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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 .

vikas meena
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    You require the roots to belong to the open interval $(1,2)$? Otherwise $x^2-3x+2$ would work with roots at $1,2$. – almagest Jun 07 '16 at 18:09
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    @almagest It seems from the work shown in the post ("the quadratic... positive values at $1$ and $2$...") that the OP is indeed looking for the roots to lie in $(1,2).$ – coffeemath Jun 07 '16 at 18:22
  • @coffeemath. Thanks. I should have read that more carefully! – almagest Jun 07 '16 at 18:26
  • vikas: Since there are two natural numbers $a,b$ and you seek the "least" choice(s), one needs a single measure such as $a+b$ for there to be just one answer. Otherwise one might, for each $a$ get the minimal $b,$ and conversely for each $b$ get the minimal $a.$ But the latter approach might yield a messy "answer"... – coffeemath Jun 07 '16 at 19:08

1 Answers1

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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 $(1,2)$.

almagest
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