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Question Number 167493 by mathlove last updated on 18/Mar/22

(√(x+6))+(√(8−x))=A  x∈Z   and A∈R          ^(    faid  Σx=?)

$$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A} \\ $$$${x}\in{Z}\:\:\:{and}\:{A}\in{R}\:\:\:\:\:\:\:\:\:\overset{\:\:\:\:{faid}\:\:\Sigma{x}=?} {\:} \\ $$

Answered by Rasheed.Sindhi last updated on 18/Mar/22

(√(x+6))+(√(8−x))=A  x∈Z   and A∈R         find Σx.  (√(x+6))+(√(8−x))=A∈R  ⇒  x+6≥0 ∧ 8−x≥0    [x∈Z]  x≥−6 ∧ x≤8  x=−6,−5,−4,...0,1,...,8  Σx=−6−5...−1+0+1...+5+6+7+8=15

$$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A} \\ $$$${x}\in{Z}\:\:\:{and}\:{A}\in{R}\:\:\:\:\:\:\:\:\:{find}\:\Sigma{x}. \\ $$$$\sqrt{{x}+\mathrm{6}}+\sqrt{\mathrm{8}−{x}}={A}\in\mathbb{R} \\ $$$$\Rightarrow \\ $$$${x}+\mathrm{6}\geqslant\mathrm{0}\:\wedge\:\mathrm{8}−{x}\geqslant\mathrm{0}\:\:\:\:\left[{x}\in\mathbb{Z}\right] \\ $$$${x}\geqslant−\mathrm{6}\:\wedge\:{x}\leqslant\mathrm{8} \\ $$$${x}=−\mathrm{6},−\mathrm{5},−\mathrm{4},...\mathrm{0},\mathrm{1},...,\mathrm{8} \\ $$$$\Sigma{x}=−\cancel{\mathrm{6}}−\cancel{\mathrm{5}}...−\cancel{\mathrm{1}}+\mathrm{0}+\cancel{\mathrm{1}}...+\cancel{\mathrm{5}}+\cancel{\mathrm{6}}+\mathrm{7}+\mathrm{8}=\mathrm{15} \\ $$

Commented by mathlove last updated on 18/Mar/22

thanks bro

$${thanks}\:{bro} \\ $$

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