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Given that the Quadratic equation $f(x)=ax^2-bx+c=0$ has two roots in $(0 \: 1)$

if $a,b,c \in \mathbb{N}$ Find Minimum values of $a$ and $b$

Since $a$ is Natural number graph of parabola will be open upwards.

Now $f(0) \gt 0$ and $f(1) \gt 0$ so we get

$c \gt 0$ and $a+c \gt b$ and since roots are in $(0 \: 1)$ product of roots also lies in $(0 \: 1)$

we get

$$\frac{c}{a} \lt 1$$ or $$a \gt c$$

Also Discriminant $$b^2 \gt 4ac \gt 4c^2$$

now how can we find minimum values of $a$ and $b$

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  • $\begingroup$ Is the interval $(0,1]$ ? $\endgroup$ – A---B Apr 14 '17 at 2:52
  • $\begingroup$ No it is $(0 \: 1)$ only $\endgroup$ – Umesh shankar Apr 14 '17 at 2:58
  • $\begingroup$ Is the answer $(5, 5,1)$ ? $\endgroup$ – A---B Apr 14 '17 at 3:19
  • $\begingroup$ @Umeshshankar has two roots Do you mean "two distinct roots", or "two roots counting multiplicities"? That's a clarification which should be edited into the question. $\endgroup$ – dxiv Apr 14 '17 at 5:38
  • $\begingroup$ two distinct roots $\endgroup$ – Umesh shankar Apr 14 '17 at 16:54
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For $f(x) = ax^2 - bx + c$ , we get $f^{'}(x) = 2ax - b$.

Thus the minima is at $\displaystyle f^{'}(x) = 0 \iff 2ax-b = 0 \iff x = {b\over 2a}$.

Therefore a root must be in $[0, b/2a]$ and one in $[b/2a, 1]$.

And since $f(0) > 0$ therefore $f(b/2a) < 0$

therefore $f(b/2a) < f(0) \implies c < b^2/4a$

Similarly, $\displaystyle f(1) > f(b/2a) \implies a-b +c > {-b^2\over4a} + c \implies b < 2a $

Since there is no zero at $1$ and $f(1) > 0$ we get $a + c > b$.

So we need $3$ three numbers that satisfy,

  1. $c < b^2/4a$
  2. $b < 2a$
  3. $a + c > b$

The smallest number $z \in \mathbb{N}$ that satisfy $z > b$ is $b+1$ therefore on comparing this with $(3)$ we get $a = b$ and $c = 1$(*).

Plugging $a = b$ and $c = 1$ in $(1)$ we get $a = b = 5$ and $c = 1$.


(*) :- If $a = 1$ and $c = b$ then from $(1)$ we get $4 < b$ and from $(2)$ we will get $b < 2$ which is not possible, thus $a = b$ and $c = 1$.

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  • $\begingroup$ This answer is incorrect, do you think you can modify your proof to correct it - I like the style... $\endgroup$ – Robin Aldabanx Apr 14 '17 at 5:43
  • $\begingroup$ @RobinAldabanx No, my proof is correct. The roots in below proof are same $x = 1/2$. I can easily change my first inequality from strict to slack to get the same result. $\endgroup$ – A---B Apr 14 '17 at 5:44
  • $\begingroup$ I see, does it say somewhere that the roots are distinct? $\endgroup$ – Robin Aldabanx Apr 14 '17 at 5:45
  • $\begingroup$ @RobinAldabanx The problem becomes trivial if we don't assume so because for same roots the form of quadratic is $( (a/b)x - 1)^2 = 0$, for natural $a,b$ we have least value as $a=2, b= 1$. $\endgroup$ – A---B Apr 14 '17 at 5:47
  • $\begingroup$ Well he said $c \in \mathbb{N}$, which usually doesn't include $0$... $\endgroup$ – Robin Aldabanx Apr 14 '17 at 5:55
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By inspection, it looks like $y = f(x) = 4x^2 - 4x + 1$ is a candidate.

Analysis:

  • $f(0) = c > 0$ because $c \in \mathbb{N}$.

Thus, the parabola is open upwards because it has to cross two points $0 < x_1, x_2 < 1$ on the $x$ axis and passing the point $(0,c)$ in the first quadrant.

  • Because the parabola is open upwards, $f(1)$ must be greater than zero:

$$ f(1) = a - b + c > 0 \implies b < a + c. \quad (1)$$

  • The product of two roots:

$$x_1\cdot x_2 < 1 \implies c/a < 1 \implies a > c. \quad (2)$$

  • Finally, to have two real roots:

$$ b^2 - 4ac \ge 0 \implies b^2 \ge 4ac. \quad (3)$$

We will tabulate all possible values that satisfy the three conditions above.

A. Case $c = 1:$

  • According to $(2)$: $\,a = 2, 3, 4, 5, 6, 7, 8, \ldots$

    • For $a=2:$

      • $(1)$ says $b < 3 \implies b = 1, 2.$
      • $(3)$ says $b^2 \ge 8 \implies b = 3, 4, 5, \ldots$
      • Thus, no values of $b$ satisfy both conditions.
    • For $a=3:$

      • $(1)$ says $b < 4 \implies b = 1, 2, 3.$
      • $(3)$ says $b^2 \ge 12 \implies b = 4, 5, 6, \ldots$
      • Thus, no values of $b$ satisfy both conditions.
    • For $a=4:$

      • $(1)$ says $b < 5 \implies b = 1, 2, 3, 4.$
      • $(3)$ says $b^2 \ge 16 \implies b = 4, 5, 6, \ldots$
      • Thus, only $b = 4$ satisfies both conditions.
      • The first solution is $a=4,\; b =4\; c=1$
    • For $a=5:$

      • $(1)$ says $b < 6 \implies b = 1, 2, 3, 4, 5.$
      • $(3)$ says $b^2 \ge 20 \implies b = 5, 6, \ldots$
      • Thus, only $b = 5$ satisfies both conditions.
      • The second solution is $a=5,\; b =5\; c=1$
    • For $a=6:$

      • $(1)$ says $b < 7 \implies b = 1, 2, 3, 4, 5, 6.$
      • $(3)$ says $b^2 \ge 24 \implies b = 5, 6, \ldots$
      • Thus, only $b = 6$ satisfies both conditions.
      • The second solution is $a=6,\; b =6\; c=1$

In general, we get a set of solutions $a = b = 4, 5, 6, \ldots$ for $c=1$. The resulting parabola is given by:

$$ y = f(x) = ax^2 -ax + 1 = a (x-1/2)^2 + 1 - a/4, \quad a = 4, 5, 6,\ldots $$

B. Case $c = 2:$

  • According to (2), $a = 3, 4, 5, 6, 7, 8, \ldots$

    • For $a=3:$

      • $(1)$ says $b < 5 \implies b = 1, 2, 3, 4.$
      • $(3)$ says $b^2 \ge 24 \implies b = 5, 6, 7, \ldots$
      • Thus, no value of $b$ satisfies both conditions.
    • For $a=4:$

      • $(1)$ says $b < 6 \implies b = 1, 2, 3, 4, 5.$
      • $(3)$ says $b^2 \ge 32 \implies b = 6, 7, 8, \ldots$
      • Thus, no value of $b$ satisfies both conditions.

We can stop here because the next values of $a$ and $b$ to check will exceed the minimum values found in case A.

Thus, the solution with minimum values of $a$ and $b$ is:

$$ y = 4x^2 - 4x +1.$$

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