Question Number 144260 by ZiYangLee last updated on 23/Jun/21
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{number} \\ $$$${x}\:\mathrm{that}\:\mathrm{satisfy}\:\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}\right)^{\mathrm{2}{x}−\mathrm{3}} =\mathrm{1} \\ $$
Answered by Ar Brandon last updated on 24/Jun/21
$$\mathrm{2x}−\mathrm{3}=\mathrm{0}\:\vee\:\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}\:\vee\:\mathrm{x}=\mathrm{0},\:\mathrm{x}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{2x}^{\mathrm{2}} +\mathrm{5x}+\mathrm{1}>\mathrm{0} \\ $$
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](Q144312.png)
$$\boldsymbol{{solution}}\:\mathrm{1}\:\left(\mathrm{1}\right)^{\alpha} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}=\mathrm{1}\:\:\:{x}\left(\mathrm{2}{x}+\mathrm{5}\right)=\mathrm{0} \\ $$$${x}=\mathrm{0}\:\:\:{x}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\boldsymbol{{solution}}\:\mathrm{2}\:\:\:\:\alpha^{\mathrm{0}} =\mathrm{1}\:\:{with}\:\alpha#\mathrm{0} \\ $$$$\mathrm{2}{x}−\mathrm{3}=\mathrm{0}\:\:\:{x}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\boldsymbol{{solution}}\:\mathrm{3}\:\:\:\left(−\mathrm{1}\right)^{\mathrm{2}\boldsymbol{{n}}} =\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{1}=−\mathrm{1}\:\: \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{4}}+\frac{\mathrm{8}}{\mathrm{4}}\right]=\mathrm{0} \\ $$$$\mathrm{2}\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{17}}{\mathrm{4}}\right]=\mathrm{0} \\ $$$$\mathrm{2}\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\right)\right]=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\:\:{and}\:{x}_{\mathrm{2}} =−\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$${then}\:{x}_{\mathrm{1}} \:{and}\:{x}_{\mathrm{2}} \:{are}\:{not}\:{peer} \\ $$$${so}\:{in}\:{case}\:\mathrm{3}\:{there}\:{is}\:{not}\:{solution} \\ $$$${S}=\left\{\mathrm{0}.\frac{\mathrm{3}}{\mathrm{2}}.−\frac{\mathrm{5}}{\mathrm{2}}\right\} \\ $$$${we}\:{have}\:\mathrm{0}+\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}=−\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$