If the equation has roots, then by quadratic formula we have $x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}$ is the larger root. We can easily see that picking smaller $a$'s gives us a larger root, so $a=1$. Meaning that the largest rational root is an integer, but now $x_1 < \frac{-b+b}{2} = 0$ and $-1$ is a root of $x^2+2x+1$, so $-1$ is the largest possible rational root.

TheMathBob wrote: $x_1 = \frac{-b+\sqrt{b^2-4ac}}{2a}$ is the larger root. We can easily see that picking smaller $a$'s gives us a larger root Not if the root is negative (and it is since $a,b,c>0$) oVlad wrote: What is the largest possible rational root of the equation $ax^2 + bx + c = 0{}$ where $a, b$ and $c{}$ are positive integers that do not exceed $100{}$? Since rational, we need $b^2-4ac=u^2$ for some integer $b>u\ge 0$ and then : $a=\frac{b^2-u^2}{4c}$ and root is $r=-\frac{2c}{u+b}$ So we need $b,c$ both odd or both even (in order $a\in\mathbb Z$) and $c$ smallest possible and $u$ greatest possible $u<b$ with same parity implies $u\le b-2$ and smallest $c$ is $1$ Choosing $u=b-2$ and $c=1$ implies $a=b-1$ and $r=-\frac 1{b-1}$ And constraint $a,b,c\le 100$ implies $b=100$ And $\boxed{r=-\frac 1{99}}$ when quadratic is $99x^2+100x+1=0$