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(−∞,∞)a+b+c=a2+b2+c2+(2ab+2bc+2ca)=a2+b2+c2+4a2b2+4b2c2+4c2a2+8abc(a+b+c)
put x=a+b+c :x=a2+b2+c2+4a2b2+4b2c2+4c2a2+8abcx
Let x=α+β+γx, then x4−2αx2−γx+(α2−β)=0−−(i). Comparing it to the original radical, ⎩⎪⎪⎪⎨⎪⎪⎪⎧α=a2+b2+c24β=a2b2+b2c2+c2a264γ2=a2b2c2. By Vieta’s, we have that a2,b2,c2 are roots of the cubic: z3−αz2+4βz−64γ2=0−−(ii). Just change (α,β,γ)=(2−α,4α2−4γ,−β),then (i) becomes P(x)=x4+αx2+βx+γ, and (ii) becomes F(z)=z3+2αz2+16α2−4γz−64β2 since x is a+b+c, and the equation in z has roots a2,b2,c2, 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 β in P. but if β is positive, it would give abc negative, which would change things. so we put x=−x, and get a negative β. This would simply result in −1 times the sum of the square roots of roots of F.
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@Pi Han Goh
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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(x2).
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find the roots of this quartic
Are you trying to construct a polynomial whose roots are squares of a depressed quartic polynomial?
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yes.
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Why don't you just do this: P(x)=x4+αx2+βx+γ⇒P(x)=x2+αx+βx+γ. Set P(x)=0 and you get another quartic polynomial with the desired roots?
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f(x2).
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....