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