Solution of a cubic

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Solution of a cubic

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Source: INMO 2000 Problem 5

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1592 posts

Rushil

#1 h Oct 10, 2005, 4:53 AM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be three real numbers such that $1 \geq a \geq b \geq c \geq 0$ . prove that if $\lambda$ is a root of the cubic equation $x^3 + ax^2 + bx + c = 0$ (real or complex), then $| \lambda | \leq 1.$

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3952 posts

#2 h Oct 10, 2005, 11:40 AM • 1 Y

Y by Adventure10

I think that it will help
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1592 posts

Rushil

#3 h Oct 10, 2005, 12:19 PM • 2 Y

Y by Adventure10, Mango247

Can you please tell waht that is? Please post the whole solution for the Resoures Section!

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393 posts

tµtµ

#4 h Oct 10, 2005, 6:15 PM • 2 Y

Y by Adventure10, Mango247

$P(x) = x^3 + ax^2 + bx + c$

If $|z| > 1$ then $|(z-1) \times P(z)| \geq |z|^4 - ( (1-a)|z^3| + (a-b)|z|^2 + (b-c)|z| + c) > 0$

So every roots $z$ are such as $|z| \leq 1$

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4 posts

#5 h Jan 13, 2011, 7:06 PM • 2 Y

Y by Adventure10, Mango247

can anyone post a solution without using complex numbers????????
It's getting complex using them.............

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147 posts

#6 h Feb 4, 2011, 6:36 AM • 3 Y

Y by Math-wiz, Adventure10, Mango247

Here's a long and class-10-student-like solution

Firstly we prove this for a real root $\lambda$
Case 1. $|\lambda|=\lambda$
Assume that $|\lambda|>1$
This implies $|\lambda|^3+a|\lambda|^2+b|\lambda|+c>1+a+b+c\ge 1$ which is a contradiction
Case 2. $|\lambda|=-\lambda$
Assume that $|\lambda|>1$
This implies $|\lambda|^2+b\ge a|\lambda|^2+c=|\lambda|(|\lambda|^2+b)>|\lambda|^2+b$ .Contradiction !

Nextly
Let the roots of the equation be $\lambda,re^{i\theta}$ and $re^{-i\theta}$
Since $\lambda r^2=-c<0$ we have $|\lambda|=-\lambda$
We also have $\lambda+r(e^{i\theta}+e^{-i\theta})=-a$ and $\lambda r(e^{i\theta}+e^{-i\theta})+r^2=b$ .
Combining these two we get
$r^2=|\lambda|^2-a|\lambda|+b$
Here we again consider 2 cases
case(I)
$|\lambda|\le a$ .This implies $r^2=|\lambda|^2-a|\lambda|+b\le a|\lambda|-a|\lambda|+b=b\le 1$ .
case(II)
$|\lambda|>a$ .
$r^2=|\lambda|^2-a|\lambda|+b=|\lambda|(|\lambda|-a)+b\le|\lambda|-a+b=|\lambda|-(a-b)\le |\lambda|\le 1$
QED

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124 posts

#7 h Sep 30, 2011, 11:49 AM • 2 Y

Y by Adventure10, Mango247

The generalisation to higher degree polynomials is also true

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9 posts

#8 h Jan 30, 2014, 1:42 PM • 4 Y

Y by SohamSchwarz119, ayan.nmath, Adventure10, Mango247

I think here is a simple proof.
Since is a root of the equation x3 + ax2 + bx + c = 0, we have
y^3 = −ay^2 − by − c.
This implies that
y^4 = −ay^3 − by^2 − cy
= (1 − a)y^3 + (a − b)y^2 + (b − c)y + c
where we have used again
−y^3 − ay^2 − by − c = 0.
Suppose |y| ≥ 1. Then we obtain
|y|^4 ≤ (1 − a)|y^|3 + (a − b)|y|^2 + (b − c)|y| + c
≤ (1 − a)|y|^3 + (a − b)|y|^3 + (b − c)|y|^3 + c|y|^3
≤ |y|^3.
This shows that |y| ≤ 1. Hence the only possibility in this case is |y| = 1. We conclude
that |y| ≤ 1 is always true.

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DeepanshuPrasad

45 posts

DeepanshuPrasad

#9 h Apr 18, 2016, 6:33 PM • 2 Y

Y by Adventure10, Mango247

$0 \le c \le b \le a \le 1$
therefore $a \le 1 , b \le 1, c \le 1$ so,
$|\lambda^3+a\lambda^2+b\lambda+c|=0\le|\lambda|^3+|\lambda|^2+|\lambda|+1 \le 1+|\lambda|+|\lambda|^2+|\lambda|^3+|\lambda|^4+.....$ upto infinity $=\frac{1}{1-|\lambda|}$
i.e. $\frac{1}{1-|\lambda|}\ge 0$ i.e. $1-|\lambda| \ge 0$
hence, $|\lambda| \le 1$

This post has been edited 1 time. Last edited by DeepanshuPrasad, Apr 18, 2016, 6:36 PM
Reason: to be more specific

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Gaurav-Nagotanekar

7 posts

Gaurav-Nagotanekar

#11 h Jan 3, 2017, 6:47 AM • 3 Y

Y by aops29, Adventure10, Mango247

DeepanshuPrasad wrote:

$0 \le c \le b \le a \le 1$
therefore $a \le 1 , b \le 1, c \le 1$ so,
$|\lambda^3+a\lambda^2+b\lambda+c|=0\le|\lambda|^3+|\lambda|^2+|\lambda|+1 \le 1+|\lambda|+|\lambda|^2+|\lambda|^3+|\lambda|^4+.....$ upto infinity $=\frac{1}{1-|\lambda|}$
i.e. $\frac{1}{1-|\lambda|}\ge 0$ i.e. $1-|\lambda| \ge 0$
hence, $|\lambda| \le 1$

But in this solution ....to sum it to infinity $|\lambda| \le 1$ if its not then you cannot sum it to infinity....so you are assuming the same thing which you have to prove.

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1 post

#12 h Apr 13, 2017, 11:25 AM • 1 Y

Y by Adventure10

First prove that it has only one real root (since f(x) is <=0 for all x<=-1 and >=0 for all x>=1).And the only real root say "£" lies between -1 and 0.

Now let the roots be £,m+in,m-in
£*(m+in)*(m-in)= (-c)
m^2+n^2= (-c)/£ <=1 since £>=-1
So √(m^2+n^2) <=1, |£|<=1 and we are done.

Somebody please convert it into LaTeX

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7 posts

#14 h Jun 6, 2024, 3:03 PM

Y by

Rushil wrote:

Let $a,b,c$ be three real numbers such that $1 \geq a \geq b \geq c \geq 0$ . prove that if $\lambda$ is a root of the cubic equation $x^3 + ax^2 + bx + c = 0$ (real or complex), then $| \lambda | \leq 1.$

If lambda is a root btw [-1,1], then there must be a sign change for f(-1) to f(1) by using rolles theorem,which gives the answer right away.

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106 posts

#15 h Nov 4, 2024, 12:06 PM

Y by

IMOasspirant wrote:

Rushil wrote:

Let $a,b,c$ be three real numbers such that $1 \geq a \geq b \geq c \geq 0$ . prove that if $\lambda$ is a root of the cubic equation $x^3 + ax^2 + bx + c = 0$ (real or complex), then $| \lambda | \leq 1.$

If lambda is a root btw [-1,1], then there must be a sign change for f(-1) to f(1) by using rolles theorem,which gives the answer right away.

f(-1) $\neq$ f(1)

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66 posts

#16 h Mar 26, 2025, 1:40 PM

Y by

IMOasspirant wrote:

Rushil wrote:

Let $a,b,c$ be three real numbers such that $1 \geq a \geq b \geq c \geq 0$ . prove that if $\lambda$ is a root of the cubic equation $x^3 + ax^2 + bx + c = 0$ (real or complex), then $| \lambda | \leq 1.$

If lambda is a root btw [-1,1], then there must be a sign change for f(-1) to f(1) by using rolles theorem,which gives the answer right away.

bro lambda can be complex too rolles is only applicable for reals

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