If the roots of the equation $$ax^2-bx+c=0$$ lie in the interval $(0,1)$, find the minimum possible value of $abc$.

Edit: I forgot to mention in the question that $a$, $b$, and $c$ are natural numbers. Sorry for the inconvenience.
Edit 2: As Hagen von Eitzen said about the double roots not allowed, I forgot to mention that too. Extremely sorry :(

I tried to use $D > 0$, where $D$ is the discriminant but I can't further analyze in terms of the coefficients. Thanks in advance!

  • 3
    $\begingroup$ My bet: $a,b,c$ must be nonnegative integers and double root is not allowed. $\endgroup$ – Hagen von Eitzen May 10 '14 at 19:10
  • $\begingroup$ @MathGod I'm sorry I forgot to mention that $a$, $b$,and $c$ are Natural numbers. $\endgroup$ – Henry Durham May 10 '14 at 19:11

Given: Roots lie in $(0,1).$

Let $f(x)=ax^2-bx+c$ and it's roots be $\alpha$ and $\beta$

$\implies f(0) \times f(1) > 0$ (Can be easily verified from the parabolic graph of $f(x)$)

or $c(a-b+c)>0$

$\implies \frac{c}{a}(1-\frac{b}{a}+\frac{c}{a})>0$

$\implies \alpha \beta (1-(\alpha + \beta) + \alpha\beta) > 0$ (Using Vieta's Formula)

$\implies \alpha\beta(1-\alpha)(1-\beta) > 0$


Consider $\alpha(1-\alpha)$,

By AM-GM inequality,



$\beta(1-\beta)< \frac{1}{4}$

By multiplying the above two inequalities, we get,


$\implies \frac{c}{a}(1-\frac{b}{a}+\frac{c}{a})<\frac{1}{16}$

$\implies c(a-b+c)<\frac{a^2}{16}$

If we let $a$=$b$ and $c=1$, clearly we are getting the minimum value of $a$, i.e.


$a>4$ or minimum $a =5$

Since $D > 0$, we have $b^2-4ac > 0$ (where $D$ is the discriminant of $f(x)=0$)

this inequality is satisfied for $a=b=5$ which we calculated above

thus at $a=b=5$ and $c=1$ the minimum value of $abc=25$ is achieved.

  • $\begingroup$ Compare with @apt1002's answer. $\endgroup$ – vadim123 May 10 '14 at 19:36
  • 1
    $\begingroup$ Awesome and rigorous proof! $\endgroup$ – Henry Durham May 10 '14 at 19:45
  • $\begingroup$ That answer is indeed correct; however it has a double root, which the OP (belatedly) forbid. $\endgroup$ – vadim123 May 10 '14 at 19:46
  • $\begingroup$ @vadim123: My bad, I didn't see it was $-b$ in the question! $\endgroup$ – rubik May 10 '14 at 19:50
  • $\begingroup$ @rubik In my answer the roots of the final equation are $0.276$ and $0.724$ (approx) which are in the interval $(0,1)$. $\endgroup$ – MathGod May 10 '14 at 19:54

The discrimimnat $D=b^2-4ac$ must be positive to ensure two distinct real roots. (If double root is not forbidden, we have $4x^2-4x+1$ with double root at $\frac12$ and $abc=16$). Next, we must have $f(1)>0$, i.e. $$a+c>b.$$ For naturals $a,c$ we also have $ a+c\le 1+ac$ and conclude $$\tag1b\le ac.$$ If $b\le 4$ we obtain $b\le ac<\frac14b^2\le b$, contradiction. (NB: If we relax the condition that the roots be distinct, the $<$ becomes a $\le$ and instead of a contradiction we find $b=ac=4$, hence $abc=16$). Hence $b\ge 5$ and by $(1)$ $$abc\ge b^2\ge 25.$$ The minimum is indeed attained as can be seen by making all iniequalities sharp, which gives: Either $(a,b,c)=(5,5,1)$ or $(a,b,c)=(1,5,5)$. The first of these indeed gives two roots in $(0,1)$.


If you multiply the equation by $k$, you get $$(ka)x^2-(bk)x+(ck)=0$$ This new equation has the same roots as the original, hence in $(0,1)$, but has the product of its coefficients $k^3abc$. By letting $k\to\pm \infty$ (depending on whether $abc>0$ or $abc<0$), you can make this product as small as you like. Hence the answer is $-\infty$.

  • $\begingroup$ typo in line 2 bkx*x $\endgroup$ – drawnonward May 10 '14 at 18:45
  • $\begingroup$ thx @drawnonward $\endgroup$ – vadim123 May 10 '14 at 18:45
  • $\begingroup$ @vadim123 I'm afraid the answer is $25$ (as given on the last page of my book). $\endgroup$ – Henry Durham May 10 '14 at 19:05
  • $\begingroup$ @boxed__l We cannot take $k=\pm\infty$, only consider $k\to\pm\infty$; whatever $M\in\mathbb R$ you pick, one can pick $k$ with $|k|$ large enough to ensure $k^3abc<M$. Strictly speaking, $-\infty$ is not make the minimum, but the infimum of all possible $abc$. Also, we need to ensure that there exists such a polynomial with $abc\ne0$ in the first place. $(x-\frac12)(x-\frac13)=x^2-\frac56x+\frac16$ shows this. $\endgroup$ – Hagen von Eitzen May 10 '14 at 19:06
  • $\begingroup$ @Samurai I have given an explcit example with $abc=\frac5{36}$ already in my previous comment. Did you perhaps forget an important condition on $a,b,c$ (such as: nonnegative integer)? $\endgroup$ – Hagen von Eitzen May 10 '14 at 19:07

The answer is $a=4$, $b=4$, $c=1$, giving $x = \frac12$ (twice), and a product $abc=16$. Exhaustive search through all $1 \leq a,b,c \leq 16$ gave no better answer.


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