# Got stuck due to the inequality not being satisfied

#### decentfellow

1. Homework Statement
Let $a,b,c$ be positive integers and consider all the quadratic equations of the form $ax^2-bx+c=0$ which have two distinct real roots in $(0,1)$. Find the least positive integers $a$ and $b$ for which such a quadratic equation exist.

2. Homework Equations

3. The Attempt at a Solution
Let $f(x)=ax^2-bx+c$. Now as the roots of $f(x)=0$ lie in $(0,1)$ hence $f(0)\cdot f(1)\gt0$.

$$D\gt0\implies b^2-4ac\gt0$$
$$f(0)=c \\ f(1)=a-b+c$$

As, $a,b,c\in\mathbb{Z}^+$, so
$$f(0)\ge1 \implies c\ge1\tag{1}$$
$$f(1)\ge1\implies a-b+c\ge1\tag{2}$$
$$f(0)\cdot f(1)\ge1\implies c(a-b+c)\ge1\implies c^2+(a-b)c-1\ge0\tag{3}$$

Now, for the quadratic inequality in $c$ that we obtained in $(3)$, for it to hold we get

$$D\le0\implies (a-b)^2+4\le0$$
The above inequality doesn't hold $\forall a,b \in \mathbb{Z}^+$.

After arriving at this conclusion I am kind of stuck on what to conclude from the above result or rather am I going in the right direction.

A little push in the right direction in the form of a hint or rather what should be the line of thought for attempting the question. Please don't write the solution, just the line of thought to arrive at the solution would be enough.

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#### decentfellow

@haruspex. Can you guide me in the right direction.

#### fresh_42

Mentor
2018 Award
1. Homework Statement
Let $a,b,c$ be positive integers and consider all the quadratic equations of the form $ax^2-bx+c=0$ which have two distinct real roots in $(0,1)$. Find the least positive integers $a$ and $b$ for which such a quadratic equation exist.

2. Homework Equations

3. The Attempt at a Solution
Let $f(x)=ax^2-bx+c$. Now as the roots of $f(x)=0$ lie in $(0,1)$ hence $f(0)\cdot f(1)\gt0$.
Why? And why don't you analyse the roots by the formula for them?
$$D\gt0\implies b^2-4ac\gt0$$
$$f(0)=c \\ f(1)=a-b+c$$

As, $a,b,c\in\mathbb{Z}^+$, so
$$f(0)\ge1 \implies c\ge1\tag{1}$$
$$f(1)\ge1\implies a-b+c\ge1\tag{2}$$
$$f(0)\cdot f(1)\ge1\implies c(a-b+c)\ge1\implies c^2+(a-b)c-1\ge0\tag{3}$$

Now, for the quadratic inequality in $c$ that we obtained in $(3)$, for it to hold we get

$$D\le0\implies (a-b)^2+4\le0$$
The above inequality doesn't hold $\forall a,b \in \mathbb{Z}^+$.

After arriving at this conclusion I am kind of stuck on what to conclude from the above result or rather am I going in the right direction.

A little push in the right direction in the form of a hint or rather what should be the line of thought for attempting the question. Please don't write the solution, just the line of thought to arrive at the solution would be enough.

#### decentfellow

Why? And why don't you analyse the roots by the formula for them?
I don't think it gives a stricter bound.

#### fresh_42

Mentor
2018 Award
I haven't done the exercise, but there is at least another condition, because you not only have $D > 0$ but also $0 < b \pm \sqrt{D} < 2a\,$.

#### decentfellow

I haven't done the exercise, but there is at least another condition, because you not only have $D > 0$ but also $0 < b \pm \sqrt{D} < 2a\,$.
Yeah and on solving bot the inequalities the cases that you get are
$$4a(a-b+c)\gt0 \\ 4ac\gt0$$ and we already have stricter bounds than these with $$f(1)\ge1\implies a-b+c\ge1 \\ f(0)\ge1\implies c\ge1$$ which I have already shown in my original post.

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#### fresh_42

Mentor
2018 Award
How did you get
$D\le0\implies (a-b)^2+4\le0$​
from $(3)$? Isn't e.g. $(15,15,1)$ one possible solution?

#### decentfellow

How did you get
from $(3)$? Isn't e.g. $(15,15,1)$ one possible solution?
Yeah that's right, and I got the solution. will be done writing it in my notes, then will post it. The inequalities that need to be solved are:
$$|a|\gt4$$
$$c\lt a$$
$$a+c\lt b$$
$$b^2\gt4ac$$

From these we get the least value of $a$ and $b$ $(a,b)=(5,5)$

Will be posting a thorough solution in about half an hour.

#### decentfellow

After a thinking a lot on the question I was able to come up with a good enough solution.

Let $f(x)=ax^2-bx+c$. Let the roots of the equation $f(x)=0$ be $\alpha$ and $\beta$. Then, $\alpha,\beta\in(0,1)$.

As the roots of the equation $f(x)=0$ are real and distinct, then $b^2\gt4ac$. Also, due to the roots lying in $(0,1)$, hence $f(0)\gt0$, $f(1)\gt0$ and $f(0)f(1)\gt0$

Now, $$f(0)=c\ge1\tag{1}$$
$$f(1)=a-b+c\ge1\tag{2}$$
$$\therefore \qquad f(0)\cdot f(1)\ge1\qquad\qquad\tag{3}$$
$$b^2\gt4ac\tag{4}$$

If $\alpha$ and $\beta$ are the zeroes of $f(x)$, then $f(x)=a(x-\alpha)(x-\beta)$
$$\therefore f(0)\cdot f(1)=a\alpha(1-\alpha)a\beta(1-\beta)\ge 1$$

Using A.M-G.M inequality we get,
$$\dfrac{b+(1-b)}{2}\ge\sqrt{b(1-b)}\implies b(1-b)\le\dfrac{1}{4}$$
equality occurs when $b=\dfrac{1}{2}$
$$\therefore \qquad\qquad\alpha(1-\alpha)\ge\dfrac{1}{4}\qquad\qquad \beta(1-\beta)\ge\dfrac{1}{4}$$

In both the cases equality occurs when $\alpha=\beta=\dfrac{1}{2}$, but as $\alpha\neq\beta$

So, $$\alpha(1-\alpha)\beta(1-\beta)\gt\dfrac{1}{16}\implies a\alpha(1-\alpha)a\beta(1-\beta)\gt\dfrac{a^2}{16}\\ \implies \dfrac{a^2}{16}\lt f(0)\cdot f(1)\implies |a|\lt4$$

$$\therefore a_{min}=5$$

Also, $$\because \qquad\qquad\alpha\cdot\beta\lt1\implies c\lt a_{min}\qquad\qquad\tag{5}$$

$$(2)\implies a+c\ge1+b\implies a_{min}+c-1\ge b\implies 4+c\ge b \\ \implies c^2+8c+16\ge b^2\implies c^2+8c+16 \gt 4a_{min}c\\ \implies c^2-12c+16\gt0\implies c\in(0,6-2\sqrt5)\cup(6+2\sqrt5,\infty)$$

But, as $c\lt a_{min}$, hence $c=1$

Now, from $$b^2\gt 4a_{min}c\implies b^2\gt20$$
Hence, $b_{min}=5$

So, the value of least positive integers $a$, $b$ and $c$ for which $f(x)$ has roots b/w $0$ and $1$ is $(a,b,c)=(5,5,1)$

#### haruspex

Homework Helper
Gold Member
2018 Award
So, $$\dfrac{a^2}{16}\lt f(0)\cdot f(1)\implies |a|\lt4$$
I don't follow that step. You had f(0)f(1)>=1, not f(0)f(1)<=1.
If you meant |a|>4, that does not follow either.

#### Buffu

$ax^2 -bx + c = f(x)$
$f(0) = c \ge 1$
$f(1) = a-b+c \ge 1$
$D \gt 0 \implies b^2 -4ac \gt 0 \implies b^2 \gt 4ac$

Putting it together
$c \ge 1$
$a-b+c \ge 1 \implies a+c \ge b+1$
$b^2 \gt 4ac$

Taking $c = 1$

We get,
$a \ge b$
$b^2 \gt 4a$

For lowest value we fix $a = b$
$b^2 \gt 4b$
From which we get, $b \gt 4$
For lowest value, $a = b = 5$
So solution is $(a,b,c)=(5,5,1)$

@haruspex
@fresh_42

Is this correct ?

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#### fresh_42

Mentor
2018 Award
Except I don't know what A.M-G.M inequality is, where you lost me in post #9 so I couldn't see why $b(1-b)\leq\frac{1}{4}$,
I have the following questions: The inequalities $b^2 > 4ac\, , \,c \geq 1\, , \,a+c \geq b+1$ you have are o.k. Now you set $c=1$ and next $a=b$ which gave you $(a,b,c)=(5,5,1)$ which is a solution and minimal under these conditions. Right.

But what happens in the cases
1. $c=1 \,\wedge\, a>b$?
2. $c > 1$?
Since it is not defined what the least value of a triple $(a,b,c)$ means (sum, average, least maximal value, least minimal value or whatever), one has to somehow show, that the cases above are either impossible or lead to higher numbers of all three values $a,b,c$. The first case is easy (by looking at $ab > b^2 > 4a$), but why can't e.g. $(a,b,c)=(3,4,2)$ or something like this be a solution?

#### haruspex

Homework Helper
Gold Member
2018 Award
Since it is not defined what the least value of a triple (a,b,c) means
The OP actually says the least a and b; c is not required to be minimised. (But as you say, that is still not well defined.)
Consequently...
Taking c=1
... is not justified.

#### Buffu

The OP actually says the least a and b; c is not required to be minimised. (But as you say, that is still not well defined.)
Consequently...

... is not justified.
I don't have a proof but condition for a, b to minimum for a given c is $a=b=1+4c$ where $c > 0$.

#### haruspex

Homework Helper
Gold Member
2018 Award
I don't have a proof but condition for a, b to minimum for a given c is $a=b=1+4c$ where $c > 0$.
If you are relying on that then it needs to be demonstrated within your proof prior to the line "Taking c=1".

#### Buffu

If you are relying on that then it needs to be demonstrated within your proof prior to the line "Taking c=1".
Yes I do rely on that but I don't have a proof.
I observed it while playing with a graphing calculator. Can you prove/disprove it ?

#### decentfellow

I don't follow that step. You had f(0)f(1)>=1, not f(0)f(1)<=1.
If you meant |a|>4, that does not follow either.
Yeah, that was a mistake on my part while writing the solution and the chain errors occured because I was not following the rough work I had done but what I had typed on the screen, so it should be
$$\alpha(1-\alpha)\beta(1-\beta)\lt\dfrac{1}{16}\implies a\alpha(1-\alpha)a\beta(1-\beta)\lt\dfrac{a^2}{16}\\ \implies \dfrac{a^2}{16}\gt f(0)\cdot f(1)\implies |a|\gt4$$

#### decentfellow

The OP actually says the least a and b; c is not required to be minimised. (But as you say, that is still not well defined.)
Consequently...

... is not justified.
I just wrote the triplet $(a,b,c)$ to make it easy to write the values that correspond to each $a,b$ and $c$ it just resembles and ordered pair.

Although the question doesn't ask for the least value of $c$ but if it is not minimised then we will not get the least value of all the possible values of $a$ and $b$ for a different $c$

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#### Mark44

Mentor
I just wrote the triplet $(a,b,c)$ to make it easy to write the values that correspond to each $a,b$ and $c$ it just resembles and ordered pair.

Although the question doesn't ask for the least value of $c$ but if it is not minimised then we will not get the least value of all the possible values of $a$ and $b$ for a different $c$
Don't write stuff like $(a, b, c)$ -- it doesn't do anything but clutter up what you wrote. We have a LaTeX tutorial here: https://www.physicsforums.com/help/latexhelp/

#### decentfellow

Don't write stuff like $(a, b, c)$ -- it doesn't do anything but clutter up what you wrote. We have a LaTeX tutorial here: https://www.physicsforums.com/help/latexhelp/
Oh, sorry I was just writing some stuff in TeX editor hence forgot that we are supposed to use "$" here instead of "" for inline. #### decentfellow Yes I do rely on that but I don't have a proof. I observed it while playing with a graphing calculator. Can you prove/disprove it ? Pretty interesting, using the conditions that you gave I did get the minimum (a,b), but I would wanna ask you how did you arrive at that, who knows maybe it would spread some light on how to prove/disprove it. #### haruspex Science Advisor Homework Helper Gold Member 2018 Award Seeing this thread has ground to a halt, here's my approach. You have three basic equations:$b^2>4ac0<b-\sqrt{b^2-4ac}b+\sqrt{b^2-4ac}<2a##
You can square up to get rid of the square root signs, and in the process make maximum use of knowing a, b and c are integers in changing the inequality signs to the "or equal" form.
That should get you to (2a-b-1)2>=b2-4ac>=1
Next, consider the cases b even, b odd, separately. "b even" lets you do a bit more with the second inequality above, while "b odd" lets you get a bit further with the first one.
Finally, you might need to do a bit of iteration of the form a > something, so reusing an earlier inequality b > something, so a > something a bit higher...

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