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.

Note by Aareyan Manzoor
3 years, 11 months ago

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find the roots of this quartic

Aareyan Manzoor - 3 years, 11 months ago

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

Pi Han Goh - 3 years, 9 months ago

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yes.

Aareyan Manzoor - 3 years, 9 months ago

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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 - 3 years, 9 months ago

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@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 - 3 years, 9 months ago

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