If $y = a \ln x + b x^2 + x$ has extreme values at $x = 1$ and $x = 2$, then $(a, b)$ is:

$y' = a/x + 2bx + 1 = 0$. At $x = 1$: $a + 2b + 1 = 0$. At $x = 2$: $a/2 + 4b + 1 = 0$. Solving: from first, $a = -2b - 1$. Sub into second: $(-2b - 1)/2 + 4b + 1 = 0$ ⇒ $-b - 0.5 + 4b + 1 = 0$ ⇒ $3b = -0.5$ ⇒ $b = -1/6$. Then $a = -2(-1/6) - 1 = 1/3 - 1 = -2/3$.