The number $a=\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying
\[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\]
is $420$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\}=x- \lfloor x \rfloor$ denotes the fractional part of $x$. What is $p+q$?
$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$
Let $w=\lfloor x \rfloor$ and $f=\{x\}$ denote the whole part and the fractional part of $x,$ respectively, for which $0\leq f0, (2a-1)w0.$ By Descartes' Rule of Signs, we deduce that $(2)$ must have two positive roots, so $f\geq0$ is always valid.
Alternatively, from $(3)$ and $(4),$ note that all values of $a$ for which $0\sqrt{1-4a}.$ We deduce that both roots in $(3)$ must be positive, so $f\geq0$ is always valid.
$f0.$ We rewrite $f$ as \[f=wk,\] in which $f0,$ we obtain $\frac{60^2}{29^2}a-\frac{4}{29}=0,$ from which \[a=\frac{4}{29}\cdot\frac{29^2}{60^2}=\frac{29}{900}.\]
The answer is $29+900=\boxed{\textbf{(C) } 929}.$
~MRENTHUSIASM (inspired by Math Jams's 2020 AMC 10/12A Discussion)