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Let $a,b,c $ be Natural Numbers, such that roots of the equation $ax^2-bx+c=0$ are distinct and both lie in the interval

  1. (0,1)
  2. (1,2)
  3. (2,3)

(Brackets signify open interval, roots are $IN BETWEEN $ the numbers in each part.)

Find minimum possible value of $a, b, c.$

On my part, I solved for part 1, i.e. for distinct roots between (0,1). But for the next two parts, the things are getting a too bit messy.

While it may have similarity in question for given part 1 in stack exchange, there is no generalized method so that we can solve for other such intervals.

So please help, I am new to stack exchange.

For part 3, I tried by taking $0<m-2, 3-m, n-2, 3-n <1$ where m, n are the roots of the equation, and then using A. M. - G. M method, but i failed.

Please check another question of this type, But please don't provide with such answers as given in the link, as this is a question of an entrance exam, to be solved by hand, and not wolfram mathematica.

Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root...

Please also tell me whether such numbers correspond to any famous known series.

Many are telling that there can be no such coefficients, then for part 1, please check $5x^2-5x+1=0$. It has it's roots between 0 and 1, and the coefficients are natural numbers to be sure, namely $a=5,b=5,c=1$.

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  • $\begingroup$ I want to say that the roots are between, say for example, in between 2 and 3, and the equation has natural numbers as coefficients. Find the least possible such natural number coefficients. $\endgroup$ – user10595795 May 5 at 12:23
  • $\begingroup$ The midpoint between the roots is $\frac{b}{2a}$. Since the midpoint should be in the interval, that gives you two linear inequalitties. For example, for $(1,2)$ it gives $2a<b<4a$. The distance from the roots to their midpoint is $\frac{\sqrt{b^2-4ac}}{2a}$. This distance should be smaller than the distance from the midpoint to the endpoints of the interval. This gives two more linear inequalities. In the same example $\frac{\sqrt{b^2-4ac}}{2a}<\frac{b}{2a}-1$ gives $0<a+c-b$, and $\frac{\sqrt{b^2-4ac}}{2a}<2-\frac{b}{2a}$ gives $0<4a+c-4b$. $\endgroup$ – logarithm May 5 at 12:30
  • $\begingroup$ I tried all this, but without results. There are no bounds for any of the integral coefficient. I would like you to please provide the entire solution, I am just stuck. $\endgroup$ – user10595795 May 5 at 12:32
  • $\begingroup$ The bounds come from the coefficients being natural numbers. Once you have one solution, there are a finite number of smaller possibilities. $\endgroup$ – Mark Bennet May 5 at 12:33
  • $\begingroup$ Those are bounds on the coefficients. The locus where each inequality is an equality are planes. You can draw those planes in $3$-D Cartesian coordinates and pick those integer points in the first octant that lie inside the approprieate half-spaces. $\endgroup$ – logarithm May 5 at 12:35
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So multiply through by $4a$ to obtain $$4a^2x^2-4abx+4ac=(2ax-b)^2+4ac-b^2$$

Now with $a$ positive the quadratic has a minimum value at $x=\cfrac b{2a}$ and the value must be negative at this point so that $4ac-b^2\lt 0$. You also need $\cfrac b{2a}$ in the relevant range of values for $x$, and if the range is the interval $(p,q)$ you need the values at $p$ and $q$ to be positive.

The question is then about controlling for these conditions whilst keeping $a,b,c$ small.


So let's try it for the interval $(2,3)$

We need $4a-2b+c\gt 0$ and $9a-3b+c\gt 0$ from the conditions at the endpoints.

We also need $2\lt \frac b{2a}\lt 3$ ie $4a\lt b\lt 6a$

The smallest possible value of $a$ is $1$ which would make $b=5$ and we'd need $-6+c\gt 0$ ie $c\gt 6$

Now $4ac-b^2=4c-25$ must be negative and that is impossible for $c\gt 6$

So try $a=2$ with $b=9,10,11$

With $b=9$ we need$-10+c\gt 0$ and $-9+c\gt 0$ ie $c\gt 10$, and $8c-81\lt 0$ which is impossible for $c\gt 10$

With $b=10$ we get $c\gt 12$ and need $8c-100 \lt 0$ which is again impossible. But there is more margin here than with $a=1$ or with the asymmetric $b=9$.

Looking to be efficient, let's try for a symmetric solution with $a$ as small as possible and $b=5a$. We can the check other possibilities later.

The edge conditions are then $4a-10a+c\gt 0$ and $9a-15a+c\gt 0$ ie $c\gt 6a$ (or for later $4c\gt 24a$)

Then $4ac-b^2=4ac-25a^2\lt 0$ so that $25a\gt 4c\gt 24a$. So we need $a$ large enough that there is a multiple of $4$ between $24a$ (which is divisible by $4$) and $25a$. The least value of $a$ is $5$ and gives $b=25$ and $c=31$

I will leave you to do the extra checking. It is tedious going through the cases, but perfectly possible to do.

Note that (as suggested in comments) this is $5(x-2)^2-5(x-2)+1$

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  • $\begingroup$ I know that, but that doesn't provide the answer, it is merely an equation $\endgroup$ – user10595795 May 5 at 12:33
  • $\begingroup$ There can be more equation like taking the higher root less than upper bound, f(x) >0 at the two bounds and so on, but non provides the answer $\endgroup$ – user10595795 May 5 at 12:35
  • $\begingroup$ @user10595795 I have added some workings to show what I intended. All the steps are straightforward. I have shown a polynomial for $(2,3)$ but left some work for you to do. $\endgroup$ – Mark Bennet May 5 at 13:05
  • $\begingroup$ I must admit that you have done it quite correctly, but this is just one of the question among 54 other questions to be solved within 3 hours, namely jee advanced. Can you provide a solution which is less tedious and can be solved in such a time crunch? $\endgroup$ – user10595795 May 5 at 13:35
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Hint:

A quadratic polynomial $ax^2-bx+c \in\mathbf R[x]$ has distinct reals in the interval $(\alpha,\beta)$ id and only if

  1. It has two distinct roots, i.e. $\;b^2>4ac$
  2. Neither $\alpha$ nor $\beta$ separate the roots, i.e. $\begin{cases}a\mkern1mu p(\alpha)>0,\\a\mkern1mup(\beta)>0.\end{cases}$
  3. The arithmetic mean of the roots is in $(\alpha,\beta)$, which is equivalent to $$(b-2a\alpha)(b-2a\beta)<0.$$
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  • $\begingroup$ This is something even i tried, but it literally provides nothing. I think i have provided what i tried in the comments and the question, kindly please read it. $\endgroup$ – user10595795 May 5 at 12:38

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