Question Number 226577 by mr W last updated on 06/Dec/25
Answered by Ghisom_ last updated on 06/Dec/25
$${x}_{\mathrm{1}} =\alpha+\sqrt{\beta}+\sqrt{\gamma}+\sqrt{\beta\gamma} \\ $$$${x}_{\mathrm{2}} =\alpha−\sqrt{\beta}−\sqrt{\gamma}+\sqrt{\beta\gamma} \\ $$$${x}_{\mathrm{3}} =\alpha−\sqrt{\beta}+\sqrt{\gamma}−\sqrt{\beta\gamma} \\ $$$${x}_{\mathrm{4}} =\alpha+\sqrt{\beta}−\sqrt{\gamma}−\sqrt{\beta\gamma} \\ $$$$\alpha=\mathrm{0}\wedge\beta=\mathrm{2}\wedge\gamma=\mathrm{3}\:\Rightarrow \\ $$$${a}=\mathrm{0}\wedge{b}=−\mathrm{22}\wedge{c}=−\mathrm{48}\wedge{d}=−\mathrm{23} \\ $$$$\mid{a}+{b}+{c}+{d}\mid=\mathrm{93} \\ $$
Answered by Spillover last updated on 06/Dec/25
$${x}=\sqrt{\mathrm{2}\:}\:+\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{6}} \\ $$$$\left({x}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{3}}\:+\sqrt{\mathrm{6}}\:\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{7}=\mathrm{8}\sqrt{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{7}\right)^{\mathrm{2}} =\left(\mathrm{8}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} −\mathrm{14}{x}^{\mathrm{2}} −\mathrm{79}=\mathrm{0} \\ $$$${from} \\ $$$${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${a}=\mathrm{0} \\ $$$${b}=−\mathrm{14} \\ $$$${c}=\mathrm{4} \\ $$$${d}=−\mathrm{79} \\ $$$$\mid{a}+{b}+{c}+{d}\mid=\mathrm{93} \\ $$