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13 Jul 2008 11:33:39 IST

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Nice Polynomial Prob

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The polynomial P(x) = x^3+bx^2+cx+d has three distinct real roots.

Q(x) = x^2+x+2001

It is given that the polynomial P(Q(x)) has no real roots.

Prove that $P(2001)>\frac{1}{64}$

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Comments (8)

abhishek sinha

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13 Jul 2008 13:06:32 IST

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That's very simple !!!

See, first of all we want to show that

$Q(x) >=2001-\frac{1}{4}$ ........................(1) ( for all real 'x')

it is very simple as Q(x)= $(x+\frac{1}{2})^2 +2001-\frac{1}{4}$

Again it is given that has no real roots for any real x . That means that

if P(x) = $(x-\alpha)(x-\beta)(x-\gamma)$ then all of $\alpha ,\beta,\gamma$ are less than 2001-1/4 , as it is given that those roots are all real ( note this argument !!!)

(further clarification :Q(x) takes all values equal or above 2001 - 1/4, and for all those values of x , P(x) has no roots ( given ) , so all the three real roots that P(x) have must be less than 2001-1/4)

so we get P(2001) = $(2001-\alpha)(2001-\beta)(2001-\gamma)$ > $(\frac{1}{4})^3$ $=\frac{1}{64}$ ( PROVED)

®µD®A

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13 Jul 2008 13:14:09 IST

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@feynmann sir can you please give us a bit more time to think. Please I am just a mere beginner and while i am thinking you post the solution. Please give us some more time. It is my humble request to you.

gaurang

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13 Jul 2008 13:20:40 IST

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sir i didnt get wht exactly u have done...plz can u elaborate over it...

abhishek sinha

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13 Jul 2008 13:22:44 IST

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Plz quote the lines u are unable to understand

gaurang

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13 Jul 2008 13:43:34 IST

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sir i didnot get the part whr u have written tht the roots ofP(x) are,2001-1/4..n every thing after tht...plz can u explain it to me..

Hari Shankar

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13 Jul 2008 13:56:53 IST

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Well it is easy to guess the procedure, but you have to clear why you are doing what you are doing. So, let me explain

Suppose the roots of P(x) are $\alpha, \beta, \gamma$

Then of course $P(x) = (x-\alpha)(x- \beta)(x- \gamma)$

Now P(Q(x)) becomes zero whenever Q(x) assumes the values $\alpha, \beta, \gamma$ .

Now, we can assume $\alpha \ge \beta \ge \gamma$

This means we must have $Q(x) > \alpha$ for all real x.

That means the discriminant of $x^2+x+2001 - \alpha$ must be negative i.e. $2001 - \alpha \ge \frac{1}{4}$ $\therefore P(x) \equiv (2001 - \alpha)(2001 - \beta)(2001 - \gamma) \ge \frac{1}{64}$

gaurang

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13 Jul 2008 14:02:51 IST

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thank you sir ...i got it..

abhishek sinha

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13 Jul 2008 14:26:23 IST

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HsBhatt , Plz be a li'l careful while postimg the soln

$\therefore P(x) \equiv (2001 - \alpha)(2001 - \beta)(2001 - \gamma) \ge \frac{1}{64}$ "

That should be P(2001) . Further ur soln is also not transparent enough ( although u have applied more or less the same procedure )

The flaw :::: U wrote

"Now P(Q(x)) becomes zero whenever Q(x) assumes the values $\alpha, \beta, \gamma$ .

Now, we can assume $\alpha \ge \beta \ge \gamma$

This means we must have $Q(x) > \alpha$ for all real x."

Why not Q(x) < gamma ?( U have to show reason for that telling that Q(x) is an increasing fn and that inequality can't occur for all x )

Now @arg121 ,

plz note that in that problem Q(x) has no role except that of shifting the argument of P(x) ( in P(Q(x))to be greater than 2001-1/4. And it says that P(x) has no real root beyond 2001-1/4.So all the real roots that P(x) has must be less than 2001-1/4.

I think it is pretty clear now .

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