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If $$ ax^2-bx+c = 0 $$ has two distinct real roots in (0,1) where a, b, c are natural numbers then find the minimum value of product abc ?

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    This was a rather wicked question they asked in my exam, the answer was 25 I think. – 2017-02-08
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    @Sawarnik What kind of exam was it? So we can have some context about where it comes from. – 2017-02-08
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    Can you precise "two distinct real roots in $(0,1)$? Are you meaning that you are looking at $y=ax^2+bx+c$ with the constraint that this curve pass through the point $(0,1)$? – 2017-02-08
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    @Maczinga I think the OP means **the interval** $\;(0,1)\;$ ... – 2017-02-08
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    Ah ok, sorry. That's my poor english interpreter ;-) – 2017-02-08
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    @leon sot Well it was a monthly test in my coaching from the quadratics chapter. However, no one was able to solve it, was too tough to be included in that test. – 2017-02-08

2 Answers 2

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Since $a,b,c $ are positive the roots are trivially greater than 0.

What remains is to solve the inequality:

$\frac{b + \sqrt{b^2-4ac}}{2a} <1$

This reduces to $ a+c>b$

But the roots being real and distinct we have $b^2 >4ac$

Combining both we have :

$a^2 + c^2 + 2ac > b^2 > 4ac$

$b^2 > 4ac$ tells us $b> 2$ (why?)

$ a^2 + c^2+ 2ac > 4ac $ tells us $a \neq c$

Checking small cases we get $(a,b,c) =(5,5,1)$ where $abc =25$

EDIT:

Checking "small" cases is not informative, so adding an explanation:

Keeping in mind $a+c>b$, the minimum value of $ac$ occurs when $a=b$ and $c=1$. So for given $b$, the minim of $abc$ is $b^2$. The smallest value of $b$ which agrees the inequality $b^2>4b$ is 5 (as $ac=b$). Hence the corresponding minimum value is $5^2$

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    What cases in "checking small cases"? Small values of $\;(a,b,c)\;$ ? Or integer polynomials with roots in the interval $\;(0,1)\;$ ? – 2017-02-08
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    @DonAntonio I have edited the answer – 2017-02-08
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    Your first line made me think of this: http://spikedmath.com/557.html :) – 2017-02-08
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Hints: If the roots are $\;\alpha,\,\beta\;$ , then

$$\alpha+\beta=\frac ba\;,\;\;\alpha\beta=\frac ca\,,\,\,\text{and we can also write}\;\; abc=a^3\cdot\frac ba\cdot\frac ca\;\ldots$$

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    I didnt get what the last step is? – 2017-02-08
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    @ArchisWelankar A way to express the product $\;abc\;$ by means of the roots $\;\alpha,\,\beta\;$ . There was a wrong two where it should be a three. Edited. Of course, it could also help to know the condition on the discriminant to have tow **different** roots... – 2017-02-08
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    The coefficients are supposed to be integral. How is that checked? – 2017-02-08
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    @N.S.JOHN Are you addressing me? I don't need to check anything. The OP must use, if he wills, the hints given and solve. He has to take into account several things, I guess. – 2017-02-08
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    Sir considering the coefficients are positive integers there must be. – 2017-02-08
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    @N.S.JOHN Yes, so I realized two minutes ago: the integers are natural, *not* merily integer. – 2017-02-08
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    The use of the double arrow makes your last equation $abc = a^3 \frac{b}{a} \frac{c}{a}$ look like a conclusion, but of course that equation is immediately true (assuming only that $a$ is nonzero). Did you mean to write, for instance, $abc = a^3(\alpha + \beta)(\alpha \beta)$? – 2017-02-08
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    @mathmandan It **is** a conclusion, of course. That doesn't mean it is the end of the work... – 2017-02-08
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    Well, agreed! I'm just saying that your use of the double arrow notation has the potential to confuse, or mislead, the reader. It makes it look like your last equation follows from, or depends on, the other equations in some way. It doesn't really, except in the sense that a true statement is implied by every statement. (You could replace the double arrow by another comma, for instance, and everything in your answer would still be correct.) – 2017-02-08
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    @mathmandan I think you're right: the end of that arrow is a trivial equality. I only tried to attract the OP's attention to the fact that $\;\alpha,\,\beta\;$ fit there. Thanks, I shall edit that. – 2017-02-08