Show all real roots are negative

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    Negative Roots
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Discussion Overview

The discussion revolves around the polynomial $f(x)=x^5-10x+38$ and the assertion that all its real roots are negative. Participants explore different approaches and reasoning related to this claim.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant poses the challenge of demonstrating that all real roots of the polynomial are negative.
  • Another participant questions the implications of rewriting the function as $f(x) = x(x^4-10) + 38$, specifically regarding the domain when $x \ge 10^{\tiny\dfrac{1}{4}}$.
  • A later reply expresses regret over asking a question, indicating a sense of uncertainty about the previous inquiry.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the assertion that all real roots are negative, and there are indications of uncertainty and differing interpretations regarding the function's behavior.

Contextual Notes

There are unresolved assumptions regarding the implications of the function's rewritten form and its domain, which may affect the analysis of the roots.

anemone
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Show that all real roots of the polynomial $f(x)=x^5-10x+38$ are negative.

Note:

I know this is a fairly easy challenge, but it's good to see how different approaches can be generated from different people so that we can learn from one another. :o (Yes)
 
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anemone said:
Show that all real roots of the polynomial $f(x)=x^5-10x+38$ are negative.

Note:

I know this is a fairly easy challenge, but it's good to see how different approaches can be generated from different people so that we can learn from one another. :o (Yes)

$f(x)= x^5 + 10(3.8-x) $
it is >0 for $0\le x\lt3.8$

further
$f(x) = x(x^4-10) + 38$

for $x\gt 2$ above is > 0 as $2^4 = 16 \gt 10$

so above is >0 for $2\le x$

we have shown that it is positive for x > 0 so no root 0 or positive hence all real roots are -ve.
 
Last edited:
kaliprasad said:
$f(x)= x^5 + 10(3.8-x) $
it is >0 for $0\le x\lt3.8$

further
$f(x) = x(x^4-10) + 38$

for $x\gt 2$ above is > 0 as $2^4 = 16 \gt 10$

so above is >0 for $2\le x$

we have shown that it is positive for x > 0 so no root 0 or positive hence all real roots are -ve.

Hi kaliprasad,

Thanks for participating.:) I am curious, is there any chance when you rewritten the function of $f$ as $f(x) = x(x^4-10) + 38$, you mean to imply the domain when $x\ge 10^{\tiny\dfrac{1}{4}}$?
 
anemone said:
Hi kaliprasad,

Thanks for participating.:) I am curious, is there any chance when you rewritten the function of $f$ as $f(x) = x(x^4-10) + 38$, you mean to imply the domain when $x\ge 10^{\tiny\dfrac{1}{4}}$?

You are right but I have shown that it is true for x < 3.8 from the 1st equation and I have chosen a suitable value < 3.8 ( that is 2) to show that it is true for x > 2. $x\ge 10^{\tiny\dfrac{1}{4}}$ condition is a superset of it. but if it is true for x > 2 it does meet the criteria
 
kaliprasad said:
You are right but I have shown that it is true for x < 3.8 from the 1st equation and I have chosen a suitable value < 3.8 ( that is 2) to show that it is true for x > 2. $x\ge 10^{\tiny\dfrac{1}{4}}$ condition is a superset of it. but if it is true for x > 2 it does meet the criteria

Oh My...I don't know what is wrong with me! How could I ask something so stupid! Sorry kali, I so wish to retract what I have asked just to appeared to be less silly...:o
 

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