Question Number 167493 by mathlove last updated on 18/Mar/22
$$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A} \\ $$$${x}\in{Z}\:\:\:{and}\:{A}\in{R}\:\:\:\:\:\:\:\:\:\overset{\:\:\:\:{faid}\:\:\Sigma{x}=?} {\:} \\ $$
Answered by Rasheed.Sindhi last updated on 18/Mar/22
$$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A} \\ $$$${x}\in{Z}\:\:\:{and}\:{A}\in{R}\:\:\:\:\:\:\:\:\:{find}\:\Sigma{x}. \\ $$$$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A}\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$${x}+\mathrm{6}\geqslant\mathrm{0}\:\wedge\:\mathrm{8}−{x}\geqslant\mathrm{0}\:\:\:\:\left[{x}\in\mathbb{Z}\right] \\ $$$${x}\geqslant−\mathrm{6}\:\wedge\:{x}\leqslant\mathrm{8} \\ $$$${x}=−\mathrm{6},−\mathrm{5},−\mathrm{4},…\mathrm{0},\mathrm{1},…,\mathrm{8} \\ $$$$\Sigma{x}=−\cancel{\mathrm{6}}−\cancel{\mathrm{5}}…−\cancel{\mathrm{1}}+\mathrm{0}+\cancel{\mathrm{1}}…+\cancel{\mathrm{5}}+\cancel{\mathrm{6}}+\mathrm{7}+\mathrm{8}=\mathrm{15} \\ $$
Commented by mathlove last updated on 18/Mar/22
$${thanks}\:{bro} \\ $$