location of roots of quadratic with natural coefficients the quadratic equation $ ax^2-bx+c=0 $ ; $a,b,c \in \mathbb{N}$, has two distinct real roots belonging to the interval $(1,2)$ , then what would be least value of $a$  and $b$?
I tried to solve these four inequalities with iterations,which seems to be a lengthy process
as roots are real and distinct $b^2-4ac>0$
abscissa of vertex  of parabola will lie between $(1,2)$ so $ 1<\frac{b}{2a} <2$
also quadratic will have positive values at $x=1$ and $x=2$ so   $a-b+c> 0 $ and $4a-2b+c>0$
please help me out .
 A: You have correctly derived the conditions $2a<b<4a$, $a-b+c>0$, $4a-2b+c>0$, $c<\frac{b^2}{4a}$. So if we take $a=1$, then $b=3$. But now the conditions on $c$ cannot be satisfied.
For $a=2$ we must have $b=5,6,7$. If $b=5$ we require $c>3$ and $c\le3$. If $b=6$ we require $c>4$ and $c\le4$. If $b=7$ we require $c>6$ and $c\le6$.
For $a=3$ we must have $b=7,8,9,10,11$. If $b=7$ we require $c>4$ and $c\le4$. If $b=8$ we require $c>5$ and $c\le5$. If $b=9$ we require $c>6$ and $c\le6$. If $b=10$ we require $c>8$ and $c\le8$. If $b=11$ we require $c>10$ and $c\le10$. 
For $a=4$ we must have $b=9,10,11,12,13,14,15$. If $b=9$ we require $c>5$ and $c\le5$. If $b=10$ we require $c>6$ and $c\le6$. If $b=11$ we require $c>7$ and $c\le7$. If $b=12$ we require $c>8$ and $c<9$ ($c=9$ gives equal real roots). For $b=13$ we require $c>10$ and $c<11$. For $b=14$ we require $c>12$ and $c\le12$. For $b=15$ we require $c>14$ and $c\le14$.
For $a=5$ we must have $b=11,12,13,14,15,16,17,18,19$. For $b=11$ we require $c>6$ and $c\le6$. For $b=12$ we require $c>7$ and $c\le7$. For $b=13$ we require $c>8$ and $c\le8$. For $b=14$ we require $c>9$ and $c\le9$. For $b=15$ we find $c=11$ gives two roots in $(1,2)$. 
