# 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$$(-\infty,\infty)$$ $a+b+c=\sqrt{a^2+b^2+c^2+(2ab+2bc+2ca)}=\sqrt{a^2+b^2+c^2+\sqrt{4a^2b^2+4b^2c^2+4c^2a^2+8abc(a+b+c)}}$ put $$x=a+b+c$$ :$x=\sqrt{a^2+b^2+c^2+\sqrt{4a^2b^2+4b^2c^2+4c^2a^2+8abcx}}$ Let $$x=\sqrt{α+\sqrt{β+γx}}$$, then $$x^4-2αx^2-γx+(α^2-β)=0--(i)$$. Comparing it to the original radical, $$\begin{cases}α=a^2+b^2+c^2\\\dfrac{β}{4}=a^2 b^2+b^2 c^2+c^2 a^2\\\dfrac{γ^2}{64}=a^2 b^2 c^2 \end{cases}$$. By Vieta’s, we have that $$a^2, b^2, c^2$$ are roots of the cubic: $$z^3-αz^2+\dfrac{β}{4} z-\dfrac{γ^2}{64}=0--(ii)$$. Just change $$(\alpha,\beta,\gamma)=(\dfrac{-\alpha}{2},\dfrac{\alpha^2-4\gamma}{4},-\beta)$$,then (i) becomes $$P(x)=x^4+\alpha x^2+\beta x+\gamma$$, and (ii) becomes $$F(z)=z^3+\dfrac{\alpha}{2}z^2+ \dfrac{\alpha^2-4\gamma}{16}z-\dfrac{\beta^2}{64}$$ since $$x$$ is $$a+b+c$$, and the equation in $$z$$ has roots $$a^2,b^2,c^2$$, we can say that the sum of the square roots of roots of F is a root of P, but wait!, we are squaring $$abc$$ to get $$\beta$$ in P. but if $$\beta$$ is positive, it would give $$abc$$ negative, which would change things. so we put $$x= -x$$, and get a negative $$\beta$$. This would simply result in $$-1$$ times the sum of the square roots of roots of F.

Note by Aareyan Manzoor
3 years, 5 months ago

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## Comments

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find the roots of this quartic

- 3 years, 5 months ago

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

- 3 years, 3 months ago

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yes.

- 3 years, 3 months ago

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

- 3 years, 3 months ago

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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.... $$f(x^2)$$.

- 3 years, 3 months ago

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