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Question Number 222885 by gabthemathguy25 last updated on 10/Jul/25
Answered by MrGaster last updated on 10/Jul/25
Commented by gabthemathguy25 last updated on 10/Jul/25
nice
Answered by AntonCWX8 last updated on 10/Jul/25
2x^3 −4x^2 −22x+24=0 x^3 −2x^2 −11x+12=0 x=t+(2/3) t^3 −((37)/3)t+((110)/(27))=0 a^3 =−((((110)/(27)) )/2)+(√((((((110)/(27)))^2 )/4)+(((−((37)/3))^3 )/(27))))=−((55)/(27))+((14(√3))/3)i b^3 =−((((110)/(27)) )/2)−(√((((((110)/(27)))^2 )/4)+(((−((37)/3))^3 )/(27))))=−((55)/(27))−((14(√3))/3)i t_1 =a+b=((−((55)/(27))+((14(√3))/3)i))^(1/3) +((−((55)/(27))−((14(√3))/3)i))^(1/3) =((10)/3) t_2 =(((−1+i(√3))/2))(((−((55)/(27))+((14(√3))/3)i))^(1/3) )+(((−1+i(√3))/2))^2 (((−((55)/(27))−((14(√3))/3)i))^(1/3) )=−((11)/3) t_3 =(((−1+i(√3))/2))^2 (((−((55)/(27))+((14(√3))/3)i))^(1/3) )+(((−1+i(√3))/2))(((−((55)/(27))−((14(√3))/3)i))^(1/3) )=(1/3) x_1 =((10)/3)+(2/3)=4 x_2 =−((11)/3)+(2/3)=−3 x_3 =(1/3)+(2/3)=1 x=−3, 1, 4
$$\mathrm{2}{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{24}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} −\mathrm{11}{x}+\mathrm{12}=\mathrm{0} \\ $$$${x}={t}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{37}}{\mathrm{3}}{t}+\frac{\mathrm{110}}{\mathrm{27}}=\mathrm{0} \\ $$$${a}^{\mathrm{3}} =−\frac{\frac{\mathrm{110}}{\mathrm{27}}\:}{\mathrm{2}}+\sqrt{\frac{\left(\frac{\mathrm{110}}{\mathrm{27}}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left(−\frac{\mathrm{37}}{\mathrm{3}}\right)^{\mathrm{3}} }{\mathrm{27}}}=−\frac{\mathrm{55}}{\mathrm{27}}+\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i} \\ $$$${b}^{\mathrm{3}} =−\frac{\frac{\mathrm{110}}{\mathrm{27}}\:}{\mathrm{2}}−\sqrt{\frac{\left(\frac{\mathrm{110}}{\mathrm{27}}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left(−\frac{\mathrm{37}}{\mathrm{3}}\right)^{\mathrm{3}} }{\mathrm{27}}}=−\frac{\mathrm{55}}{\mathrm{27}}−\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i} \\ $$$$ \\ $$$${t}_{\mathrm{1}} ={a}+{b}=\sqrt[{\mathrm{3}}]{−\frac{\mathrm{55}}{\mathrm{27}}+\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{55}}{\mathrm{27}}−\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i}}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$$${t}_{\mathrm{2}} =\left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\sqrt[{\mathrm{3}}]{−\frac{\mathrm{55}}{\mathrm{27}}+\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i}}\right)+\left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{−\frac{\mathrm{55}}{\mathrm{27}}−\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i}}\right)=−\frac{\mathrm{11}}{\mathrm{3}} \\ $$$${t}_{\mathrm{3}} =\left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{−\frac{\mathrm{55}}{\mathrm{27}}+\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i}}\right)+\left(\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\left(\sqrt[{\mathrm{3}}]{−\frac{\mathrm{55}}{\mathrm{27}}−\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}{i}}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{10}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{4} \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{11}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}=−\mathrm{3} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}}=\mathrm{1} \\ $$$$ \\ $$$${x}=−\mathrm{3},\:\mathrm{1},\:\mathrm{4} \\ $$
Commented by AntonCWX8 last updated on 10/Jul/25
Bear in mind that this is the Cardano′s Formula. It′s similar with the Cubic Formula, just that the process is simpler to be conducted. I′m lazy to show the calculations for the cube of complex.
$${Bear}\:{in}\:{mind}\:{that}\:{this}\:{is}\:{the}\:{Cardano}'{s}\:{Formula}. \\ $$$${It}'{s}\:{similar}\:{with}\:{the}\:{Cubic}\:{Formula},\:{just}\:{that}\:{the}\:{process}\:{is}\:{simpler}\:{to}\:{be}\:{conducted}. \\ $$$${I}'{m}\:{lazy}\:{to}\:{show}\:{the}\:{calculations}\:{for}\:{the}\:{cube}\:{of}\:{complex}. \\ $$
Commented by gabthemathguy25 last updated on 10/Jul/25
perfect
Answered by Tawa11 last updated on 10/Jul/25
2x^3 − 4x^2 − 22x + 24 = 0 ∴ 2x^3 − 2x^2 − 2x^2 + 2x − 24x + 24 = 0 ∴ (2x^3 − 2x^2 ) − (2x^2 − 2x) − (24x − 24) = 0 ∴ 2x^2 (x − 1) − 2x(x − 1) − 24(x − 1) = 0 ∴ (x − 1)(2x^2 − 2x − 24) = 0 ∴ (x − 1)(2x^2 − 8x + 6x − 24) = 0 ∴ (x − 1)[(2x^2 − 8x) + (6x − 24)] = 0 ∴ (x − 1)[2x(x − 4) + 6(x − 4)] = 0 ∴ (x − 1)[(x − 4)(2x + 6)] = 0 ∴ (x − 1)(x − 4)(2x + 6) = 0 ∴ x − 1 = 0 or x − 4 = 0 or 2x + 6 = 0 ∴ x = 1 or x = 4 or 2x = − 6 ∴ x = 1 or x = 4 or x = − (6/2) ∴ x = 1 or x = 4 or x = − 3
$$\mathrm{2x}^{\mathrm{3}} \:\:−\:\:\mathrm{4x}^{\mathrm{2}} \:\:−\:\:\mathrm{22x}\:\:+\:\:\mathrm{24}\:\:\:=\:\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\mathrm{2x}^{\mathrm{3}} \:\:−\:\:\mathrm{2x}^{\mathrm{2}} \:\:−\:\:\mathrm{2x}^{\mathrm{2}} \:\:+\:\:\mathrm{2x}\:\:−\:\:\mathrm{24x}\:\:+\:\:\mathrm{24}\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\left(\mathrm{2x}^{\mathrm{3}} \:\:−\:\:\mathrm{2x}^{\mathrm{2}} \right)\:\:−\:\:\left(\mathrm{2x}^{\mathrm{2}} \:\:−\:\:\mathrm{2x}\right)\:\:−\:\:\left(\mathrm{24x}\:\:−\:\:\mathrm{24}\right)\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\mathrm{2x}^{\mathrm{2}} \left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\:\:−\:\:\mathrm{2x}\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\:\:−\:\:\mathrm{24}\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\left(\mathrm{2x}^{\mathrm{2}} \:\:−\:\:\mathrm{2x}\:\:−\:\:\mathrm{24}\right)\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\left(\mathrm{2x}^{\mathrm{2}} \:\:−\:\:\mathrm{8x}\:\:\:+\:\:\:\mathrm{6x}\:\:−\:\:\mathrm{24}\right)\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\left[\left(\mathrm{2x}^{\mathrm{2}} \:\:−\:\:\mathrm{8x}\right)\:\:\:+\:\:\:\left(\mathrm{6x}\:\:−\:\:\mathrm{24}\right)\right]\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\left[\mathrm{2x}\left(\mathrm{x}\:\:−\:\:\mathrm{4}\right)\:\:\:+\:\:\:\mathrm{6}\left(\mathrm{x}\:\:−\:\:\mathrm{4}\right)\right]\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\left[\left(\mathrm{x}\:\:−\:\:\mathrm{4}\right)\left(\mathrm{2x}\:\:+\:\:\mathrm{6}\right)\right]\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\left(\mathrm{x}\:\:−\:\:\mathrm{1}\right)\left(\mathrm{x}\:\:−\:\:\mathrm{4}\right)\left(\mathrm{2x}\:\:+\:\:\mathrm{6}\right)\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\:\:\mathrm{x}\:\:−\:\:\mathrm{1}\:\:=\:\:\mathrm{0}\:\:\:\mathrm{or}\:\:\:\mathrm{x}\:\:−\:\:\mathrm{4}\:\:=\:\:\mathrm{0}\:\:\:\mathrm{or}\:\:\:\mathrm{2x}\:\:+\:\:\mathrm{6}\:\:\:=\:\:\mathrm{0} \\ $$$$\therefore\:\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1}\:\:\:\mathrm{or}\:\:\:\mathrm{x}\:\:=\:\:\mathrm{4}\:\:\:\mathrm{or}\:\:\:\mathrm{2x}\:\:\:=\:\:−\:\:\mathrm{6} \\ $$$$\therefore\:\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1}\:\:\:\mathrm{or}\:\:\:\mathrm{x}\:\:=\:\:\mathrm{4}\:\:\:\mathrm{or}\:\:\:\mathrm{x}\:\:\:=\:\:−\:\:\frac{\mathrm{6}}{\mathrm{2}} \\ $$$$\therefore\:\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1}\:\:\:\mathrm{or}\:\:\:\mathrm{x}\:\:=\:\:\mathrm{4}\:\:\:\mathrm{or}\:\:\:\mathrm{x}\:\:\:=\:\:−\:\:\mathrm{3} \\ $$
Commented by gabthemathguy25 last updated on 11/Jul/25
solid. ��

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