IIT JEE , AIEEE Forum - Mathematics , Algebra

little_genius

if has two distindt roots in(0,1) where a,b,c are natural numbers then prove that abc=25

hsbhatt

Please clarify whether you mean abc 25.

little_genius

sorry for my mistake.........

its abc>=25

hsbhatt

Since both roots are real we must have b²4ac.

$\text{Since} \ ac>0 \ \text{we have} \ \sqrt {b^2 -4ac} < b$

This ensures that both roots are positive.

$\text{If the larger of the roots is} \ \alpha \\ \\<br/>\alpha = \frac{b+\sqrt{b^2-4ac}} {2a} < \frac{2a}{2b} = \frac{b}{a} \\ \\<br/>b<a \Rightarrow \alpha<1 \\ \\<br/>\text{So, we can impose the condition} \ b \le a \\ \\<br/>$

We need not have b<a as ac>0 still ensure that the root is less than 1

Product of roots is less than one forces c<a

Thus, we have 4ac<b²a²

For abc to be minimum, we can set c=1

Hence a²>4a a>4

So the minimum value a can take is 5.

Now 20<b²25.

So, b = 5 for this minimal choice of a and c.

So, now abc = 25

This also happens to be the minimal value as for c = 2, a>8 and abc exceeds 25

Hence we have abc 25.

oneyeartogo

Please can you explain this step:

We need not have b<a as ac>0 still ensure that the root is less than 1

hsbhatt

This is because $\sqrt{b^2 - 4ac}$ is strictly less than b. So $\frac{b+\sqrt{b^2 - 4ac}}{2a}$ is strictly less than $\frac{b}{a}$

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