Quintic Quandary

By Vieta's formulas , the coefficient of $x^2$ the sum of all triple products of roots. Thus $\begin{aligned} {3 \choose 3}{2 \choose 0}p^3 + {3\choose 2}{2\choose 1}p^2q + {3\choose 1}{2\choose 2}pq^2 &= 0 && \color{#3D99F6}\text{All ways to choose three roots} \\ p^3 + 6p^2q+3pq^2 &= 0 \\ \left(\frac{p}{q}\right)^2 + 6\frac{p}{q} + 3 &= 0 && \color{#3D99F6}\text{Multiply by }\frac{1}{pq^2}\end{aligned}$ $\begin{aligned} \frac{p}{q} &= \frac{-6\pm\sqrt{6^2-12}}{2} && \color{#3D99F6}\text{Apply the quadratic formula} \\ \frac{p}{q} &= -3\pm\sqrt{6} \end{aligned}$ Thus $a=3$ and $b=6$ , giving us $a+b=3+6=\boxed{9}$ .

Let $f(x)$ denote the monic polynomial equation in question, then $f(x) = (x-p)^3 (x-q)^2$ , where $p \ne q$ .

Since the coefficient of (the expanded form of) $x^2$ in $f(x)$ is 0, then the coefficient of the constant in $f''(x)$ must be 0 as well. As such, $f''(x)$ has a root of 0.

By chain rule , $f'(x) = 3(x-p)^2 (x-q)^2 + 2(x-p)^3 (x-q) \quad \implies \quad f''(x) = 6(x-p)(x-q)^2 + 6(x-p)^2 (x-q) + 6(x-p)^2 (x-q) + 2(x-p)^3.$

With $f''(0) = 0$ , the equation above simplifies to $-2p(p^2 + 6pq + 3q^2) = 0$ . Since $p\ne 0$ , $p^2 + 6pq + 3q^2 = 0 \quad \implies \quad \left(\frac pq\right)^2 + 6\left( \frac pq\right) + 3 = 0.$ Using the quadratic formula , $\frac pq = -3\pm \sqrt 6$ . Our answer is $3 + 6 = \boxed9$ .

The quintic equation is therefore,

$\begin{aligned} (x-p)^3(x-q)^2 & = 0 \\ (x^3-3px^2+3p^2x - p^3)(x^2-2qx+q^2) & = 0 \end{aligned}$

Then the coefficient of $x^2$ is:

$\begin{aligned} - 3pq^2 -6p^2q - p^3 & = 0 & \small \color{#3D99F6} \text{Multiply both sides by }-\frac 1{pq^2} \\ 3 + 6\frac pq + \frac {p^2}{q^2} & = 0 & \small \color{#3D99F6} \text{Rearrange} \\ \left(\frac pq\right)^2 + 6 \frac pq + 3 & = 0 & \small \color{#3D99F6} \text{Solve the quadratic for }\frac pq \\ \implies \frac pq & = - 3 \pm \sqrt 6 \end{aligned}$

Therefore, $a+b=3+6 = \boxed 9$ .