<s>[INST] Find (a,b,c) if $ax^2-bx+c=0$ have roots lying in $(0,1)$ where $a,b,c\in\mathbb{Z_+}$ 
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 ?
 [/INST] You have that $b^2-4ac\gt0$, since the roots are real.  Only $(5,5,1)$ meets this condition. 
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