\begin{aligned} (ab)c+(ab)d+a(cd)+b(cd) & = -200 \\ 62(c+d)-32(a+b) & = -200 \\ 62[18-(a+b)]-32(a+b) & =-200 \\ 1116 - 94(a+b) & = -200 \\ 95(a+b) & = 1316 \\ \Rightarrow \quad a+b & = 14 \quad ... (3a) \end{aligned}$

From equation 1a: $\quad \Rightarrow c+d = 18-(a+b) = 18 - 14 = 4$

From equation 2:

$\begin{aligned} k & = ab+ac+ad+bc+bd+cd \\ & = 62 +ac+ad+bc+bd-32 \\ & = 30 + (a+b)(c+d) \\ & = 30 + (14)(4) \\ & = 30 + 56 \\ & = \boxed{86} \end{aligned}$

Let roots be a,b,c,d.let s=a+b , s'=c+d , p=ab, p'=cd . the. From data put ab=-32 . by viete relations we get s+s'=18 , ss'+p'+p=k , sp'+s'p=-200 , pp'=-1984 After sub p=-32 we cab easily solve for s and s' . then k=30+4*14=86