A new way to solve Quartics (Reposted)

Since I was unable to explain this properly last time,I have reposted.Hope this clears the concept. We see an easy but rather useful radical here. It works over only positives,I will show you how to use this over(,)(-\infty,\infty) a+b+c=a2+b2+c2+(2ab+2bc+2ca)=a2+b2+c2+4a2b2+4b2c2+4c2a2+8abc(a+b+c)a+b+c=\sqrt{a^2+b^2+c^2+(2ab+2bc+2ca)}=\sqrt{a^2+b^2+c^2+\sqrt{4a^2b^2+4b^2c^2+4c^2a^2+8abc(a+b+c)}} put x=a+b+cx=a+b+c :x=a2+b2+c2+4a2b2+4b2c2+4c2a2+8abcxx=\sqrt{a^2+b^2+c^2+\sqrt{4a^2b^2+4b^2c^2+4c^2a^2+8abcx}} Let x=α+β+γxx=\sqrt{α+\sqrt{β+γx}} , then x42αx2γx+(α2β)=0(i)x^4-2αx^2-γx+(α^2-β)=0--(i). Comparing it to the original radical, {α=a2+b2+c2β4=a2b2+b2c2+c2a2γ264=a2b2c2\begin{cases}α=a^2+b^2+c^2\\\dfrac{β}{4}=a^2 b^2+b^2 c^2+c^2 a^2\\\dfrac{γ^2}{64}=a^2 b^2 c^2 \end{cases}. By Vieta’s, we have that a2,b2,c2a^2, b^2, c^2 are roots of the cubic: z3αz2+β4zγ264=0(ii)z^3-αz^2+\dfrac{β}{4} z-\dfrac{γ^2}{64}=0--(ii). Just change (α,β,γ)=(α2,α24γ4,β)(\alpha,\beta,\gamma)=(\dfrac{-\alpha}{2},\dfrac{\alpha^2-4\gamma}{4},-\beta),then (i) becomes P(x)=x4+αx2+βx+γP(x)=x^4+\alpha x^2+\beta x+\gamma, and (ii) becomes F(z)=z3+α2z2+α24γ16zβ264F(z)=z^3+\dfrac{\alpha}{2}z^2+ \dfrac{\alpha^2-4\gamma}{16}z-\dfrac{\beta^2}{64} since xx is a+b+ca+b+c, and the equation in zz has roots a2,b2,c2a^2,b^2,c^2, we can say that the sum of the square roots of roots of F is a root of P, but wait!, we are squaring abcabc to get β\beta in P. but if β\beta is positive, it would give abcabc negative, which would change things. so we put x=xx= -x, and get a negative β\beta. This would simply result in 1-1 times the sum of the square roots of roots of F.

#Algebra #Polynomials #QuarticPolynomial #Root(Equation) #Radicals

Note by Aareyan Manzoor
5 years, 7 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

find the roots of this quartic

Aareyan Manzoor - 5 years, 7 months ago

Are you trying to construct a polynomial whose roots are squares of a depressed quartic polynomial?

Pi Han Goh - 5 years, 5 months ago

Log in to reply

yes.

Aareyan Manzoor - 5 years, 5 months ago

Log in to reply

Why don't you just do this: P(x)=x4+αx2+βx+γP(x)=x2+αx+βx+γP(x)= x^4 + \alpha x^2 + \beta x + \gamma \Rightarrow P(\sqrt x) = x^2 + \alpha x + \beta \sqrt x + \gamma . Set P(x)=0P(\sqrt x) = 0 and you get another quartic polynomial with the desired roots?

Pi Han Goh - 5 years, 5 months ago

Log in to reply

@Pi Han Goh hmm... the sum of the squares roots of the roots of the cubic is one roots of the quartic... if we do that we get for each sqrt we get + and minus cancel out and sum zero. that doesnot work.... f(x2)f(x^2).

Aareyan Manzoor - 5 years, 5 months ago
×

Problem Loading...

Note Loading...

Set Loading...