Let $P(x) = x^2 - 3x - 7$, and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$. David computes each of the three sums $P + Q$, $P + R$, and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$, then $R(0) = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$.  We can write the following:
\[P + Q = 2x^2 + (a - 3)x - 5\]
\[P + R = 2x^2 + (b - 3)x + (c - 7)\]
\[Q + R = 2x^2 + (a + b)x + (c + 2)\]
Let the common root of $P+Q,P+R$ be $r$; $P+R,Q+R$ be $s$; and $P+Q,Q+R$ be $t$.  We then have that the roots of $P+Q$ are $r,t$, the roots of $P + R$ are $r, s$, and the roots of $Q + R$ are $s,t$.
By Vieta's, we have:
\[r + t = \dfrac{3 - a}{2}\tag{1}\]
\[r + s = \dfrac{3 - b}{2}\tag{2}\]
\[s + t = \dfrac{-a - b}{2}\tag{3}\]
\[rt = \dfrac{-5}{2}\tag{4}\]
\[rs = \dfrac{c - 7}{2}\tag{5}\]
\[st = \dfrac{c + 2}{2}\tag{6}\]
Subtracting $(3)$ from $(1)$, we get $r - s = \dfrac{3 + b}{2}$.  Adding this to $(2)$, we get $2r = 3 \implies r = \dfrac{3}{2}$.  This gives us that $t = \dfrac{-5}{3}$ from $(4)$.  Substituting these values into $(5)$ and $(6)$, we get $s = \dfrac{c-7}{3}$ and $s = \dfrac{-3c - 6}{10}$.  Equating these values, we get $\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)$.  Thus, our answer is $52 + 19 = \boxed{071}$. ~ TopNotchMath