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Question Number 171922 by Mikenice last updated on 21/Jun/22
if α and β are the positive roots of the eqn  x^2 +px+q=0, find the sum M=(α)^(1/4)  +(β)^(1/4)
$${if}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{positive}\:{roots}\:{of}\:{the}\:{eqn} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0},\:{find}\:{the}\:{sum}\:{M}=\sqrt[{\mathrm{4}}]{\alpha}\:+\sqrt[{\mathrm{4}}]{\beta} \\ $$
Commented by kaivan.ahmadi last updated on 22/Jun/22
we now S=α+β=−p , P=αβ=q  M^2 =(√α)+(√β)+2((αβ))^(1/4)   on the othere hand if K=(√α)+(√(β ))then  K^2 =α+β+2(√(αβ))=S+2(√P)  ⇒K=(√(S+2(√P)))=(√(−p+2(√q)))  so we have  M^2 =K+2(P)^(1/4) =(√(−p+2(√q)))+2(q)^(1/4)
$${we}\:{now}\:{S}=\alpha+\beta=−{p}\:,\:{P}=\alpha\beta={q} \\ $$$${M}^{\mathrm{2}} =\sqrt{\alpha}+\sqrt{\beta}+\mathrm{2}\sqrt[{\mathrm{4}}]{\alpha\beta} \\ $$$${on}\:{the}\:{othere}\:{hand}\:{if}\:{K}=\sqrt{\alpha}+\sqrt{\beta\:}{then} \\ $$$${K}^{\mathrm{2}} =\alpha+\beta+\mathrm{2}\sqrt{\alpha\beta}={S}+\mathrm{2}\sqrt{{P}} \\ $$$$\Rightarrow{K}=\sqrt{{S}+\mathrm{2}\sqrt{{P}}}=\sqrt{−{p}+\mathrm{2}\sqrt{{q}}} \\ $$$${so}\:{we}\:{have} \\ $$$${M}^{\mathrm{2}} ={K}+\mathrm{2}\sqrt[{\mathrm{4}}]{{P}}=\sqrt{−{p}+\mathrm{2}\sqrt{{q}}}+\mathrm{2}\sqrt[{\mathrm{4}}]{{q}} \\ $$
Commented by Mikenice last updated on 23/Jun/22
thanks sir
$${thanks}\:{sir} \\ $$

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