Let the equation $ax^2-bx+c=0$ have distinct real roots both lying in the open interval $(0,1)$ where $a,b,c$ are given to be positive integers. Then the value of the ordered triplet $(a,b,c)$ can be $$ a)\quad (5,3,1)\;,\quad b)\quad(4,3,2)\;,\quad c)\quad(5,5,1)\;,\quad d)\quad(6,4,1) $$ The solution given in my reference is $(5,5,1)$.
My Attempt
$$ a,b,c\in\mathbb{Z_+}\\ \alpha,\beta=\frac{b\pm\sqrt{b^2-4ac}}{2a}\in(0,1)\\ b+\sqrt{b^2-4ac}<2a\implies\sqrt{b^2-4ac}<2a-b\\ b^2-4ac<4a^2+b^2-4ab\implies-ac<a^2-ab\\ a=0\text{ (or) }-c<a-b\implies a+c>b\\ \Big[a=0\text{ (or) } a+c>b\Big]\; \&\;b^2\geq4ac $$
Does that constitute the complete condition for the given problem or am I missing something in my attempt ?
You have that $b^2-4ac\gt0$, since the roots are real. Only $(5,5,1)$ meets this condition.