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Question Number 144260 by ZiYangLee last updated on 23/Jun/21
Find the sum of all the real number  x that satisfy (2x^2 +5x+1)^(2x−3) =1
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{number} \\ $$$${x}\:\mathrm{that}\:\mathrm{satisfy}\:\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\right)^{\mathrm{2}{x}−\mathrm{3}} =\mathrm{1} \\ $$
Answered by Ar Brandon last updated on 24/Jun/21
2x−3=0 ∨ 2x^2 +5x+1=1  x=(3/2) ∨ x=0, x=−(5/2)  2x^2 +5x+1>0
$$\mathrm{2x}−\mathrm{3}=\mathrm{0}\:\vee\:\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}\:\vee\:\mathrm{x}=\mathrm{0},\:\mathrm{x}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{1}>\mathrm{0} \\ $$
Answered by ArielVyny last updated on 24/Jun/21
solution 1 (1)^α =1  2x^2 +5x+1=1   x(2x+5)=0  x=0   x=−(5/2)  solution 2    α^0 =1  with α#0  2x−3=0   x=(3/2)  solution 3   (−1)^(2n) =1  2x^2 +5x+1=−1    2x^2 +5x+2=0  2[(x+(5/2))^2 −((25)/4)+(8/4)]=0  2[(x+(5/2))^2 −((17)/4)]=0  2[(x+(5/2)−((√(17))/2))(x+(5/2)+((√(17))/2))]=0  x_1 =−(5/2)+((√(17))/2)  and x_2 =−(5/2)−((√(17))/2)  then x_1  and x_2  are not peer  so in case 3 there is not solution  S={0.(3/2).−(5/2)}  we have 0+(3/2)−(5/2)=−1
$$\boldsymbol{{solution}}\:\mathrm{1}\:\left(\mathrm{1}\right)^{\alpha} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}=\mathrm{1}\:\:\:{x}\left(\mathrm{2}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\:\:{x}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\boldsymbol{{solution}}\:\mathrm{2}\:\:\:\:\alpha^{\mathrm{0}} =\mathrm{1}\:\:{with}\:\alpha#\mathrm{0} \\ $$$$\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\:\:\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{{solution}}\:\mathrm{3}\:\:\:\left(−\mathrm{1}\right)^{\mathrm{2}\boldsymbol{{n}}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}=−\mathrm{1}\:\: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{4}}+\frac{\mathrm{8}}{\mathrm{4}}\right]=\mathrm{0} \\ $$$$\mathrm{2}\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{17}}{\mathrm{4}}\right]=\mathrm{0} \\ $$$$\mathrm{2}\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\right)\right]=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\:\:{and}\:{x}_{\mathrm{2}} =−\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${then}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{are}\:{not}\:{peer} \\ $$$${so}\:{in}\:{case}\:\mathrm{3}\:{there}\:{is}\:{not}\:{solution} \\ $$$${S}=\left\{\mathrm{0}.\frac{\mathrm{3}}{\mathrm{2}}.−\frac{\mathrm{5}}{\mathrm{2}}\right\} \\ $$$${we}\:{have}\:\mathrm{0}+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}=−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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