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Question Number 120475 by Jamshidbek2311 last updated on 31/Oct/20

Answered by mathmax by abdo last updated on 31/Oct/20

$$\mathrm{if}\mathrm{we}\mathrm{consider}\mathrm{congruence}\mathrm{modulo}2\mathrm{e}\Rightarrow \stackrel{{-}^2}{\mathrm{x}}+\stackrel{-}{\mathrm{x}}=\stackrel{-}{0}\Rightarrow$$$$\stackrel{-}{\mathrm{x}}(\stackrel{-}{\mathrm{x}}+1)=\stackrel{-}{0}\Rightarrow \stackrel{-}{\mathrm{x}}=\stackrel{-}{0}\mathrm{or}\stackrel{-}{\mathrm{x}}=-1\left(\mathrm{Z}/2\mathrm{Z}\mathrm{is}\mathrm{a}\mathrm{corps}\right)$$$$\stackrel{-}{\mathrm{x}}=0\Rightarrow \mathrm{x}=2\mathrm{k}\mathrm{e}\Rightarrow {\left(2\mathrm{k}\right)}^2+2\mathrm{k}-2=6^{\mathrm{y}}\Rightarrow {4\mathrm{k}}^2+2\mathrm{k}-2=6^{\mathrm{y}}\Rightarrow$$$${\mathrm{e}}^{\mathrm{yln}6}={4\mathrm{k}}^2+2\mathrm{k}-2\Rightarrow \mathrm{yln}6=\ln ({4\mathrm{k}}^2+2\mathrm{k}-2)\Rightarrow \mathrm{y}=\left[\frac{\ln ({4\mathrm{k}}^2+2\mathrm{k}-2)}{6}\right]$$$$\mathrm{we}\mathrm{chose}\mathrm{k}/{4\mathrm{k}}^2+2\mathrm{k}-2>0$$$$\stackrel{-}{\mathrm{x}}=-1\Rightarrow \mathrm{x}=2\mathrm{k}-1\mathrm{so}\mathrm{e}\Rightarrow {(2\mathrm{k}-1)}^2+2\mathrm{k}-1-2=6^{\mathrm{y}}\Rightarrow$$$${4\mathrm{k}}^2-4\mathrm{k}+2\mathrm{k}-2=6^{\mathrm{y}}\Rightarrow {4\mathrm{k}}^2-2\mathrm{k}-2=6^{\mathrm{y}}\Rightarrow$$$${\mathrm{e}}^{\mathrm{yln}6}={4\mathrm{k}}^2-2\mathrm{k}-2\Rightarrow \mathrm{yln}6=\ln ({4\mathrm{k}}^2-2\mathrm{k}-2)\Rightarrow \mathrm{y}=\left[\frac{\ln ({4\mathrm{k}}^2-2\mathrm{k}-2)}{\ln 6}\right]$$$$\mathrm{we}\mathrm{chose}\mathrm{k}/{4\mathrm{k}}^2-2\mathrm{k}-2>0\left(\mathrm{k}\mathrm{from}\mathrm{Z}\right)$$

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