Quartic real roots - factor part into quadratic

  • Context: Undergrad 
  • Thread starter Thread starter binbagsss
  • Start date Start date
  • Tags Tags
    Quadratic Roots
Click For Summary

Discussion Overview

The discussion revolves around the conditions under which a quartic polynomial can be concluded to have real roots based on its factorization into a cubic term and a quadratic term. Participants explore specific examples and counterexamples related to the existence of real roots in quartic functions.

Discussion Character

  • Exploratory, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant proposes that a quartic polynomial of the form ##f(x)=x^4+(x+2)(x+1)## could have real roots if the associated quadratic has a non-negative discriminant.
  • Another participant counters this by providing an example, ##f(x) = x^4 + (x+1)(x+1)##, asserting that it has no real roots because it is always positive.
  • A follow-up question is posed regarding the polynomial ##x^4-(x+1)(x+1)##, suggesting a change in sign to explore the implications for real roots.
  • A participant seeks clarification on whether the last part refers to ##(x+1)(x+2)## or ##(x+1)^2##, indicating potential confusion in the expressions used.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views on the conditions for real roots in quartic polynomials based on their factorization.

Contextual Notes

The discussion includes assumptions about the discriminant of the quadratic and the behavior of quartic functions, which may not be universally applicable without further context.

binbagsss
Messages
1,291
Reaction score
12
If I have ##f(x)=x^4+(x+2)(x+1)##

basically a quartic without a cubic term for which it can be written as above : ##x^3## + some quadratic which has discrimant ##\geq 0 ##, so that the quadratic has real roots, can one ocnclude that ##f(x)## has real roots too?

thanks
 
Mathematics news on Phys.org
##f(x) = x^4 + (x+1)(x+1)##

has no real roots, since ##f > 0## everywhere (because ##f(x) = x^4 + (x+1)^2 \geq 0)##).

So, no, in general you can't conclude that.
 
Math_QED said:
##f(x) = x^4 + (x+1)(x+1)##

has no real roots, since ##f > 0## everywhere (because ##f(x) = x^4 + (x+1)^2 \geq 0)##).

So, no, in general you can't conclude that.
how about ##x^4-(x+1)(x+1)## ?
changing the sign to minus a quadratic?
 
Did you mean ##(x+1)(x+2)## or ##(x+1)^2## for the last part?
 

Similar threads

  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K