5
$\begingroup$

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!

$\endgroup$
  • 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 May 10 '14 at 19:11
6
$\begingroup$

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)

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

Now,

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

By AM-GM inequality,

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

Similarly,

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

By multiplying the above two inequalities, we get,

$\alpha\beta(1-\alpha)(1-\beta)<\frac{1}{16}$

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

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

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

$\frac{a^2}{16}>1$

$a>4$ 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.

$\endgroup$
  • $\begingroup$ Compare with @apt1002's answer. $\endgroup$ – vadim123 May 10 '14 at 19:36
  • 1
    $\begingroup$ Awesome and rigorous proof! $\endgroup$ – Henry 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
7
$\begingroup$

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 $(0,1)$.

$\endgroup$
1
$\begingroup$

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 $-\infty$.

$\endgroup$
  • $\begingroup$ 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 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.