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Question Number 192186 by Tawa11 last updated on 11/May/23

If α, β and γ are the roots of x^3 + px + q = 0, find Σα^4 .

$$\mathrm{If}\:\:\alpha,\:\beta\:\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\:\:\mathrm{x}^{\mathrm{3}} \:\:+\:\:\mathrm{px}\:\:+\:\:\mathrm{q}\:\:=\:\:\mathrm{0},\:\:\:\:\mathrm{find}\:\:\:\Sigma\alpha^{\mathrm{4}} . \\ $$

Answered by BaliramKumar last updated on 10/May/23

2p^2

$$\mathrm{2p}^{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 10/May/23

Please workings sir?

$$\mathrm{Please}\:\mathrm{workings}\:\mathrm{sir}? \\ $$

Answered by manxsol last updated on 11/May/23

α+β+γ=0 αβ+αγ+βγ=p αβγ=−q α^4 +pα^2 +qα=0 β^4 +pβ^2 +qβ=0 γ^4 +pγ^2 +qγ=0 α^4 +β^4 +γ^4 +p(α^2 +β^2 +γ^2 )+q(α+β+γ=0 (α+β+γ)^2 =α^2 +β^2 +γ^2 +2(αβ+αγ+βγ)=0 (0)^2 =α^2 +β^2 +γ^2 +2(p) −2p=α^2 +β^2 +γ^2 α^4 +β^4 +γ^4 +p(−2p)+q(0)=0 α^4 +β^4 +γ^4 =2p^2

$$\alpha+\beta+\gamma=\mathrm{0} \\ $$$$\alpha\beta+\alpha\gamma+\beta\gamma={p} \\ $$$$\alpha\beta\gamma=−{q} \\ $$$$\alpha^{\mathrm{4}} +{p}\alpha^{\mathrm{2}} +{q}\alpha=\mathrm{0} \\ $$$$\beta^{\mathrm{4}} +{p}\beta^{\mathrm{2}} +{q}\beta=\mathrm{0} \\ $$$$\gamma^{\mathrm{4}} +{p}\gamma^{\mathrm{2}} +{q}\gamma=\mathrm{0} \\ $$$$\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} +\gamma^{\mathrm{4}} +{p}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)+{q}\left(\alpha+\beta+\gamma=\mathrm{0}\right. \\ $$$$\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\mathrm{2}\left(\alpha\beta+\alpha\gamma+\beta\gamma\right)=\mathrm{0} \\ $$$$\:\left(\mathrm{0}\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\mathrm{2}\left({p}\right) \\ $$$$−\mathrm{2}{p}=\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \\ $$$$\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} +\gamma^{\mathrm{4}} +{p}\left(−\mathrm{2}{p}\right)+{q}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} +\gamma^{\mathrm{4}} =\mathrm{2}{p}^{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 11/May/23

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by BaliramKumar last updated on 11/May/23

α+β+γ = 0 αβ+βγ+γα = p αβγ = −q Σα^4 = α^4 +β^4 +γ^4 = (α^2 )^2 +(β^2 )^2 +(γ^2 )^2 ⇒(α^2 +β^2 +γ^2 )^2 −2(α^2 β^2 +β^2 γ^2 +γ^2 α^2 ) ⇒{(α+β+γ)^2 −2(αβ+βγ+γα)}^2 −2{(αβ+βγ+γα)^2 −2(αβ^2 γ+αβγ^2 +α^2 βγ)} ⇒{(α+β+γ)^2 −2(αβ+βγ+γα)}^2 −2{(αβ+βγ+γα)^2 −2αβγ(α+β+γ)} ⇒{(0)^2 −2(p)}^2 −2{(p)^2 −2(−q)(0)} ⇒ {−2p}^2 −2{p^2 −0} ⇒ 4p^2 −2p^2 ⇒ determinant (((2p^2 )))

$$\alpha+\beta+\gamma\:=\:\mathrm{0} \\ $$$$\alpha\beta+\beta\gamma+\gamma\alpha\:=\:\mathrm{p} \\ $$$$\alpha\beta\gamma\:=\:−\mathrm{q} \\ $$$$\Sigma\alpha^{\mathrm{4}} \:=\:\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} +\gamma^{\mathrm{4}} \:=\:\left(\alpha^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\beta^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\gamma^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left(\alpha^{\mathrm{2}} \beta^{\mathrm{2}} +\beta^{\mathrm{2}} \gamma^{\mathrm{2}} +\gamma^{\mathrm{2}} \alpha^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left\{\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\left\{\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta^{\mathrm{2}} \gamma+\alpha\beta\gamma^{\mathrm{2}} +\alpha^{\mathrm{2}} \beta\gamma\right)\right\}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left\{\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} −\mathrm{2}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)\right\}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\left\{\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\gamma\left(\alpha+\beta+\gamma\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\left\{\left(\mathrm{0}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{p}\right)\right\}^{\mathrm{2}} −\mathrm{2}\left\{\left(\mathrm{p}\right)^{\mathrm{2}} −\mathrm{2}\left(−\mathrm{q}\right)\left(\mathrm{0}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\left\{−\mathrm{2p}\right\}^{\mathrm{2}} −\mathrm{2}\left\{\mathrm{p}^{\mathrm{2}} −\mathrm{0}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{4p}^{\mathrm{2}} −\mathrm{2p}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\begin{array}{|c|}{\mathrm{2p}^{\mathrm{2}} }\\\hline\end{array}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Commented by Tawa11 last updated on 11/May/23

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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