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Question Number 10143 by konen last updated on 26/Jan/17

x∈Z (x+4)^(x^2 −16) =1 ⇒Σx=?

$$\mathrm{x}\in\mathrm{Z} \\ $$$$\left(\mathrm{x}+\mathrm{4}\right)^{\mathrm{x}^{\mathrm{2}} −\mathrm{16}} =\mathrm{1}\:\:\:\Rightarrow\Sigma\mathrm{x}=? \\ $$

Answered by mrW1 last updated on 27/Jan/17

x+4=1 ...(i) or x^2 −16=0 and x+4≠0 ...(ii) from (i): x=−3 from (ii): x=4 Σx=−3+4=1

$${x}+\mathrm{4}=\mathrm{1}\:\:\:\:\:\:...\left({i}\right) \\ $$$${or} \\ $$$${x}^{\mathrm{2}} −\mathrm{16}=\mathrm{0}\:\:{and}\:\:{x}+\mathrm{4}\neq\mathrm{0}\:\:\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$${x}=−\mathrm{3} \\ $$$$ \\ $$$${from}\:\left({ii}\right): \\ $$$${x}=\mathrm{4} \\ $$$$ \\ $$$$\Sigma{x}=−\mathrm{3}+\mathrm{4}=\mathrm{1} \\ $$

Answered by sandy_suhendra last updated on 27/Jan/17

(1) x^2 −16=0 (x+4)(x−4)=0 x=4 or x=−4 (not feasible) (2) x+4=1 x=−3 (3) x+4=−1 if x^2 −16=even number x=−5 ⇒ (−5)^2 −16=9 (not feasible) so Σx=4−3=1

$$\left(\mathrm{1}\right)\:\mathrm{x}^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{x}+\mathrm{4}\right)\left(\mathrm{x}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}=\mathrm{4}\:\mathrm{or}\:\mathrm{x}=−\mathrm{4}\:\left(\mathrm{not}\:\mathrm{feasible}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}+\mathrm{4}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{x}=−\mathrm{3} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{x}+\mathrm{4}=−\mathrm{1}\:\mathrm{if}\:\mathrm{x}^{\mathrm{2}} −\mathrm{16}=\mathrm{even}\:\mathrm{number} \\ $$$$\:\:\:\:\:\:\:\mathrm{x}=−\mathrm{5}\:\Rightarrow\:\left(−\mathrm{5}\right)^{\mathrm{2}} −\mathrm{16}=\mathrm{9}\:\left(\mathrm{not}\:\mathrm{feasible}\right) \\ $$$$\mathrm{so}\:\Sigma\mathrm{x}=\mathrm{4}−\mathrm{3}=\mathrm{1} \\ $$

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